NCERT Solutions For Class 10 Maths Chapter 5 Arithmetic Progressions

Q: How do you find the nth term of an arithmetic progression?

The nth term of an AP is given by the formula, $a_n=a+(n-1) d$ where a = First term of the sequence. n = Term's position in the sequence. d = Common difference.

Q: How to find the missing terms in an arithmetic progression?

The missing terms in an arithmetic progression can be found using the formula of nth term of the arithmetic progression. The nth term of an AP is given by the formula, $a_n=a+(n-1) d$ where a = First term of the sequence n = Term's position in the sequence d = Common difference.

Q: What is the formula for the sum of the first n natural numbers?

The sum of first n natural numbers can be found using the formula $\frac{n(n+1)}{2}$.

NCERT Solutions For Class 10 Maths Chapter 5 Arithmetic Progressions

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Arithmetic progression (AP) is a fundamental concept in algebra, forming the basis for various advanced topics. An arithmetic progression is a sequence of numbers in which the difference between consecutive terms remains constant. This common difference plays a key role in determining the properties of the sequence. Arithmetic progressions have numerous real-life applications in fields like finance, physics, and computer science.

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NCERT Solutions For Class 10 Maths Chapter 5 Arithmetic Progressions

This article on Solutions of Arithmetic Progressions class 10 Maths NCERT Chapter 5 provides clear and step-by-step solutions for exercise problems in NCERT Class 10 Maths Book. These solutions of Arithmetic Progression Class 10 are designed by Subject Matter Experts according to the latest CBSE syllabus, ensuring that students grasp the concepts effectively. NCERT solutions for other subjects and classes can be downloaded in NCERT solutions.

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NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions PDF Free Download

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NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions - Notes

Arithmetic Progression

Arithmetic Progressions (AP) involve a sequence of terms denoted as $a_{1}, a_{2}, a_{3}, \dots, a_{n}$ representing a series of integers. A sequence $a_{1}, a_{2}, a_{3}, \dots, a_{n}, \dots$ is called an arithmetic sequence if $a_{n + 1} = a_{n} + d, n \in N$ where $d$ is a constant. Here $a_{1}$ is called the first term and the constant $d$ is called the common difference. An arithmetic sequence is also called an Arithmetic Progression (A.P.).

Common Difference of Arithmetic Progression

An AP maintains a constant difference between consecutive terms, known as the common difference. For terms $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ , and $a_{6}$ in an AP, the common difference can be expressed as $D = a_{2} - a_{1} = a_{3} - a_{2} = a_{4} - a_{3} = \dots$

nth Term of Arithmetic Progression (AP)

The nth term of an AP is given by the formula, $a_{n} = a + (n - 1) d$ .

Where

a = First term of the sequence.

n = Term's position in the sequence.

d = Common difference.

Sum of First n Terms in Arithmetic Progression (A.P)

The sum of the first 'n’ terms in an AP is calculated using the formula, $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$

Where

S_n denotes the sum of the terms.

'n' is the number of terms being summed.

'a' is the first term.

'd' stands for the common difference.

Arithmetic Progressions Class 10 NCERT Solutions (Intext Questions and Exercise)

Below are the NCERT class 10 maths chapter 5 solutions for exercise questions.

Arithmetic Progressions Class 10 Solutions Exercise: 5.1

Q1 (i) In which of the following situations, does the list of numbers involved make an arithmetic progression, and why? (i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km .

Answer:

It is given that

Fare for $1^{st} km =$ Rs. 15
And after that Rs 8 for each additional $k m$
Now,
Fare for $2^{nd} km =$ Fare of first $km +$ Additional fare for 1 km

$= Rs. 15 + 8 = Rs 23$

Fare for $3^{rd} km =$ Fare of first $km +$ Fare of additional second $km +$ Fare of additional third km

$= Rs. 23 + 8 = Rs 31$

Fare of $nkm = 15 + 8 \times (n - 1)$
( We multiplied by $n - 1$ because the first km was fixed and for rest, we are adding additional fare.

In this, each subsequent term is obtained by adding a fixed number (8) to the previous term.)

Now, we can clearly see that this is an A.P. with the first term (a) = 15 and common difference (d) = 8

Q1 (ii) In which of the following situations, does the list of numbers involved make an arithmetic progression, and why? (ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.

Answer:

It is given that

vacum pump removes $\frac{1}{4}$ of the air remaining in the cylinder at a time
Let us take initial quantity of air $= 1$
Now, the quantity of air removed in first step $= 1 / 4$
Remaining quantity after $1^{st}$ step

$= 1 - \frac{1}{4} = \frac{3}{4}$

Similarly, Quantity removed after $2^{nd}$ step $=$ Quantity removed in first step $\times$ Remaining quantity after $1^{st}$ step

$= \frac{3}{4} \times \frac{1}{4} = \frac{3}{16}$

Now,
Remaining quantity after $2^{nd}$ step would be = Remaining quantity after $1^{st}$ step - Quantity removed after $2^{nd}$ step

$= \frac{3}{4} - \frac{3}{16} = \frac{12 - 3}{16} = \frac{9}{16}$

Now, we can clearly see that

After the second step the difference between second and first and first and initial step is not the same, hence

the common difference (d) is not the same after every step

Therefore, it is not an AP

Q1 (iii) In which of the following situations, does the list of numbers involved make an arithmetic progression, and why? (iii) The cost of digging a well after every meter of digging, when it costs Rs 150 for the first metre and rises by Rs 50 for each subsequent meter.

Answer:

It is given that
Cost of digging of 1st meter = Rs 150
and
rises by Rs 50 for each subsequent meter
Therefore,

Cost of digging of first 2 meters = cost of digging of first meter + cost of digging additional meter

Cost of digging of first 2 meters = 150 + 50

= Rs 200

Similarly,
Cost of digging of first 3 meters = cost of digging of first 2 meters + cost of digging of additional meter

Cost of digging of first 3 meters = 200 + 50

= Rs 250

We can clearly see that 150, 200,250, ... is in AP with each subsequent term is obtained by adding a fixed number (50) to the previous term.

Therefore, it is an AP with first term (a) = 150 and common difference (d) = 50

Q1 (iv) In which of the following situations, does the list of numbers involved make an arithmetic progression, and why? (iv) The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8 % per annum .

Answer:

Amount in the beginning = Rs. 10000

Interest at the end of 1 st year at the rate of $8 %$

$is 8 % of 10000 = \frac{8 \times 10000}{100} = 800$

Therefore, amount at the end of 1st year will be

$\begin{aligned} = 10000 + 800 \\ = 10800 \end{aligned}$

Now,
Interest at the end of 2 nd year at rate of $8 %$

$is 8 % of 10800 = \frac{8 \times 10800}{100} = 864$

Therefore,, amount at the end of 2 ^nd year

= 10800 + 864 = 11664

Since each subsequent term is not obtained by adding a unique number to the previous term; hence, it is not an AP

Q2 (i) Write first four terms of the AP, when the first term a and the common difference d are given as follows a = 10, d = 10

Answer:

It is given that

$a = 10, d = 10$

Now,

$\begin{aligned} a_{1} = a = 10 \\ a_{2} = a_{1} + d = 10 + 10 = 20 \\ a_{3} = a_{2} + d = 20 + 10 = 30 \\ a_{4} = a_{3} + d = 30 + 10 = 40 \end{aligned}$

Therefore, the first four terms of the given series are 10,20,30,40

Q2 (ii) Write first four terms of the AP when the first term a and the common difference d are given as follows: a=-2, d=0
Answer:

It is given that

$a = - 2, d = 0$

Now,

$\begin{aligned} a_{1} = a = - 2 \\ a_{2} = a_{1} + d = - 2 + 0 = - 2 \\ a_{3} = a_{2} + d = - 2 + 0 = - 2 \\ a_{4} = a_{3} + d = - 2 + 0 = - 2 \end{aligned}$

Therefore, the first four terms of the given series are -2,-2,-2,-2

Q2 (iii) Write first four terms of the AP when the first term a and the common difference d are given as follows a=4, d=-3

Answer:

It is given that

$a = 4, d = - 3$

Now,

$\begin{aligned} a_{1} = a = 4 \\ a_{2} = a_{1} + d = 4 - 3 = 1 \\ a_{3} = a_{2} + d = 1 - 3 = - 2 \\ a_{4} = a_{3} + d = - 2 - 3 = - 5 \end{aligned}$

Therefore, the first four terms of the given series are 4,1,-2,-5

Q2 (iv) Write first four terms of the AP when the first term a and the common difference d are given as follows $a = - 1, d = \frac{1}{2}$

Answer:

It is given that

$a = - 1, d = \frac{1}{2}$

Now,

$\begin{aligned} a_{1} = a = - 1 \\ a_{2} = a_{1} + d = - 1 + \frac{1}{2} = - \frac{1}{2} \\ a_{3} = a_{2} + d = - \frac{1}{2} + \frac{1}{2} = 0 \\ a_{4} = a_{3} + d = 0 + \frac{1}{2} = \frac{1}{2} \end{aligned}$

Therefore, the first four terms of the given series are $- 1, - \frac{1}{2}, 0, \frac{1}{2}$

Q2 (v) Write first four terms of the AP when the first term a and the common difference d are given as follows a=-1.25, d=-0.25

Answer:

It is given that

$a = - 1.25, d = - 0.25$

Now,

$\begin{aligned} a_{1} = a = - 1.25 \\ a_{2} = a_{1} + d = - 1.25 - 0.25 = - 1.50 \\ a_{3} = a_{2} + d = - 1.50 - 0.25 = - 1.75 \\ a_{4} = a_{3} + d = - 1.75 - 0.25 = - 2 \end{aligned}$

Therefore, the first four terms of the given series are -1.25,-1.50,-1.75,-2

Q3 (i) For the following APs, write the first term and the common difference: $3, 1, - 1, - 3, \dots$

Answer:

Given AP series is

$3, 1, - 1, - 3, \dots$

Now, first term of this AP series is 3
Therefore,
First-term of AP series (a)=3
Now,

$a_{1} = 3 and a_{2} = 1$

And common difference $(d) = a_{2} - a_{1} = 1 - 3 = - 2$

Therefore, first term and common difference is 3 and -2 respectively

Q3 (ii) For the following APs, write the first term and the common difference: $- 5, - 1, 3, 7, \dots$

Answer:

Given AP series is

$- 5, - 1, 3, 7, \dots$

Now, the first term of this AP series is $- 5$
Therefore,
First-term of AP series $(a) = - 5$
Now,

$a_{1} = - 5 and a_{2} = - 1$

And common difference $(d) = a_{2} - a_{1} = - 1 - (- 5) = 4$

Therefore, the first term and the common difference is -5 and 4 respectively

Q3 (iii) For the following APs, write the first term and the common difference: $\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \dots$

Answer:

Given AP series is

$\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \dots$
Now, the first term of this AP series is $\frac{1}{3}$
Therefore,
Thle first term of AP series $(a) = \frac{1}{3}$

Now,

$a_{1} = \frac{1}{3} and a_{2} = \frac{5}{3}$

And common difference (d) $= a_{2} - a_{1} = \frac{5}{3} - \frac{1}{3} = \frac{5 - 1}{3} = \frac{4}{3}$
Therefore, the first term and the common difference is $\frac{1}{3}$ and $\frac{4}{3}$ respectively

Q3 (iv) For the following APs, write the first term and the common difference: $0.6, 1.7, 2.8, 3.9, \dots$

Answer:

Given AP series is

$0.6, 1.7, 2.8, 3.9, \dots$

Now, the first term of this AP series is $0.6$
Therefore,
First-term of AP series $(a) = 0.6$
Now,

$a_{1} = 0.6 and a_{2} = 1.7$

And common difference $(d) = a_{2} - a_{1} = 1.7 - 0.6 = 1.1$

Therefore, the first term and the common difference is 0.6 and 1.1 respectively.

Q4 (i) Which of the following are APs? If they form an AP, find the common difference d and write three more terms. 2, 4, 8, 12...

