NCERT Exemplar Class 10 Maths Solutions Chapter 5 Arithmetic Progressions 2021

NCERT Exemplar Class 10 Maths Solutions Chapter 5 Arithmetic Progressions

Edited By Komal Miglani | Updated on Mar 31, 2025 10:52 AM IST | #CBSE Class 10th

Picture yourself sitting in a huge stadium with thousands of people surrounding you, each row piled higher and higher in perfect symmetry. Or imagine borrowing money from a bank, where every instalment comes in the form of a fixed pattern. Ever thought about what’s behind all these situations? The solution is Arithmetic Progressions (AP)! This chapter helps students understand and solve sequences in which every term has a constant difference from the preceding one. From calculating the nth term to adding n terms, this topic is useful in various areas, ranging from financial planning to building designs.

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In this article, you will find detailed solutions for all the exemplar problems of NCERT Class 10 Maths Chapter 5 designed by the subject matter experts at Careers360. The NCERT Exemplar Class 10 Maths solutions are carefully prepared by our experienced mathematics team, ensuring in-depth concept clarity while aligning with the NCERT Syllabus Class 10 Maths.

NCERT Exemplar Class 10 Maths Solutions Chapter 5

Class 10 Maths Chapter option 5 exemplar solutions Exercise: 5.1
Page number: 45-47
Total questions: 18

Question:1

In an AP, if d = –4, n = 7, a_n=4, then a is

(A) 6

(B) 7

(D) 28

Answer:

Given
$a_{n} = 4, d = - 4, n = 7$

$U s e a_{n} = a + (n - 1) . d$

$a + (7 - 1) . (- 4) = 4$

$a + (6) . (- 4) = 4$

$a - 24 = 4$

$a = 4 + 24$

$a = 28$
Hence, option D is correct

Question:2

In an AP, if a = 3.5, d = 0, n = 101, then a_n will be
(A) 0
(B) 3.5
(C) 103.5
(D) 104.5

Answer:

We know that
$a_{n} = a + (n - 1) . d$

$G i v e n, a = 3.5, d = 0, n = 101$

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

$a_{n} = 3.5 + (101 - 1) .0$

$a_{n} = 3.5 + 0$

$a_{n} = 3.5$
Hence, option B is correct.

Question:3

The list of numbers – 10, – 6, – 2, 2... is
(A) an AP with d = – 16
(B) an AP with d = 4
(C) an AP with d = – 4
(D) not an AP

Answer:

$a_{1} = - 10, a_{2} = - 6, a_{3} = - 2, a_{4} = 2$

$a_{2} - a_{1} = - 6 - (- 10)$

$= - 6 + 10$

$= 4$

$a_{3} - a_{2} = - 2 - (- 6)$

$= - 2 + 6$

$= 4$

$a_{4} - a_{3} = 2 - (- 2)$

$= 2 + 2$

$= 4$

Since the given sequence is an AP with d = 4
ANS - (B)

Question:4

The 11th term of the AP:
$- 5, \frac{- 5}{2}, 0, \frac{5}{2},$ \ldots $. . i s$
(A) –20
(B) 20
(C) –30
(D) 30

Answer:

Here AP is

$- 5, - \frac{5}{2}, 0, \frac{5}{2}, \dots \dots \dots$
Here

$a_{1} = - 5, a_{2} = - \frac{5}{2}$

(Common difference) $d = a_{2} - a_{1}$

$\begin{aligned} = - \frac{5}{2} - (- 5) \\ = - \frac{5}{2} + 5 \\ = \frac{- 5 + 10}{2} \\ = \frac{5}{2} \end{aligned}$
$\begin{aligned} a_{n} = a_{1} + (n - 1) \cdot d \\ Put n = 11 \\ a_{11} = a_{1} + (11 - 1) \cdot d \\ a_{11} = - 5 + (10) \cdot (5 / 2) \\ = - 5 + 5 \cdot (5) \\ = - 5 + 25 \\ = 20 \end{aligned}$

Hence, option B is correct.

Question:5

The first four terms of an AP, whose first term is –2 and the common difference is –2, are
(A) – 2, 0, 2, 4
(B) – 2, 4, –8, 16
(C) – 2, – 4, – 6, – 8
(D) – 2, – 4, – 8, –16

Answer:

Given,
$a = - 2, d = - 2$

$a + d = - 2 + (- 2) = - 4$

$a + 2 d = - 2 + 2 (- 2) = - 2 - 4 = - 6$

$a + 3 d = - 2 + 3 (- 2) = - 2 - 6 = - 8$
The given AP is -2, -4, -6, 8
Hence, option (C) is correct.

Question:6

The 21st term of the AP, whose first two terms are 3 and 4, is
(A) 17
(B) 137
(C) 143
(D) –143

Answer:

Given

$a_{1} = - 3$
$a_{2} = 4$

(Common difference) $d = a_{2} - a_{1}$
$= 4 - (- 3)$
$= 4 + 3$
$= 7$

We know that the general term of an arithmetic progression is:
$a_{n} = a + (n - 1) d$

Put $n = 21$ :
$a_{21} = a + (21 - 1) d$
$= - 3 + 20 \cdot 7$
$= - 3 + 140$
$= 137$

$a_{21} = 137$

Hence, option B is correct.

Question:7

If the 2nd term of an AP is 13 and the 5th term is 25, what is its 7th term?
(A) 30
(B) 33
(C) 37
(D) 38

Answer:

We know that

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

Given: $a_{2} = 13$ , $a_{5} = 25$

From equation (1), put $n = 2$ :
$a + (2 