Answer:

Given series is

$2, 4, 8, 12, \dots$

Now, the first term to this series is = $2$

Now,

$\begin{aligned} a_{1} = 2 and a_{2} = 4 and a_{3} = 8 \\ a_{2} - a_{1} = 4 - 2 = 2 \\ a_{3} - a_{2} = 8 - 4 = 4 \end{aligned}$

We can clearly see that the difference between terms are not equal
Hence, given series is not an AP

Q4 (ii) Which of the following are APs ? If they form an AP, find the common difference d and write three more terms. $2, \frac{5}{2}, 3, \frac{7}{2}, \dots$

Answer:

Given series is

$2, \frac{5}{2}, 3, \frac{7}{2}, \dots$

Now,
first term to this series is $= 2$
Now,

$\begin{aligned} a_{1} = 2 and a_{2} = \frac{5}{2} and a_{3} = 3 and a_{4} = \frac{7}{2} \\ a_{2} - a_{1} = \frac{5}{2} - 2 = \frac{5 - 4}{2} = \frac{1}{2} \\ a_{3} - a_{2} = 3 - \frac{5}{2} = \frac{6 - 5}{2} = \frac{1}{2} \\ a_{4} - a_{3} = \frac{7}{2} - 3 = \frac{7 - 6}{2} = \frac{1}{2} \end{aligned}$

We can clearly see that the difference between terms are equal and equal to $\frac{1}{2}$ Hence, given series is in AP
Now, the next three terms are

$a_{5} = a_{4} + d = \frac{7}{2} + \frac{1}{2} = \frac{8}{2} = 4$

$\begin{aligned} a_{6} = a_{5} + d = 4 + \frac{1}{2} = \frac{8 + 1}{2} = \frac{9}{2} \\ a_{7} = a_{6} + d = \frac{9}{2} + \frac{1}{2} = \frac{10}{2} = 5 \end{aligned}$

Therefore, next three terms of given series are $4, \frac{9}{2}, 5$

Q4 (iii) Which of the following are APs ? If they form an AP, find the common difference d and write three more terms. $- 1.2, - 3.2, - 5.2, - 7.2, \dots$

Answer:

Given series is

$- 1.2, - 3.2, - 5.2, - 7.2, \dots$

Now,
the first term to this series is $= - 1.2$
Now,

$\begin{aligned} a_{1} = - 1.2 and a_{2} = - 3.2 and a_{3} = - 5.2 and a_{4} = - 7.2 \\ a_{2} - a_{1} = - 3.2 - (- 1.2) = - 3.2 + 1.2 = - 2 \\ a_{3} - a_{2} = - 5.2 - (- 3.2) = - 5.2 + 3.2 = - 2 \\ a_{4} - a_{3} = - 7.2 - (- 5.2) = - 7.2 + 5.2 = - 2 \end{aligned}$

We can clearly see that the difference between terms are equal and equal to -2 Hence, given series is in AP
Now, the next three terms are

$\begin{aligned} a_{5} = a_{4} + d = - 7.2 - 2 = - 9.2 a_{6} = a_{5} + d = - 9.2 - 2 = - 11.2 \\ a_{7} = a_{6} + d = - 11.2 - 2 = - 13.2 \end{aligned}$

Therefore, next three terms of given series are -9.2,-11.2,-13.2

Q4 (iv) Which of the following are APs? If they form an AP, find the common difference d and write three more terms. $- 10, - 6, - 2, 2, \dots$

Answer:

Given series is

$- 10, - 6, - 2, 2, \dots$

Now,
the first term to this series is $= - 10$
Now,

$\begin{aligned} a_{1} = - 10 and a_{2} = - 6 and a_{3} = - 2 and a_{4} = 2 \\ a_{2} - a_{1} = - 6 - (- 10) = - 6 + 10 = 4 \\ a_{3} - a_{2} = - 2 - (- 6) = - 2 + 6 = 4 \\ a_{4} - a_{3} = 2 - (- 2) = 2 + 2 = 4 \end{aligned}$

We can clearly see that the difference between terms are equal and equal to 4 Hence, given series is in AP
Now, the next three terms are

$\begin{aligned} a_{5} = a_{4} + d = 2 + 4 = 6 \\ a_{6} = a_{5} + d = 6 + 4 = 10 \\ a_{7} = a_{6} + d = 10 + 4 = 14 \end{aligned}$

Therefore, next three terms of given series are 6,10,14

Q4 (v) Which of the following are APs? If they form an AP, find the common difference d and write three more terms. $3, 3 + \sqrt{2}, 3 + 2 \sqrt{2}, 3 + 3 \sqrt{2}, \dots$

Answer:

Given series is

$3, 3 + \sqrt{2}, 3 + 2 \sqrt{2}, 3 + 3 \sqrt{2}, \dots$

Now,
the first term to this series is $= 3$
Now,

$\begin{aligned} a_{1} = 3 and a_{2} = 3 + \sqrt{2} and a_{3} = 3 + 2 \sqrt{2} and a_{4} = 3 + 3 \sqrt{2} \\ a_{2} - a_{1} = 3 + \sqrt{2} - 3 = \sqrt{2} \\ a_{3} - a_{2} = 3 + 2 \sqrt{2} - 3 - \sqrt{2} = \sqrt{2} \\ a_{4} - a_{3} = 3 + 3 \sqrt{2} - 3 - 2 \sqrt{2} = \sqrt{2} \end{aligned}$

We can clearly see that the difference between terms are equal and equal to $\sqrt{2}$ Hence, given series is in AP
Now, the next three terms are

$\begin{aligned} a_{5} = a_{4} + d = 3 + 3 \sqrt{2} + \sqrt{2} = 3 + 4 \sqrt{2} \\ a_{6} = a_{5} + d = 3 + 4 \sqrt{2} + \sqrt{2} = 3 + 5 \sqrt{2} \\ a_{7} = a_{6} + d = 3 + 5 \sqrt{2} + \sqrt{2} = 3 + 6 \sqrt{2} \end{aligned}$

Therefore, next three terms of given series are $3 + 4 \sqrt{2}, 3 + 5 \sqrt{2}, 3 + 6 \sqrt{2}$

Q4 (vi) Which of the following are APs? If they form an AP, find the common difference d and write three more terms. $0.2, 0.22, 0.222, 0.2222, \dots$

Answer:

Given series is

$0.2, 0.22, 0.222, 0.2222, \dots$

Now, the first term to this series is $= 0.2$
Now,

$\begin{aligned} a_{1} = 0.2 and a_{2} = 0.22 and a_{3} = 0.222 and a_{4} = 0.2222 \\ a_{2} - a_{1} = 0.22 - 0.2 = 0.02 \\ a_{3} - a_{2} = 0.222 - 0.22 = 0.002 \end{aligned}$

We can clearly see that the difference between terms are not equal
Hence, given series is not an AP

Q4 (vii) Which of the following are APs? If they form an AP, find the common difference d and write three more terms. $0, - 4, - 8, - 12, \dots$

Answer:

Given series is

Now,
first term to this series is = 0

$0, - 4, - 8, - 12, \dots$

Now,
first term to this series is $= 0$
Now,

$\begin{aligned} a_{1} = 0 and a_{2} = - 4 and a_{3} = - 8 and a_{4} = - 12 \\ a_{2} - a_{1} = - 4 - 0 = - 4 \\ a_{3} - a_{2} = - 8 - (- 4) = - 8 + 4 = - 4 \\ a_{4} - a_{3} = - 12 - (- 8) = - 12 + 8 = - 4 \end{aligned}$

We can clearly see that the difference between terms are equal and equal to -4
Hence, given series is in AP
Now, the next three terms are

$\begin{aligned} a_{5} = a_{4} + d = - 12 - 4 = - 16 \\ a_{6} = a_{5} + d = - 16 - 4 = - 20 \\ a_{7} = a_{6} + d = - 20 - 4 = - 24 \end{aligned}$

We can clearly see that the difference between terms are equal and equal to -4
Hence, given series is in AP
Now, the next three terms are

Therefore, the next three terms of given series are -16,-20,-24

Q4 (viii) Which of the following are APs? If they form an AP, find the common difference d and write three more terms. $- \frac{1}{2}, - \frac{1}{2}, - \frac{1}{2}, - \frac{1}{2}, \dots$

Answer:

Given series is

$- \frac{1}{2}, - \frac{1}{2}, - \frac{1}{2}, - \frac{1}{2}, \dots$

Now, the first term to this series is $= - \frac{1}{2}$
Now,

$\begin{aligned} a_{1} = - \frac{1}{2} and a_{2} = - \frac{1}{2} and a_{3} = - \frac{1}{2} and a_{4} = - \frac{1}{2} \\ a_{2} - a_{1} = - \frac{1}{2} - (- \frac{1}{2}) = - \frac{1}{2} + \frac{1}{2} = 0 \\ a_{3} - a_{2} = - \frac{1}{2} - (- \frac{1}{2}) = - \frac{1}{2} + \frac{1}{2} = 0 \\ a_{4} - a_{3} = - \frac{1}{2} - (- \frac{1}{2}) = - \frac{1}{2} + \frac{1}{2} = 0 \end{aligned}$

We can clearly see that the difference between terms are equal and equal to 0
Hence, given series is in AP
Now, the next three terms are

$\begin{aligned} a_{5} = a_{4} + d = - \frac{1}{2} + 0 = - \frac{1}{2} \\ a_{6} = a_{5} + d = - \frac{1}{2} + 0 = - \frac{1}{2} \\ a_{7} = a_{6} + d = - \frac{1}{2} + 0 = - \frac{1}{2} \end{aligned}$

Therefore, the next three terms of given series are $- \frac{1}{2}, - \frac{1}{2}, - \frac{1}{2}$

Q4 (ix) Which of the following are APs? If they form an AP, find the common difference d and write three more terms. $1, 3, 9, 27, \dots$

Answer:

Given series is

$1, 3, 9, 27, \dots$

Now, the first term to this series is $= 1$
Now,

$\begin{aligned} a_{1} = 1 and a_{2} = 3 and a_{3} = 9 and a_{4} = 27 \\ a_{2} - a_{1} = 3 - 1 = 2 \\ a_{3} - a_{2} = 9 - 3 = 6 \end{aligned}$

We can clearly see that the difference between terms are not equal
Hence, given series is not an AP

Q4 (x) Which of the following are APs ? If they form an AP, find the common difference d and write three more terms. $a, 2 a, 3 a, 4 a, \dots$

Answer:

Given series is

$a, 2 a, 3 a, 4 a, \dots$

Now, the first term to this series is = a
Now,

$\begin{aligned} a_{1} = a and a_{2} = 2 a and a_{3} = 3 a and a_{4} = 4 a \\ a_{2} - a_{1} = 2 a - a = a \\ a_{3} - a_{2} = 3 a - 2 a = a \\ a_{4} - a_{3} = 4 a - 3 a = a \end{aligned}$

We can clearly see that the difference between terms are equal and equal to a Hence, given series is in AP
Now, the next three terms are

$\begin{aligned} a_{5} = a_{4} + d = 4 a + a = 5 a \\ a_{6} = a_{5} + d = 5 a + a = 6 a \\ a_{7} = a_{6} + d = 6 a + a = 7 a \end{aligned}$

Therefore, next three terms of given series are 5a,6a,7a

Q4 (xi) Which of the following are APs ? If they form an AP, find the common difference d and write three more terms. $a, a^{2}, a^{3}, a^{4}, \dots$
Answer:

Given series is

$a, a^{2}, a^{3}, a^{4}, \dots$

Now, the first term to this series is = a
Now,

$\begin{aligned} a_{1} = a and a_{2} = a^{2} and a_{3} = a^{3} and a_{4} = a^{4} \\ a_{2} - a_{1} = a^{2} - a = a (a - 1) \\ a_{3} - a_{2} = a^{3} - a^{2} = a^{2} (a - 1) \end{aligned}$

We can clearly see that the difference between terms are not equal
Hence, given series is not in AP

Q4 (xii) Which of the following are APs ? If they form an AP, find the common difference d and write three more terms. $\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \dots$

Answer:

Given series is

$\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \dots$

We can rewrite it as

$\sqrt{2}, 2 \sqrt{2}, 3 \sqrt{2}, 4 \sqrt{2}, \dots$

Now,
first term to this series is = a
Now,

$\begin{aligned} a_{1} = \sqrt{2} and a_{2} = 2 \sqrt{2} and a_{3} = 3 \sqrt{2} and a_{4} = 4 \sqrt{2} \\ a_{2} - a_{1} = 2 \sqrt{2} - \sqrt{2} = \sqrt{2} \\ a_{3} - a_{2} = 3 \sqrt{2} - 2 \sqrt{2} = \sqrt{2} \\ a_{4} - a_{3} = 4 \sqrt{2} - 3 \sqrt{2} = \sqrt{2} \end{aligned}$

We can clearly see that difference between terms are equal and equal to $\sqrt{2}$
Hence, given series is in AP
Now, the next three terms are

$\begin{aligned} a_{5} = a_{4} + d = 4 \sqrt{2} + \sqrt{2} = 5 \sqrt{2} \\ a_{6} = a_{5} + d = 5 \sqrt{2} + \sqrt{2} = 6 \sqrt{2} \\ a_{7} = a_{6} + d = 6 \sqrt{2} + \sqrt{2} = 7 \sqrt{2} \end{aligned}$

Therefore, next three terms of given series are $5 \sqrt{2}, 6 \sqrt{2}, 7 \sqrt{2}$
That is the next three terms are $\sqrt{50}, \sqrt{72}, \sqrt{98}$

Q4 (xiii) Which of the following are APs? If they form an AP, find the common difference d and write three more terms. $\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \dots$

Answer:

Given series is

$\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \dots$

Now, the first term to this series is $= \sqrt{3}$
Now,

$\begin{aligned} a_{1} = \sqrt{3} and a_{2} = \sqrt{6} and a_{3} = \sqrt{9} and a_{4} = \sqrt{12} \\ a_{2} - a_{1} = \sqrt{6} - \sqrt{3} = \sqrt{3} (\sqrt{2} - 1) \\ a_{3} - a_{2} = 3 - \sqrt{3} = \sqrt{3} (\sqrt{3} - 1) \end{aligned}$

We can clearly see that the difference between terms are not equal
Hence, given series is not in AP

Q4 (xiv) Which of the following are APs ? If they form an AP, find the common difference d and write three more terms. $1^{2}, 3^{2}, 5^{2}, 7^{2}, \dots$

Answer:

Given series is

$1^{2}, 3^{2}, 5^{2}, 7^{2}, \dots$

we can rewrite it as

$1, 9, 25, 49, \dots$

Now, the first term to this series is =1
Now,

$\begin{aligned} a_{1} = 1 and a_{2} = 9 and a_{3} = 25 and a_{4} = 49 \\ a_{2} - a_{1} = 9 - 1 = 8 \\ a_{3} - a_{2} = 25 - 9 = 16 \end{aligned}$

We can clearly see that the difference between terms are not equal
Hence, given series is not in AP

Q4 (xv) Which of the following are APs ? If they form an AP, find the common difference d and write three more terms. $1^{2}, 5^{2}, 7^{2}, 73, \dots$

Answer:

Given series is

$1^{2}, 5^{2}, 7^{2}, 73, \dots$

we can rewrite it as

$1, 25, 49, 73 \dots$

Now,
the first term to this series is =1
Now,

$\begin{aligned} a_{1} = 1 and a_{2} = 25 and a_{3} = 49 and a_{4} = 73 \\ a_{2} - a_{1} = 25 - 1 = 24 \\ a_{3} - a_{2} = 49 - 25 = 24 \\ a_{4} - a_{3} = 73 - 49 = 24 \end{aligned}$

We can clearly see that the difference between terms are equal and equal to 24
Hence, given series is in AP
Now, the next three terms are

$\begin{aligned} a_{5} = a_{4} + d = 73 + 24 = 97 \\ a_{6} = a_{5} + d = 97 + 24 = 121 \\ a_{7} = a_{6} + d = 121 + 24 = 145 \end{aligned}$

Therefore, the next three terms of given series are 97,121,145

Arithmetic Progressions Class 10 Solutions Exercise: 5.2

Q1 Fill in the blanks in the following table, given that a is the first term, d the common difference and $a_{n}$ the $n$ th term of the AP:

a_{n}

(i)

(ii)

(iii)

(iv)

(v)

-18

...

-18.9

3.5

...

-3

2.5

...

105

...

-5

3.6

...

Answer:

(i)

It is given that

$a = 7, d = 3, n = 8$

Now, we know that

$\begin{aligned} a_{n} = a + (n - 1) d \\ a_{8} = 7 + (8 - 1) 3 = 7 + 7 \times 3 = 7 + 21 = 28 \end{aligned}$

Therefore,

$a_{8} = 28$

(ii) It is given that

$a = - 18, n = 10, a_{10} = 0$

Now, we know that

$\begin{aligned} a_{n} = a + (n - 1) d \\ a_{10} = - 18 + (10 - 1) d \\ 0 + 18 = 9 d \\ d = \frac{18}{9} = 2 \end{aligned}$

(iii) It is given that

$d = - 3, n = 18, a_{18} = - 5$

Now, we know that

$\begin{aligned} a_{n} = a + (n - 1) d \\ a_{18} = a + (18 - 1) (- 3) \\ - 5 = a + 17 \times (- 3) \\ a = 51 - 5 = 46 \end{aligned}$

Therefore,

$a = 46$

(iv) It is given that

$a = - 18.9, d = 2.5, a_{n} = 3.6$

Now, we know that

$\begin{aligned} a_{n} = a + (n - 1) d \\ a_{n} = - 18.9 + (n - 1) 2.5 \\ 3.6 + 18.9 = 2.5 n - 2.5 \\ n = \frac{22.5 + 2.5}{2.5} = \frac{25}{2.5} = 10 \end{aligned}$

Therefore,

$n = 10$

(v) It is given that

$a = 3.5, d = 0, n = 105$

Now, we know that

$\begin{aligned} a_{n} = a + (n - 1) d \\ a_{105} = 3.5 + (105 - 1) 0 \\ a_{105} = 3.5 \end{aligned}$

Therefore,

$a_{105} = 3.5$

Q2 (i) Choose the correct choice in the following and justify:30 th term of the AP: $10, 7, 4, \dots$ , is

(A) 97 (B) 77 (C) -77 (D) -87

Answer:

$10, 7, 4, \dots$

Here, $a = 10$
and

$d = 7 - 10 = - 3$

Now, we know that

$a_{n} = a + (n - 1) d$

It is given that $n = 30$
Therefore,

$\begin{aligned} a_{30} = 10 + (30 - 1) (- 3) \\ a_{30} = 10 + (29) (- 3) \\ a_{30} = 10 - 87 = - 77 \end{aligned}$

Therefore, 30 th term of the AP: $10, 7, 4, \dots$ , is $- 77$

Hence, Correct answer is (C)

Q2 (ii) Choose the correct choice in the following and justify : 11 th term of the AP: $- 3, - \frac{1}{2}, 2, \dots$ , is

(A) 28 (B) 22 (C) -38 (D) $- 48 \frac{1}{2}$

Answer:

Given series is

$- 3, - \frac{1}{2}, 2, \dots$

Here, $a = - 3$
and

$d = - \frac{1}{2} - (- 3) = - \frac{1}{2} + 3 = \frac{- 1 + 6}{2} = \frac{5}{2}$

Now, we know that

$a_{n} = a + (n - 1) d$

It is given that $n = 11$
Therefore,

$\begin{aligned} a_{11} = - 3 + (11 - 1) (\frac{5}{2}) \\ a_{11} = - 3 + (10) (\frac{5}{2}) \\ a_{11} = - 3 + 5 \times 5 = - 3 + 25 = 22 \end{aligned}$

Therefore, 11 th term of the AP: $- 3, - \frac{1}{2}, 2, \dots$ , is 22

Hence, the Correct answer is (B)

Q3 (i) In the following APs, find the missing terms in the boxes : $2, ◻, 26$

Answer:

Given AP series is
2, $◻$ 26
Here, $a = 2, n = 3$ and $a_{3} = 26$
Now, we know that

$\begin{aligned} a_{n} = a + (n - 1) d \\ \Rightarrow a_{3} = 2 + (3 - 1) d \\ \Rightarrow 26 - 2 = (2) d \\ \Rightarrow d = \frac{24}{2} = 12 \end{aligned}$

Now,

$\begin{aligned} a_{2} = a_{1} + d \\ a_{2} = 2 + 12 = 14 \end{aligned}$

Therefore, the missing term is 14

Q3 (ii) In the following APs, find the missing terms in the boxes:

$◻, 13, ◻, 3$

Answer:

Given AP series is

13 3
Here, $a_{2} = 13, n = 4$ and $a_{4} = 3$
Now,

$\begin{aligned} a_{2} = a_{1} + d \\ a_{1} = a = 13 - d \end{aligned}$

Now, we know that

$\begin{aligned} a_{n} = a + (n - 1) d \\ \Rightarrow a_{4} = 13 - d + (4 - 1) d \\ \Rightarrow 3 - 13 = - d + 3 d \\ \Rightarrow d = - \frac{10}{2} = - 5 \end{aligned}$

Now,

$\begin{aligned} a_{2} = a_{1} + d \\ a_{1} = a = 13 - d = 13 - (- 5) = 18 \end{aligned}$

And

$\begin{aligned} a_{3} = a_{2} + d \\ a_{3} = 13 - 5 = 8 \end{aligned}$

Therefore, missing terms are 18 and 8
AP series is 18,13,8,3

Q3 (iii) In the following APs, find the missing terms in the boxes : $5, ◻, ◻, 9 \frac{1}{2}$

Answer:

Given AP series is

$5, ◻, ◻, 9 \frac{1}{2}$

Here,
$a = 5, n = 4$ and $a_{4} = 9 \frac{1}{2} = \frac{19}{2}$
Now, we know that

$\begin{aligned} a_{n} = a + (n - 1) d \\ \Rightarrow a_{4} = 5 + (4 - 1) d \\ \Rightarrow \frac{19}{2} - 5 = 3 d \\ \Rightarrow d = \frac{19 - 10}{2 \times 3} = \frac{9}{6} = \frac{3}{2} \end{aligned}$

Now,

$\begin{aligned} a_{2} = a_{1} + d \\ a_{2} = 5 + \frac{3}{2} = \frac{13}{2} \end{aligned}$

And

$\begin{aligned} a_{3} = a_{2} + d \\ a_{3} = \frac{13}{2} + \frac{3}{2} = \frac{16}{2} = 8 \end{aligned}$

Therefore, missing terms are $\frac{13}{2}$ and 8

$AP series is 5, \frac{13}{2}, 8, \frac{19}{2}$

Q3 (iv) In the following APs, find the missing terms in the boxes : $- 4, ◻, ◻, ◻, ◻, 6$

Answer:

Given AP series is

$- 4, ◻, ◻, ◻, ◻, 6$

Here, $a = - 4, n = 6$ and $a_{6} = 6$
Now, we know that

$\begin{aligned} a_{n} = a + (n - 1) d \\ \Rightarrow a_{6} = - 4 + (6 - 1) d \\ \Rightarrow 6 + 4 = 5 d \\ \Rightarrow d = \frac{10}{5} = 2 \end{aligned}$

Now,

$\begin{aligned} a_{2} = a_{1} + d \\ a_{2} = - 4 + 2 = - 2 \end{aligned}$

And

$\begin{aligned} a_{3} = a_{2} + d \\ a_{3} = - 2 + 2 = 0 \end{aligned}$

And

$\begin{aligned} a_{4} = a_{3} + d \\ a_{4} = 0 + 2 = 2 \end{aligned}$

And

$\begin{aligned} a_{5} = a_{4} + d \\ a_{5} = 2 + 2 = 4 \end{aligned}$

Therefore, missing terms are -2,0,2,4
AP series is -4,-2,0,2,4,6

Q3 (v) In the following APs, find the missing terms in the boxes : $38, ◻, ◻, ◻, - 22$

Answer:

Given AP series is

$◻, 38, ◻, ◻, ◻, - 22$

Here, $a_{2} = 38, n = 6$ and $a_{6} = - 22$
Now,

$\begin{aligned} a_{2} = a_{1} + d \\ a_{1} = a = 38 - d \end{aligned}$

Now, we know that

$\begin{aligned} a_{n} = a + (n - 1) d \\ \Rightarrow a_{6} = 38 - d + (6 - 1) d \\ \Rightarrow - 22 - 38 - = - d + 5 d \\ \Rightarrow d = - \frac{60}{4} = - 15 \end{aligned}$

Now,

$\begin{aligned} a_{2} = a_{1} + d \\ a_{1} = 38 - (- 15) = 38 + 15 = 53 \end{aligned}$

And

$\begin{aligned} a_{3} = a_{2} + d \\ a_{3} = 38 - 15 = 23 \end{aligned}$

And

$\begin{aligned} a_{4} = a_{3} + d \\ a_{4} = 23 - 15 = 8 \end{aligned}$

And

$\begin{aligned} a_{5} = a_{4} + d \\ a_{5} = 8 - 15 = - 7 \end{aligned}$

Therefore, missing terms are 53,23,8,-7
AP series is 53,38,23,8,-7,-22

Q4 Which term of the AP : $3, 8, 13, 18, \dots$ , is $78 ?$
Answer:

Given AP is

$3, 8, 13, 18, \dots$

Let suppose that nth term of AP is 78
Here, $a = 3$
And

$d = a_{2} - a_{1} = 8 - 3 = 5$

Now, we know that that

$\begin{aligned} a_{n} = a + (n - 1) d \\ \Rightarrow 78 = 3 + (n - 1) 5 \\ \Rightarrow 78 - 3 = 5 n - 5 \\ \Rightarrow n = \frac{75 + 5}{5} = \frac{80}{5} = 16 \end{aligned}$

Therefore, value of 16th term of given AP is 78

Q5 (i) Find the number of terms in each of the following APs : $7, 13, 19, \dots, 205$

Answer:

Given AP series is

$7, 13, 19, \dots, 205$

Let's suppose there are n terms in given AP
Then,

$a = 7, a_{n} = 205$

And

$d = a_{2} - a_{1} = 13 - 7 = 6$

Now, we know that

$\begin{aligned} a_{n} = a + (n - 1) d \\ \Rightarrow 205 = 7 + (n - 1) 6 \\ \Rightarrow 205 - 7 = 6 n - 6 \\ \Rightarrow n = \frac{198 + 6}{6} = \frac{204}{6} = 34 \end{aligned}$

Therefore, there are 34 terms in given AP

Q5 (ii) Find the number of terms in each of the following APs : $18, 15 \frac{1}{2}, 13, \dots, - 47$

Answer:

$18, 15 \frac{1}{2}, 13, \dots, - 47$
suppose there are n terms in given AP
Then,

$a = 18, a_{n} = - 47$

And

$d = a_{2} - a_{1} = \frac{31}{2} - 18 = \frac{31 - 36}{2} = - \frac{5}{2}$

Now, we know that

$\begin{aligned} a_{n} = a + (n - 1) d \\ \Rightarrow - 47 = 18 + (n - 1) (- \frac{5}{2}) \\ \Rightarrow - 47 - 18 = - \frac{5 n}{2} + \frac{5}{2} \\ \Rightarrow - \frac{5 n}{2} = - 65 - \frac{5}{2} \\ \Rightarrow - \frac{5 n}{2} = - \frac{135}{2} \\ \Rightarrow n = 27 \end{aligned}$

Therefore, there are 27 terms in given AP

Q6 Check whether -150 is a term of the AP : $11, 8, 5, 2 \dots$

Answer:

Given AP series is

$11, 8, 5, 2 \dots$

Here, $a = 11$
And

$d = a_{2} - a_{1} = 8 - 11 = - 3$

Now, suppose - 150 is $n$ th term of the given AP
Now, we know that

$\begin{aligned} a_{n} = a + (n - 1) d \\ \Rightarrow - 150 = 11 + (n - 1) (- 3) \\ \Rightarrow - 150 - 11 = - 3 n + 3 \\ \Rightarrow= n = \frac{161 + 3}{3} = \frac{164}{3} = 54.66 \end{aligned}$

Value of $n$ is not an integer
Therefore, -150 is not a term of AP $11, 8, 5, 2 \dots$

Q7 Find the 31 st term of an AP whose 11 th term is 38 and the 16 th term is 73 .

Answer:

It is given that
11 th term of an AP is 38 and the 16 th term is 73
Now,

Now,

$a_{11} = 38 = a + 10 d$

And

$a_{16} = 73 = a + 15 d$

On solving equation (i) and (ii) we will get

$a = - 32 and d = 7$

Now,

$a_{31} = a + 30 d = - 32 + 30 \times 7 = - 32 + 210 = 178$

Therefore, 31st terms of given AP is 178

Q8 An AP consists of 50 terms of which 3 rd term is 12 and the last term is 106 . Find the 29 th term.
Answer:

It is given that

AP consists of 50 terms of which 3 rd term is 12 and the last term is 106
Now,

$a_{3} = 12 = a + 2 d$

And

$a_{50} = 106 = a + 49 d$

On solving equation (i) and (ii) we will get

$a = 8 and d = 2$

Now,

$a_{29} = a + 28 d = 8 + 28 \times 2 = 8 + 56 = 64$

Therefore, 29th term of given AP is 64

Q9 If the 3 rd and the 9 th terms of an AP are 4 and -8 respectively, which term of this AP is zero?
Answer:

It is given that

3 rd and the 9 th terms of an AP are 4 and -8 respectively
Now,

$a_{3} = 4 = a + 2 d$

And

$a_{9} = - 8 = a + 8 d$

On solving equation (i) and (ii) we will get

$a = 8 and d = - 2$

Now,
Let nth term of given AP is 0
Then,

$\begin{aligned} a_{n} = a + (n - 1) d \\ 0 = 8 + (n - 1) (- 2) \\ 2 n = 8 + 2 = 10 \\ n = \frac{10}{2} = 5 \end{aligned}$

Therefore, 5th term of given AP is 0

Q10 The 17 th term of an AP exceeds its 10 th term by 77 . Find the common difference.
Answer:

It is given that

17 th term of an AP exceeds its 10 th term by 7
i.e.

$\begin{aligned} a_{17} = a_{10} + 7 \\ \Rightarrow a + 16 d = a + 9 d + 7 \\ \Rightarrow a + 16 d - a - 9 d = 7 \\ \Rightarrow 7 d = 7 \\ \Rightarrow d = 1 \end{aligned}$

Therefore, the common difference of AP is 1

Q11 Which term of the AP : $3, 15, 27, 39, \dots$ will be 132 more than its 54 th term?
Answer:

Given AP is

$3, 15, 27, 39, \dots$

Here, $a = 3$
And

$d = a_{2} - a_{1} = 15 - 3 = 12$

Now, let's suppose nth term of given AP is 132 more than its 54 th term
Then,

$\begin{aligned} a_{n} = a_{54} + 132 \\ \Rightarrow a + (n - 1) d = a + 53 d + 132 \\ \Rightarrow 3 + (n - 1) 12 = 3 + 53 \times 12 + 132 \\ \Rightarrow 12 n = 3 + 636 + 132 + 12 \\ \Rightarrow 12 n = 636 + 132 + 12 \\ \Rightarrow n = \frac{780}{12} = 65 \end{aligned}$

Therefore, 65th term of given AP is 132 more than its 54 th term

Q12 Two APs have the same common difference. The difference between their [100 th terms is 100 what is the difference between their 1000 th terms?
Answer:

It is given that
Two APs have the same common difference and difference between their 100 th terms is 100
i.e.

$a_{100} - a_{100}^{'} = 100$

Let common difference of both the AP's is d

$\begin{aligned} \Rightarrow a + 99 d - a^{'} - 99 d = 100 \\ \Rightarrow a - a^{'} = 100 \end{aligned}$

Now, difference between 1000th term is

$\begin{aligned} a_{1000} - a_{1000}^{'} \\ \Rightarrow a + 999 d - a^{'} - 999 d \\ \Rightarrow a - a^{'} \\ \Rightarrow 100 \end{aligned}$

Therefore, difference between 1000th term is 100

Q 13 How many three-digit numbers are divisible by 7 ?
Answer:

We know that the first three digit number divisible by 7 is 105 and last three-digit number divisible by 7 is 994
Therefore,

$a = 105, d = 7 and a_{n} = 994$

Let there are n three digit numbers divisible by 7
Now, we know that

$\begin{aligned} a_{n} = a + (n - 1) d \\ \Rightarrow 994 = 105 + (n - 1) 7 \\ \Rightarrow 7 n = 896 \\ \Rightarrow n = \frac{896}{7} = 128 \end{aligned}$

Therefore, there are 128 three-digit numbers divisible by 7

Q14 How many multiples of 4 lie between 10 and 250 ?
Answer:

We know that the first number divisible by 4 between 10 to 250 is 12 and last number divisible by 4 is $248$
Therefore,

$a = 12, d = 4 and a_{n} = 248$

Let there are $n$ numbers divisible by 4
Now, we know that

$\begin{aligned} a_{n} = a + (n - 1) d \\ \Rightarrow 248 = 12 + (n - 1) 4 \\ \Rightarrow 4 n = 240 \\ \Rightarrow n = \frac{240}{4} = 60 \end{aligned}$

Therefore, there are 60 numbers between 10 to 250 that are divisible by 4

Q15 For what value of $π_{- i}$ , are the $π$ th terms of two APs: $63, 65, 67, \dots$ and $3, 10, 17, \dots$ equal?

Answer:

Given two AP's are

$63, 65, 67, \dots and 3, 10, 17, \dots$

Let first term and the common difference of two AP's are $a, a^{'}$ and $d, d^{'}$

$a = 63, d = a_{2} - a_{1} = 65 - 63 = 2$

And

$a^{'} = 3, d^{'} = a_{2}^{'} - a_{1}^{'} = 10 - 3 = 7$

Now,
Let nth term of both the AP's are equal

$\begin{aligned} a_{n} = a_{n}^{'} \\ \Rightarrow a + (n - 1) d = a^{'} + (n - 1) d^{'} \\ \Rightarrow 63 + (n - 1) 2 = 3 + (n - 1) 7 \\ \Rightarrow 5 n = 65 \\ \Rightarrow n = \frac{65}{5} = 13 \end{aligned}$

Therefore, the 13th term of both the AP's are equal

Q16 Determine the AP whose third term is 16 and the 7 th term exceeds the 5 th term by 12 .
Answer:

It is given that
3rd term of AP is 16 and the 7 th term exceeds the 5 th term by 12
i.e.

$a_{3} = a + 2 d = 16$

And

$\begin{aligned} a_{7} = a_{5} + 12 \\ a + 6 d = a + 4 d + 12 \\ 2 d = 12 \\ d = 6 \end{aligned}$

Put the value of $d$ in equation (i) we will get

$a = 4$

Now, AP with first term $= 4$ and common difference $= 6$ is

$4, 10, 16, 22, \dots \dots$

Q17 Find the 20 th term from the last term of the $AP : 3, 8, 13, \dots, 253$

Answer:

Given AP is

$3, 8, 13, \dots, 253$

Here, $a = 3$ and $a_{n} = 253$
And

$d = a_{2} - a_{1} = 8 - 3 = 5$

Let suppose there are $n$ terms in the AP
Now, we know that

$\begin{aligned} a_{n} = a + (n - 1) d \\ 253 = 3 + (n - 1) 5 \\ 5 n = 255 \\ n = 51 \end{aligned}$

So, there are 51 terms in the given AP and 20th term from the last will be 32th term from the starting Therefore,

$\begin{aligned} a_{32} = a + 31 d \\ a_{32} = 3 + 31 \times 5 = 3 + 155 = 158 \end{aligned}$

Therefore, 20th term from the of given AP is 158

Q18 The sum of the 4 th and 8 th terms of an AP is 24 and the sum of the 6 th and 10 th terms is 44 Find the first three terms of the AP.
Answer:

It is given that
sum of the 4 th and 8 th terms of an AP is 24 and the sum of the 6 th and 10 th terms is 44
i.e.

$\begin{aligned} a_{4} + a_{8} = 24 \\ \Rightarrow a + 3 d + a + 7 d = 24 \\ \Rightarrow 2 a + 10 d = 24 \\ \Rightarrow a + 5 d = 12 \end{aligned}$

And

$\begin{aligned} a_{6} + a_{10} = 44 \\ \Rightarrow a + 5 d + a + 9 d = 44 \\ \Rightarrow 2 a + 14 d = 44 \\ \Rightarrow a + 7 d = 22 \end{aligned}$

On solving equation (i) and (ii) we will get

$a = - 13 and d = 5$

Therefore, first three of AP with $a = - 13$ and $d = 5$ is

$- 13, - 8, - 3$

Q19 Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000?

Answer:

It is given that
Subba Rao started work at an annual salary of Rs 5000 and received an increment of Rs 200 each year
Therefore, $a = 5000$ and $d = 200$
Let's suppose after n years his salary will be Rs 7000
Now, we know that

$\begin{aligned} a_{n} = a + (n - 1) d \\ \Rightarrow 7000 = 5000 + (n - 1) 200 \\ \Rightarrow 2000 = 200 n - 200 \\ \Rightarrow 200 n = 2200 \\ \Rightarrow n = 11 \end{aligned}$

Therefore, after 11 years his salary will be Rs 7000 after 11 years, starting from 1995, his salary will reach to 7000 , so we have to add 10 in 1995 , because these numbers are in years Thus, $1995 + 10 = 2005$

Q20 Ramkali saved Rs 5 in the first week of a year and then increased her weekly savings by Rs 1.75 . If in the n th week, her weekly savings become Rs 20.75 , find n

Answer:

It is given that
Ramkali saved Rs 5 in the first week of a year and then increased her weekly savings by Rs 1.75
Therefore, $a = 5$ and $d = 1.75$
after $n$ th week, her weekly savings become Rs 20.75
Now, we know that

$\begin{aligned} a_{n} = a + (n - 1) d \\ \Rightarrow 20.75 = 5 + (n - 1) 1.75 \\ \Rightarrow 15.75 = 1.75 n - 1.75 \\ \Rightarrow 1.75 n = 17.5 \\ \Rightarrow n = 10 \end{aligned}$

Therefore, after 10 weeks her saving will become Rs 20.75

Arithmetic Progressions Class 10 Solution Exercise: 5.3

Q1 (i) Find the sum of the following APs: $2, 7, 12, \dots$ , to 10 terms.

Given AP is
$2, 7, 12, \dots$ , to 10 terms
Here, $a = 2$ and $n = 10$
And

$d = a_{2} - a_{1} = 7 - 2 = 5$

Now, we know that

$\begin{aligned} S = \frac{n}{2} {2 a + (n - 1) d} \\ \Rightarrow S = \frac{10}{2} {2 \times 2 + (10 - 1) 5} \\ \Rightarrow S = 5 {4 + 45} \\ \Rightarrow S = 5 {49} \\ \Rightarrow S = 245 \end{aligned}$

Therefore, the sum of AP $2, 7, 12, \dots$ , to 10 terms is $245$

Q1 (ii) Find the sum of the following APs: $- 37, - 33, - 29, \dots$ , to 12 terms.

Answer:

Given AP is
$- 37, - 33, - 29, \dots$ , to 12 terms.
Here, $a = - 37$ and $n = 12$
And

$d = a_{2} - a_{1} = - 33 - (- 37) = 4$

Now, we know that

$\begin{aligned} S = \frac{n}{2} {2 a + (n - 1) d} \\ \Rightarrow S = \frac{12}{2} {2 \times (- 37) + (12 - 1) 4} \\ \Rightarrow S = 6 {- 74 + 44} \\ \Rightarrow S = 5 {- 30} \\ \Rightarrow S = - 180 \end{aligned}$

Therefore, the sum of $AP - 37, - 33, - 29, \dots$ , to 12 terms. is $- 180$

Q1 (iii) Find the sum of the following APs: $0.6, 1.7, 2.8, \dots$ , to 100 terms.

Answer:

Given AP is
$0.6, 1.7, 2.8, \dots$ , to 100 terms..
Here, $a = 0.6$ and $n = 100$
And

$d = a_{2} - a_{1} = 1.7 - 0.6 = 1.1$

Now, we know that

$\begin{aligned} S = \frac{n}{2} {2 a + (n - 1) d} \\ \Rightarrow S = \frac{100}{2} {2 \times (0.6) + (100 - 1) (1.1)} \\ \Rightarrow S = 50 {1.2 + 108.9} \\ \Rightarrow S = 50 {110.1} \\ \Rightarrow S = 5505 \end{aligned}$

Therefore, the sum of $AP 0.6, 1.7, 2.8, \dots$ , to 100 terms. is $5505$

Q1 (iv) Find the sum of the following APs: $\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \dots$ , to 11 terms.

Answer:

Given AP is
$\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \dots$ , to 11 terms.
Here, $a = \frac{1}{15}$ and $n = 11$
And

$d = a_{2} - a_{1} = \frac{1}{12} - \frac{1}{15} = \frac{5 - 4}{60} = \frac{1}{60}$

Now, we know that

$\begin{aligned} S = \frac{n}{2} {2 a + (n - 1) d} \\ \Rightarrow S = \frac{11}{2} {2 \times \frac{1}{15} + (11 - 1) (\frac{1}{60})} \\ \Rightarrow S = \frac{11}{2} {\frac{2}{15} + \frac{1}{6}} \\ \Rightarrow S = \frac{11}{2} {\frac{9}{30}} \\ \Rightarrow S = \frac{99}{60} = \frac{33}{20} \end{aligned}$

Therefore, the sum of AP $\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \dots$ , to 11 terms. is $\frac{33}{20}$

Q2 (i) Find the sums given below : $7 + 10 \frac{1}{2} + 14 + \dots + 84$

Answer:

Given AP is

$7 + 10 \frac{1}{2} + 14 + \dots + 84$

We first need to find the number of terms
Here, $a = 7$ and $a_{n} = 84$
And

$d = a_{2} - a_{1} = \frac{21}{2} - 7 = \frac{21 - 14}{2} = \frac{7}{2}$

Let suppose there are $n$ terms in the AP
Now, we know that

$\begin{aligned} a_{n} = a + (n - 1) d \\ \Rightarrow 84 = 7 + (n - 1) \frac{7}{2} \\ \Rightarrow \frac{7 n}{2} = 77 + \frac{7}{2} \\ \Rightarrow n = 23 \end{aligned}$

Now, we know that

$\begin{aligned} S = \frac{n}{2} {2 a + (n - 1) d} \\ \Rightarrow S = \frac{23}{2} {2 \times 7 + (23 - 1) (\frac{7}{2})} \\ \Rightarrow S = \frac{23}{2} {14 + 77} \\ \Rightarrow S = \frac{23}{2} {91} \\ \Rightarrow S = \frac{2093}{2} = 1046 \frac{1}{2} \end{aligned}$

Therefore, the sum of $AP^{7 + 10 \frac{1}{2} + 14 + \dots + 84}$ is $1046 \frac{1}{2}$

Q2 (ii) Find the sums given below : $34 + 32 + 30 + \dots + 10$

Answer:

Given AP is

$34 + 32 + 30 + \dots + 10$

We first need to find the number of terms
Here, $a = 34$ and $a_{n} = 10$
And

$d = a_{2} - a_{1} = 32 - 34 = - 2$

Let suppose there are $n$ terms in the AP
Now, we know that

$\begin{aligned} a_{n} = a + (n - 1) d \\ \Rightarrow 10 = 34 + (n - 1) (- 2) \\ \Rightarrow - 26 = - 2 n \\ \Rightarrow n = 13 \end{aligned}$

Now, we know that

$\begin{aligned} S = \frac{n}{2} {a + a_{n}} \\ \Rightarrow S = \frac{13}{2} {44} \\ \Rightarrow S = 13 \times 22 = 286 \end{aligned}$

Therefore, the sum of AP $34 + 32 + 30 + \dots + 10$ is $286$

Q2 (iii) Find the sums given below : $- 5 + (- 8) + (- 11) + \dots + (- 230)$

Answer:

Given AP is

$- 5 + (- 8) + (- 11) + \dots + (- 230)$

We first need to find the number of terms
Here, $a = - 5$ and $a_{n} = - 230$
And

$d = a_{2} - a_{1} = - 8 - (- 5) = - 3$

Let suppose there are $n$ terms in the AP
Now, we know that

$\begin{aligned} a_{n} = a + (n - 1) d \\ \Rightarrow - 230 = - 5 + (n - 1) (- 3) \\ \Rightarrow - 228 = - 3 n \\ \Rightarrow n = 76 \end{aligned}$

Now, we know that

$\begin{aligned} S = \frac{n}{2} {a + a_{n}} \\ \Rightarrow S = \frac{76}{2} {(- 5 - 230)} \\ \Rightarrow S = 38 {- 235} \\ \Rightarrow S = - 8930 \end{aligned}$

Therefore, the sum of $AP - 5 + (- 8) + (- 11) + \dots + (- 230)$ is -8930

Q3 (i) In an AP: given $a = 5, d = 3, a_{n} = 50$ , find $n$ and $S_{n}$

Answer:

It is given that

$a = 5, d = 3 and a_{n} = 50$

Let suppose there are n terms in the AP
Now, we know that

$\begin{aligned} a_{n} = a + (n - 1) d \\ \Rightarrow 50 = 5 + (n - 1) 3 \\ \Rightarrow 48 = 3 n \\ \Rightarrow n = 16 \end{aligned}$

Now, we know that

$\begin{aligned} S = \frac{n}{2} {2 a + (n - 1) d} \\ \Rightarrow S = \frac{16}{2} {2 \times (5) + (16 - 1) (3)} \\ \Rightarrow S = 8 {10 + 45} \\ \Rightarrow S = 8 {55} \\ \Rightarrow S = 440 \end{aligned}$

Therefore, the sum of the given AP is 440

Q3 (ii) In an AP: given $a = 7_{- i} a_{13} = 35$ find $d$ and $S_{13}$

Answer:

It is given that

$\begin{aligned} a = 7 and a_{13} = 35 \\ a_{13} = a + 12 d = 35 \\ = 12 d = 35 - 7 = 28 \\ d = \frac{28}{12} = \frac{7}{3} \end{aligned}$

Now, we know that

$\begin{aligned} S_{n} = \frac{n}{2} {2 a + (n - 1) d} \\ \Rightarrow S_{13} = \frac{13}{2} {2 \times (7) + (13 - 1) (\frac{7}{3})} \\ \Rightarrow S_{13} = \frac{13}{2} {14 + 28} \\ \Rightarrow S_{13} = \frac{13}{2} {42} \\ \Rightarrow S_{13} = 13 \times 21 = 273 \end{aligned}$

Therefore, the sum of given AP is $273$

Q3 (iii) In an AP: given $a_{12} = 37, d = 3$ , find $a$ and $S_{12}$

Answer:

It is given that

$\begin{aligned} d = 3 and a_{12} = 37 \\ a_{12} = a + 11 d = 37 \\ = a = 37 - 11 \times 3 = 37 - 33 = 4 \end{aligned}$

Now, we know that

$\begin{aligned} S_{n} = \frac{n}{2} {2 a + (n - 1) d} \\ \Rightarrow S_{12} = \frac{12}{2} {2 \times (4) + (12 - 1) 3} \\ \Rightarrow S_{12} = 6 {8 + 33} \\ \Rightarrow S_{12} = 6 {41} \\ \Rightarrow S_{12} = 246 \end{aligned}$

Therefore, the sum of given AP is $246$

Q3 (iv) In an AP: given $a_{3} = 15, S_{10} = 125$ , find $d$ and $S_{10}$
Answer:

It is given that

$\begin{aligned} a_{3} = 15, S_{10} = 125 \\ a_{3} = a + 2 d = 15 \end{aligned}$

Now, we know that

$\begin{aligned} S_{n} = \frac{n}{2} {2 a + (n - 1) d} \\ \Rightarrow S_{10} = \frac{10}{2} {2 \times (a) + (10 - 1) d} \\ \Rightarrow 125 = 5 {2 a + 9 d} \\ \Rightarrow 2 a + 9 d = 25 \end{aligned}$

On solving equation (i) and (ii) we will get

$a = 17 and d = - 1$

Now,

$a_{10} = a + 9 d = 17 + 9 (- 1) = 17 - 9 = 8$

Therefore, the value of $d$ and 10 th terms is -1 and 8 respectively

Q3 (v) In an AP: given $d = 5, S_{9} = 75$ find $a$ and $a_{9}$

Answer:

It is given that

$d = 5, S_{9} = 75$

Now, we know that

$\begin{aligned} S_{n} = \frac{n}{2} {2 a + (n - 1) d} \\ \Rightarrow S_{9} = \frac{9}{2} {2 \times (a) + (9 - 1) 5} \\ \Rightarrow 75 = \frac{9}{2} {2 a + 40} \\ \Rightarrow 150 = 18 a + 360 \\ \Rightarrow a = - \frac{210}{18} = - \frac{35}{3} \end{aligned}$

Now,

$a_{9} = a + 8 d = - \frac{35}{3} + 8 (5) = - \frac{35}{3} + 40 = \frac{- 35 + 120}{3} = \frac{85}{3}$

Q3 (vi) In an AP: given $a = 2, d = 8, S_{n} = 90$ , find $n$ and $a_{n}$

Answer:

It is given that

$a = 2, d = 8, S_{n} = 90$

Now, we know that

$\begin{aligned} S_{n} = \frac{n}{2} {2 a + (n - 1) d} \\ \Rightarrow 90 = \frac{n}{2} {2 \times (2) + (n - 1) 8} \\ \Rightarrow 180 = n {4 + 8 n - 8} \\ \Rightarrow 8 n^{2} - 4 n - 180 = 0 \\ \Rightarrow 4 (2 n^{2} - n - 45) = 0 \\ \Rightarrow 2 n^{2} - n - 45 = 0 \\ \Rightarrow 2 n^{2} - 10 n + 9 n - 45 = 0 \\ \Rightarrow (n - 5) (2 n + 9) = 0 \\ \Rightarrow n = 5 and n = - \frac{9}{2} \end{aligned}$

n can not be negative so the only the value of n is 5
Now,

$a_{5} = a + 4 d = 2 + 4 \times 8 = 2 + 32 = 34$

Therefore, value of $n$ and $n$ th term is 5 and 34 respectively

Q3 (vii) In an AP: given $a = 8, a_{n} = 62, S_{n} = 210$ , find $n$ and $d$

Answer:

It is given that

$a = 8, a_{n} = 62, S_{n} = 210$

Now, we know that

$\begin{aligned} a_{n} = a + (n - 1) d \\ 62 = 8 + (n - 1) d \\ (n - 1) d = 54 \end{aligned}$

Now, we know that

$\begin{aligned} S_{n} = \frac{n}{2} {2 a + (n - 1) d} \\ \Rightarrow 210 = \frac{n}{2} {2 \times (8) + (n - 1) d} \\ \Rightarrow 420 = n {16 + 54} \\ \Rightarrow 420 = n {70} \\ \Rightarrow n = 6 \end{aligned}$

Now, put this value in (i) we will get

$d = \frac{54}{5}$

Therefore, value of n and d are 6 and $\frac{54}{5}$ respectively

Q3 (viii) In an AP: given $a_{n} = 4, d = 2, S_{n} = - 14$ , find $n$ and $a_{-}$

Answer:

It is given that

$a_{n} = 4, d = 2, S_{n} = - 14$

Now, we know that

$\begin{aligned} a_{n} = a + (n - 1) d \\ 4 = a + (n - 1) 2 \\ a + 2 n = 6 \Rightarrow a = 6 - 2 n \end{aligned}$

Now, we know that

$\begin{aligned} S_{n} = \frac{n}{2} {2 a + (n - 1) d} \\ \Rightarrow - 14 = \frac{n}{2} {2 \times (a) + (n - 1) 2} \\ \Rightarrow - 28 = n {2 (6 - 2 n) + 2 n - 2} \\ \Rightarrow - 28 = n {10 - 2 n} \\ \Rightarrow 2 n^{2} - 10 n - 28 = 0 \\ \Rightarrow 2 (n^{2} - 5 n - 14) = 0 \\ \Rightarrow n^{2} - 7 n + 2 n - 14 = 0 \\ \Rightarrow (n + 2) (n - 7) = 0 \\ \Rightarrow n = - 2 and n = 7 \end{aligned}$

Value of n cannot be negative so the only the value of n is 7
Now, put this value in (i) we will get
a = -8
Therefore, the value of n and a are 7 and -8 respectively

Q3 (ix) In an AP: given $a = 3, n = 8, S = 192$ , find $d$

Answer:

It is given that

$a = 3, n = 8, S = 192$

Now, we know that

$\begin{aligned} S_{n} = \frac{n}{2} {2 a + (n - 1) d} \\ \Rightarrow 192 = \frac{8}{2} {2 \times (3) + (8 - 1) d} \\ \Rightarrow 192 = 4 {6 + 7 d} \\ \Rightarrow 7 d = 48 - 6 \\ \Rightarrow d = \frac{42}{7} = 6 \end{aligned}$

Therefore, the value of $d$ is 6

Q3 (x) In an AP: given $l = 28, S = 144$ and $n = 9$ and there are total $9 \underset{―}{terms. Find} a_{-}$

Answer:

It is given that

$l = 28, S = 144 and n = 9$

Now, we know that

$\begin{aligned} l = a_{n} = a + (n - 1) d \\ 28 = a_{n} = a + (n - 1) d \end{aligned}$

Now, we know that

$\begin{aligned} S_{n} = \frac{n}{2} {2 a + (n - 1) d} \\ \Rightarrow 144 = \frac{9}{2} {a + a + (n - 1) d} \\ \Rightarrow 288 = 9 {a + 28} \\ \Rightarrow a + 28 = 32 \\ \Rightarrow a = 4 \end{aligned}$

Therefore, the value of $a$ is 4

Q4 How many terms of the AP: $9, 17, 25, \dots$ must be taken to give a sum of 636 ?

Answer:

Given AP is

$9, 17, 25, \dots$

Here, $a = 9$ and $d = 8$
And $S_{n} = 636$
Now, we know that

$\begin{aligned} S_{n} = \frac{n}{2} {2 a + (n - 1) d} \\ \Rightarrow 636 = \frac{n}{2} {18 + (n - 1) 8} \\ \Rightarrow 1272 = n {10 + 8 n} \\ \Rightarrow 8 n^{2} + 10 n - 1272 = 0 \\ \Rightarrow 2 (4 n^{2} + 5 n - 636) = 0 \\ \Rightarrow 4 n^{2} + 53 n - 48 n - 636 = 0 \\ \Rightarrow (n - 12) (4 n + 53) = 0 \\ \Rightarrow n = 12 and n = - \frac{53}{4} \end{aligned}$

Value of $n$ can not be negative so the only the value of $n$ is 12
Therefore, the sum of $12$ terms of $AP 9, 17, 25, \dots$ must be taken to give a sum of 636 .

Q5 The first term of an AP is 5, the last term is 45 and the sum is 400 . Find the number of terms and the common difference.

Answer:

It is given that

$a = 5, a_{n} = 45, S_{n} = 400$

Now, we know that

$\begin{aligned} a_{n} = a + (n - 1) d \\ 45 = 5 + (n - 1) d \\ (n - 1) d = 40 \end{aligned}$

Now, we know that

$\begin{aligned} S_{n} = \frac{n}{2} {2 a + (n - 1) d} \\ \Rightarrow 400 = \frac{n}{2} {2 \times (5) + (n - 1) d} \\ \Rightarrow 800 = n {10 + 40} \\ \Rightarrow 800 = n {50} \\ \Rightarrow n = 16 \end{aligned}$

Now, put this value in (i) we will get

$d = \frac{40}{15} = \frac{8}{3}$

Therefore, value of n and d are 16 and $\frac{8}{3}$ respectively

Q6 The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9 , how many terms are there and what is their sum?

Answer:

It is given that

$a = 17, l = 350, d = 9$

Now, we know that

$\begin{aligned} a_{n} = a + (n - 1) d \\ 350 = 17 + (n - 1) 9 \\ (n - 1) 9 = 333 \\ (n - 1) = 37 \\ n = 38 \end{aligned}$

Now, we know that

$\begin{aligned} S_{n} = \frac{n}{2} {2 a + (n - 1) d} \\ \Rightarrow S_{38} = \frac{38}{2} {2 \times (17) + (38 - 1) 9} \\ \Rightarrow S_{38} = 19 {34 + 333} \\ \Rightarrow S_{38} = 19 {367} \\ \Rightarrow S_{38} = 6973 \end{aligned}$

Therefore, there are 38 terms and their sun is 6973

Q7 Find the sum of first [22] terms of an AP in which $d = 7$ and 22 nd term is 149 .

Answer:

It is given that

$a_{22} = 149, d = 7, n = 22$

Now, we know that

$\begin{aligned} a_{22} = a + 21 d \\ 149 = a + 21 \times 7 \\ a = 149 - 147 = 2 \end{aligned}$

Now, we know that

$\begin{aligned} S_{n} = \frac{n}{2} {2 a + (n - 1) d} \\ \Rightarrow S_{22} = \frac{22}{2} {2 \times (2) + (22 - 1) 7} \\ \Rightarrow S_{22} = 11 {4 + 147} \\ \Rightarrow S_{22} = 11 {151} \\ \Rightarrow S_{22} = 1661 \end{aligned}$

Therefore, there are 22 terms and their sum is 1661

Q8 Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively,

Answer:

It is given that

$a_{2} = 14, a_{3} = 18, n = 51$

And $d = a_{3} - a_{2} = 18 - 14 = 4$
Now,

$\begin{aligned} a_{2} = a + d \\ a = 14 - 4 = 10 \end{aligned}$

Now, we know that

$\begin{aligned} S_{n} = \frac{n}{2} {2 a + (n - 1) d} \\ \Rightarrow S_{51} = \frac{51}{2} {2 \times (10) + (51 - 1) 4} \\ \Rightarrow S_{51} = \frac{51}{2} {20 + 200} \\ \Rightarrow S_{51} = \frac{51}{2} {220} \\ \Rightarrow S_{51} = 51 \times 110 \\ \Rightarrow S_{51} = 5304 \end{aligned}$

Therefore, there are 51 terms and their sum is 5610

Q9 If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289 ，find the sum of first $n$ terms．

Answer:

It is given that

$S_{7} = 49 and S_{17} = 289$

Now, we know that

$\begin{aligned} S_{n} = \frac{n}{2} {2 a + (n - 1) d} \\ \Rightarrow S_{7} = \frac{7}{2} {2 \times (a) + (7 - 1) d} \\ \Rightarrow 98 = 7 {2 a + 6 d} \\ \Rightarrow a + 3 d = 7 - (i) \end{aligned}$

Similarly,

$\begin{aligned} \Rightarrow S_{17} = \frac{17}{2} {2 \times (a) + (17 - 1) d} \\ \Rightarrow 578 = 17 {2 a + 16 d} \\ \Rightarrow a + 8 d = 17 - (i i) \end{aligned}$

On solving equation (i) and (ii) we will get

$a = 1 and d = 2$

Now, the sum of first $n$ terms is

$\begin{aligned} S_{n} = \frac{n}{2} {2 \times 1 + (n - 1) 2} \\ S_{n} = \frac{n}{2} {2 + 2 n - 2} \\ S_{n} = n^{2} \end{aligned}$

Therefore, the sum of n terms is $n^{2}$

Q10 (i) Show that $a_{1}, a_{2}, \dots, a_{n}, \dots$ form an AP where an is defined as below : $a_{n} = 3 + 4 n$ Also find the sum of the first 15 terms.

Answer:

It is given that

$a_{n} = 3 + 4 n$

We will check values of $a_{n}$ for different values of $n$

$\begin{aligned} a_{1} = 3 + 4 (1) = 3 + 4 = 7 \\ a_{2} = 3 + 4 (2) = 3 + 8 = 11 \\ a_{3} = 3 + 4 (3) = 3 + 12 = 15 \end{aligned}$

and so on.
From the above, we can clearly see that this is an AP with the first term(a) equals to 7 and common difference (d) equals to 4
Now, we know that

$\begin{aligned} S_{n} = \frac{n}{2} {2 a + (n - 1) d} \\ \Rightarrow S_{15} = \frac{15}{2} {2 \times (7) + (15 - 1) 4} \\ \Rightarrow S_{15} = \frac{15}{2} {14 + 56} \\ \Rightarrow S_{15} = \frac{15}{2} {70} \\ \Rightarrow S_{15} = 15 \times 35 \\ \Rightarrow S_{15} = 525 \end{aligned}$

Therefore, the sum of 15 terms is 525

Q10 (ii) Show that $a_{1}, a_{2}, \dots, a_{n}, \dots$ form an AP where an is defined as below : $a_{n} = 9 - 5 n$ . Also find the sum of the first 15 terms in each case.

Answer:

It is given that

$a_{n} = 9 - 5 n$

We will check values of $a_{n}$ for different values of n

$\begin{aligned} a_{1} = 9 - 5 (1) = 9 - 5 = 4 \\ a_{2} = 9 - 5 (2) = 9 - 10 = - 1 \\ a_{3} = 9 - 5 (3) = 9 - 15 = - 6 \end{aligned}$

and so on.
From the above, we can clearly see that this is an AP with the first term(a) equals to $4$ and common difference (d) equals to -5
Now, we know that

$\begin{aligned} S_{n} = \frac{n}{2} {2 a + (n - 1) d} \\ \Rightarrow S_{15} = \frac{15}{2} {2 \times (4) + (15 - 1) (- 5)} \\ \Rightarrow S_{15} = \frac{15}{2} {8 - 70} \\ \Rightarrow S_{15} = \frac{15}{2} {- 62} \\ \Rightarrow S_{15} = 15 \times (- 31) \\ \Rightarrow S_{15} = - 465 \end{aligned}$

Therefore, the sum of $15$ terms is $- 465$

Q11 If the sum of the first n terms of an AP is $4 n - n_{2}^{2}$ , what is the first term (that is $(S_{1})$ ) What is the sum of first two terms? What is the second term? similarly, find the 3rd, the 10 th and the n th terms

Answer:

It is given that the sum of the first $n$ terms of an AP is $4 n - n^{2}$
Now,

$\Rightarrow S_{n} = 4 n - n^{2}$

Now, first term is

$\Rightarrow S_{1} = 4 (1) - 1^{2} = 4 - 1 = 3$

Therefore, first term is 3
Similarly,

$\Rightarrow S_{2} = 4 (2) - 2^{2} = 8 - 4 = 4$

Therefore, sum of first two terms is 4
Now, we know that

$\begin{aligned} \Rightarrow S_{n} = \frac{n}{2} {2 a + (n - 1) d} \\ \Rightarrow S_{2} = \frac{2}{2} {2 \times 3 + (2 - 1) d} \\ \Rightarrow 4 = {6 + d} \\ \Rightarrow d = - 2 \end{aligned}$

Now,

$a_{2} = a + d = 3 + (- 2) = 1$

Similarly,

$\begin{aligned} a_{3} = a + 2 d = 3 + 2 (- 2) = 3 - 4 = - 1 \\ a_{10} = a + 9 d = 3 + 9 (- 2) = 3 - 18 = - 15 \\ a_{n} = a + (n - 1) d = 3 + (n - 1) (- 2) = 5 - 2 n \end{aligned}$

Q12 Find the sum of the first 40 positive integers divisible by 6 .

Answer:

Positive integers divisible by 6 are

$6, 12, 18, \dots$

This is an AP with
here, $a = 6$ and $d = 6$
Now, we know that

$\begin{aligned} S_{n} = \frac{n}{2} {2 a + (n - 1) d} \\ \Rightarrow S_{40} = \frac{40}{2} {2 \times 6 + (40 - 1) 6} \\ \Rightarrow S_{40} = 20 {12 + 234} \\ \Rightarrow S_{40} = 20 {246} \\ \Rightarrow S_{40} = 4920 \end{aligned}$

Therefore, sum of the first 40 positive integers divisible by 6 is 4920

Q13 Find the sum of the first 15 multiples of 8 .

Answer:

First 15 multiples of 8 are

$8, 16, 24, \dots$

This is an AP with
here, $a = 8$ and $d = 8$
Now, we know that

$\begin{aligned} S_{n} = \frac{n}{2} {2 a + (n - 1) d} \\ \Rightarrow S_{15} = \frac{15}{2} {2 \times 8 + (15 - 1) 8} \\ \Rightarrow S_{15} = \frac{15}{2} {16 + 112} \\ \Rightarrow S_{15} = \frac{15}{2} {128} \\ \Rightarrow S_{15} = 15 \times 64 = 960 \end{aligned}$

Therefore, sum of the first 15 multiple of 8 is $960$

Q14 Find the sum of the odd numbers between 0 and 50.

Answer:

The odd number between 0 and 50 are

$1, 3, 5, \dots 49$

This is an AP with
here, $a = 1$ and $d = 2$
There are total $25$ odd number between 0 and 50
Now, we know that

$\begin{aligned} S_{n} = \frac{n}{2} {2 a + (n - 1) d} \\ \Rightarrow S_{25} = \frac{25}{2} {2 \times 1 + (25 - 1) 2} \\ \Rightarrow S_{25} = \frac{25}{2} {2 + 48} \\ \Rightarrow S_{25} = \frac{25}{2} \times 50 \\ \Rightarrow S_{25} = 25 \times 25 = 625 \end{aligned}$

Therefore, sum of the odd numbers between 0 and 50625

Q15 A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs 200 for the first day, Rs 250 for the second day, Rs 300 for the third day, etc., the penalty for each succeeding day being Rs 50 more than for the preceding day. How much money the contractor has to pay a penalty, if he has delayed the work by 30 days?

Answer:

It is given that
Penalty for delay of completion beyond a certain date is Rs 200 for the first day, Rs 250 for the second day, Rs 300 for the third day and penalty for each succeeding day being Rs 50 more than for the preceding day We can clearly see that
$200, 250, 300, \dots .$ . is an AP with

$a = 200 and d = 50$

Now, the penalty for 30 days is given by the expression

$\begin{aligned} S_{30} = \frac{30}{2} {2 \times 200 + (30 - 1) 50} \\ S_{30} = 15 (400 + 1450) \\ S_{30} = 15 \times 1850 \\ S_{30} = 27750 \end{aligned}$

Therefore, the penalty for 30 days is $27750$

Q16 A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each of the prizes.

Answer:

It is given that
Each price is decreased by 20 rupees,
Therefore, $d = - 20$ and there are total 7 prizes so $n = 7$ and sum of prize money is Rs 700 so $S_{7} = 700$
Let a be the prize money given to the 1 st student
Then,

$\begin{aligned} S_{7} = \frac{7}{2} {2 a + (7 - 1) (- 20)} \\ 700 = \frac{7}{2} {2 a - 120} \\ 2 a - 120 = 200 \\ a = \frac{320}{2} = 160 \end{aligned}$

Therefore, the prize given to the first student is Rs 160
Now,
Let $a_{2}, a_{2}, \dots, a_{7}$ is the prize money given to the next 6 students then,

$\begin{aligned} a_{2} = a + d = 160 + (- 20) = 160 - 20 = 140 \\ a_{3} = a + 2 d = 160 + 2 (- 20) = 160 - 40 = 120 \\ a_{4} = a + 3 d = 160 + 3 (- 20) = 160 - 60 = 100 \\ a_{5} = a + 4 d = 160 + 4 (- 20) = 160 - 80 = 80 \\ a_{6} = a + 5 d = 160 + 5 (- 20) = 160 - 100 = 60 \\ a_{7} = a + 6 d = 160 + 6 (- 20) = 160 - 120 = 40 \end{aligned}$

Therefore, prize money given to 1 to 7 student is $160, 140, 120, 100, 80, 60, 40$

Q17 In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections in each class. How many trees will be planted by the students?

Answer:

First there are 12 classes and each class has 3 sections
Since each section of class 1 will plant 1 tree, so 3 trees will be planted by 3 sections of class 1 . Thus every class will plant 3 times the number of their class Similarly,

No. of trees planted by 3 sections of class $1 = 3$
No. of trees planted by 3 sections of class $2 = 6$
No. of trees planted by 3 sections of class $3 = 9$
No. of trees planted by 3 sections of class $4 = 12$
Its clearly an AP with first term $(a) = 3$ and common difference $(d) = 3$ and total number of classes $(n) = 12$

Now, number of trees planted by 12 classes is given by

$\begin{aligned} S_{12} = \frac{12}{2} {2 \times 3 + (12 - 1) \times 3} \\ S_{12} = 6 (6 + 33) \\ S_{12} = 6 \times 39 = 234 \end{aligned}$

Therefore, number of trees planted by 12 classes is $234$

Q18 A spiral is made up of successive semicircles, with centres alternately at $[A$ and $B]$ , starting with centre at $A$ ,of radii $0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, \dots$ as shown in Fig, 5.4 . What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take $π = \frac{22}{7}$ )

1635921529757

[ Hint : Length of successive semicircles is $l_{1}, l_{2}, l_{3}, l_{4}, \dots$ with centres at $A, B, A, B, \dots$ , respectively.]
Answer:
From the above-given figure

Circumference of 1 st semicircle $l_{1} = π r_{1} = 0.5 π$

Similarly,
Circumference of 2nd semicircle $l_{2} = π r_{2} = π$
Circumference of 3rd semicircle $l_{3} = π r_{3} = 1.5 π$
It is clear that this is an AP with $a = 0.5 π$ and $d = 0.5 π$

Now, sum of length of 13 such semicircles is given by

$\begin{aligned} S_{13} = \frac{13}{2} {2 \times 0.5 π + (13 - 1) 0.5 π} \\ S_{13} = \frac{13}{2} (π + 6 π) \\ S_{13} = \frac{13}{2} \times 7 π \\ S_{13} = \frac{91 π}{2} = \frac{91}{2} \times \frac{22}{7} = 143 \end{aligned}$

Therefore, sum of length of 13 such semicircles is $143 c m$

Q19 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (see Fig_ 5.5 ). In how many rows are the 200 logs placed and how many_logs are in the top row?

1635921556744

Answer:

As the rows are going up, the no of logs are decreasing,
We can clearly see that $20, 19, 18, \dots$ , is an AP.
and here $a = 20$ and $d = - 1$
Let suppose 200 logs are arranged in ' $n$ ' rows,
Then,

$\begin{aligned} S_{n} = \frac{n}{2} {2 \times 20 + (n - 1) (- 1)} \\ 200 = \frac{n}{2} {41 - n} \\ \Rightarrow n^{2} - 41 n + 400 = 0 \\ \Rightarrow n^{2} - 16 n - 25 n + 400 = 0 \\ \Rightarrow (n - 16) (n - 25) = 0 \\ \Rightarrow n = 16 and n = 25 \end{aligned}$

Now,

$\begin{aligned} case (i) n = 25 \\ a_{25} = a + 24 d = 20 + 24 \times (- 1) = 20 - 24 = - 4 \end{aligned}$

But number of rows can not be in negative numbers
Therefore, we will reject the value $n = 25$
case (ii) $n = 16$

$a_{16} = a + 15 d = 20 + 15 \times (- 1) = 20 - 15 = 5$

Therefore, the number of rows in which 200 logs are arranged is equal to 5

Q20 In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see Fig. 5.6 ).

A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?

[ Hint : To pick up the first potato and the second potato, the total distance (in metres) run by a competitor is $2 \times 5 + 2 \times (5 + 3)]$

Answer:

Distance travelled by the competitor in picking and dropping 1st potato $= 2 \times 5 = 10 m$
Distance travelled by the competitor in picking and dropping 2nd potato $= 2 \times (5 + 3) = 2 \times 8 = 16 m$
Distance travelled by the competitor in picking and dropping 3rd potato $= 2 \times (5 + 3 + 3) = 2 \times 11 = 22 m$
and so on
we can clearly see that it is an AP with first term $(a) = 10$ and common difference $(d) = 6$
There are 10 potatoes in the line
Therefore, total distance travelled by the competitor in picking and dropping potatoes is

$\begin{aligned} S_{10} = \frac{10}{2} {2 \times 10 + (10 - 1) 6} \\ S_{10} = 5 (20 + 54) \\ S_{10} = 5 \times 74 = 370 \end{aligned}$

Therefore, the total distance travelled by the competitor in picking and dropping potatoes is $370 m$

Solutions of Arithmetic Progressions Class 10 Exercise: 5.4

Q1 Which term of the AP: is its first negative term? [ Hint :Find $n$ for $a_{n} < 0$ ]

Answer:

Given AP is

$121, 117, 113, \dots$

Here $a = 121$ and $d = - 4$
Let suppose nth term of the AP is first negative term
Then,

$a_{n} = a + (n - 1) d$

If $n$ nth term is negative then $a_{n} < 0$

$\begin{aligned} \Rightarrow 121 + (n - 1) (- 4) < 0 \\ \Rightarrow 125 < 4 n \\ \Rightarrow n > \frac{125}{4} = 31.25 \end{aligned}$

Therefore, first negative term must be 32nd term

Q2 The sum of the third and the seventh terms of an AP is 6 and their product is 8 . Find the sum of first sixteen terms of the AP.

Answer:

It is given that sum of third and seventh terms of an AP are and their product is 8

$\begin{aligned} a_{3} = a + 2 d \\ a_{7} = a + 6 d \end{aligned}$

Now,

$\begin{aligned} a_{3} + a_{7} = a + 2 d + a + 6 d = 6 \\ \Rightarrow 2 a + 8 d = 6 \\ \Rightarrow a + 4 d = 3 \Rightarrow a = 3 - 4 d \end{aligned}$

And

$a_{3} \cdot a_{7} = (a + 2 d) \cdot (a + 6 d) = a^{2} + 8 a d + 12 d^{2} = 8 - (i i)$

put value from equation (i) in (ii) we will get

$\begin{aligned} \Rightarrow (3 - 4 d)^{2} + 8 (3 - 4 d) d + 12 d^{2} = 8 \\ \Rightarrow 9 + 16 d^{2} - 24 d + 24 d - 32 d^{2} + 12 d^{2} = 8 \\ \Rightarrow 4 d^{2} = 1 \\ \Rightarrow d = \pm \frac{1}{2} \end{aligned}$

Now,

$case (i) d = \frac{1}{2}$

$a = 3 - 4 \times \frac{1}{2} = 1$

Then,

$S_{16} = \frac{16}{2} {2 \times 1 + (16 - 1) \frac{1}{2}}$

$S_{16} = 76$

$\begin{aligned} case (ii) d = - \frac{1}{2} \\ a = 3 - 4 \times (- \frac{1}{2}) = 5 \end{aligned}$

Then,

$\begin{aligned} S_{16} = \frac{16}{2} {2 \times 1 + (16 - 1) (- \frac{1}{2})} \\ S_{16} = 20 \end{aligned}$

Q3 A ladder has rungs $[25] cm$ apart. (see Fig_ $[5.7$ ). The rungs decrease uniformly in length from 450 cm at the bottom to $[55 cm$ at the top. If the top and the bottom rungs are $^{2} 2]$ m apart, what is the length of the wood required for the rungs? [ Hint: Number of $gs = \frac{250}{25} + 1]$

1635921645173

Answer:

It is given that
The total distance between the top and bottom rung $= 2 \frac{1}{2} m = 250 cm$
Distance between any two rungs $= 25 cm$
Total number of rungs $= \frac{250}{25} + 1 = 11$
And it is also given that bottom-most rungs is of 45 cm length and topmost is of 25 cm length. As it is given that the length of rungs decrease uniformly, it will form an AP with $a = 25, a_{11} = 45$ and $n = 11$
Now, we know that

$a_{11} = a + 10 d$

$\begin{aligned} \Rightarrow 45 = 25 + 10 d \\ \Rightarrow d = 2 \end{aligned}$

Now, total length of the wood required for the rungs is equal to

$\begin{aligned} S_{11} = \frac{11}{2} {2 \times 25 + (11 - 1) 2} \\ S_{11} = \frac{11}{2} {50 + 20} \\ S_{11} = \frac{11}{2} \times 70 \\ S_{11} = 385 cm \end{aligned}$

Therefore, the total length of the wood required for the rungs is equal to 385 cm

Q4 The houses of a row are numbered consecutively from [1] to 49 . Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it Find this value $x . [$ Hint $: S_{x - 1} = S_{49} - S_{x}]$

Answer:

It is given that the sum of the numbers of the houses preceding the house numbered $x$ is equal to the sum of the numbers of the houses following it
And $1, 2, 3, \dots ., 49$ form an AP with $a = 1$ and $d = 1$
Now, we know that

$S_{n} = \frac{n}{2} {2 a + (n - 1) d}$

Suppose their exist an $n$ term such that ( $n < 49$ )
Now, according to given conditions
Sum of first $n - 1$ terms of AP = Sum of terms following the nth term
Sum of first $n - 1$ term of AP $=$ Sum of whole AP - Sum of first $m$ terms of AP
i.e.
$S_{n - 1} = S_{49} - S_{n}$
$\frac{n - 1}{2} {2 a + ((n - 1) - 1) d} = \frac{49}{2} {2 a + (49 - 1) d} - \frac{n}{2} {2 a + (n - 1) d}$
$\frac{n - 1}{2} {2 + (n - 2)} = \frac{49}{2} {2 + 48} - \frac{n}{2} {2 + (n - 1)}$
$\frac{n - 1}{2} {n} = \frac{49}{2} {50} - \frac{n}{2} {n + 1}$

$\frac{n^{2}}{2} - \frac{n}{2} = 1225 - \frac{n^{2}}{2} - \frac{n}{2}$

$\begin{aligned} n^{2} = 1225 \\ n = \pm 35 \end{aligned}$

Given House number are not negative so we reject n = -35

Therefore, the sum of no of houses preceding the house no 35 is equal to the sum of no of houses following the house no 35

Q5A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of $\frac{1}{4} m$ and a tread of $\frac{1}{2} m$ (see Fig , 5.8 ). Calculate the total volume of concrete required to build the terrace. Hint: Volume of concrete required to build the first step $= \frac{1}{4} \times \frac{1}{2} \times 50 m^{3}$ ,

1635921665136

Answer:

It is given that

football ground comprises of 15 steps each of which is 50 m long and Each step has a rise of $\frac{1}{4} m$ and a tread of $\frac{1}{2} m$
Now,
The volume required to make the first step $= \frac{1}{4} \times \frac{1}{2} \times 50 = 6.25 m^{3}$

Similarly,
The volume required to make 2nd step $= (\frac{1}{4} + \frac{1}{4}) \times \frac{1}{2} \times 50 = \frac{1}{2} \times \frac{1}{2} \times 50 = 12.5 m^{3}$
And
The volume required to make 3 rd step $= (\frac{1}{4} + \frac{1}{4} + \frac{1}{4}) \times \frac{1}{2} \times 50 = \frac{3}{4} \times \frac{1}{2} \times 50 = 18.75 m^{3}$

And so on
We can clearly see that this is an AP with $a = 6.25$ and $d = 6.25$
Now, the total volume of concrete required to build the terrace of 15 such step is

$\begin{aligned} S_{15} = \frac{15}{2} {2 \times 6.25 + (15 - 1) 6.25} \\ S_{15} = \frac{15}{2} {12.5 + 87.5} \\ S_{15} = \frac{15}{2} \times 100 \\ S_{15} = 15 \times 50 = 750 \end{aligned}$

Therefore, the total volume of concrete required to build the terrace of 15 such steps is $750 m^{3}$

If interested, students can also check exercises here:

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NCERT Solutions for Class 10 Maths - Chapter Wise

Chapter No.	Chapter Name
Chapter 1	Real Numbers
Chapter 2	Polynomials
Chapter 3	Pair of Linear Equations in Two Variables
Chapter 4	Quadratic Equations
Chapter 5	Arithmetic Progressions
Chapter 6	Triangles
Chapter 7	Coordinate Geometry
Chapter 8	Introduction to Trigonometry
Chapter 9	Some Applications of Trigonometry
Chapter 10	Circles
Chapter 11	Constructions
Chapter 12	Areas Related to Circles
Chapter 13	Surface Areas and Volumes
Chapter 14	Statistics
Chapter 15	Probability

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Benefits of Solutions of Arithmetic Progressions Class 10 Maths NCERT Chapter 5

As NCERT Class 10 Maths Chapter 5 solutions are solved by the subject matter experts, the answers to all the questions are reliable. NCERT Class 10 Maths Chapter 5 Solutions gives the step-by-step explanations to all the questions which makes it easy for the students to understand. Using the NCERT Class 10 Maths chapter 5 Solutions, students will be able to confirm the right answers once they are done solving the questions themselves.

How to Use Solutions of Arithmetic Progressions Class 10 Maths NCERT Chapter 5?

In this article, you have gone through the solutions of Arithmetic Progressions Class 10 Maths NCERT Chapter 5 and have a good knowledge of answering structurally. It's time to practice various kinds of problems based on Arithmetic Progression.

After the completion of the NCERT syllabus, you can check the past 5-year papers of board exams. Class 10 Maths Chapter 5 Test Paper with Solution will increase your dealing ability with a variety of questions.

NCERT syllabus coverage and previous year papers are enough tools to get a good score in the board examination. After covering NCERT and the previous year papers of this chapter, you can jump to the next chapters.

NCERT Solutions of Class 10 - Subject Wise

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Frequently Asked Questions (FAQs)

1. How do you find the nth term of an arithmetic progression?

The nth term of an AP is given by the formula, $a_{n} = a + (n - 1) d$
where
a = First term of the sequence.
n = Term's position in the sequence.
d = Common difference.

2. What is the sum of n terms in an arithmetic progression?

The sum of the first 'n’ terms in an AP is calculated using the formula, $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$
Where
Sn denotes the sum of the terms
'n' is the number of terms being summed
'a' is the first term
'd' stands for the common difference.

3. What is the common difference in an arithmetic sequence?

4. How to find the missing terms in an arithmetic progression?

The missing terms in an arithmetic progression can be found using the formula of nth term of the arithmetic progression. The nth term of an AP is given by the formula, $a_{n} = a + (n - 1) d$

where
a = First term of the sequence
n = Term's position in the sequence
d = Common difference.

5. What is the formula for the sum of the first n natural numbers?

The sum of first n natural numbers can be found using the formula $\frac{n (n + 1)}{2}$ .

Also Read

NCERT Exemplar Class 10 Science Solutions Chapter 1 Chemical Reactions and Equations

NCERT Exemplar Class 10 Science Solutions - Chapter Wise Problems with Solutions

NCERT Solutions for Class 10 Science Chapter 5 Life Processes

NCERT Solutions For Class 10 Maths Chapter 4 Quadratic Equations

NCERT Solutions for Exercise 12.3 Class 10 Maths Chapter 12 - Areas related to circles

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i studied 1 to 10th CBSE board abroad and 11th 12th in karnataka state board . i am a domicile of karnataka and comes under minority, income under 5 lakhs. how neet councillng will allocate my seat

Hello

Since you are a domicile of Karnataka and have studied under the Karnataka State Board for 11th and 12th , you are eligible for Karnataka State Quota for admission to various colleges in the state.

1. KCET (Karnataka Common Entrance Test): You must appear for the KCET exam, which is required for admission to undergraduate professional courses like engineering, medical, and other streams. Your exam score and rank will determine your eligibility for counseling.

2. Minority Income under 5 Lakh : If you are from a minority community and your family's income is below 5 lakh, you may be eligible for fee concessions or other benefits depending on the specific institution. Some colleges offer reservations or other advantages for students in this category.

3. Counseling and Seat Allocation:

After the KCET exam, you will need to participate in online counseling.

You need to select your preferred colleges and courses.

Seat allocation will be based on your rank , the availability of seats in your chosen colleges and your preferences.

4. Required Documents :

Domicile Certificate (proof that you are a resident of Karnataka).

Income Certificate (for minority category benefits).

Marksheets (11th and 12th from the Karnataka State Board).

KCET Admit Card and Scorecard.

This process will allow you to secure a seat based on your KCET performance and your category .

check link for more details

https://medicine.careers360.com/neet-college-predictor

Hope this helps you .

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Mitali Sharma 30 January,2025

show the previous year exams ?

Hello Aspirant, Hope your doing great, your question was incomplete and regarding what exam your asking.

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SWETCHCHA MISKA 10 July,2024

as an icse student in class 10 is above 80 percent good enough to get into commerce in 11th cbse

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

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Manasvini Gupta 23 July,2024

i had cleared 10th from cbse in 2022 2 years ago now i wish to apply for 12th as private candidate in 2025. can i do so?

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

Read Complete Answer

Ashvadha 03 July,2024

i had cleared 10th in 2022 from cbse can i apply for 12th as private candidate this year?

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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NCERT Solutions For Class 10 Maths Chapter 5 Arithmetic Progressions

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions PDF Free Download

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions - Notes

Arithmetic Progression

Common Difference of Arithmetic Progression

nth Term of Arithmetic Progression (AP)

Sum of First n Terms in Arithmetic Progression (A.P)

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Answer: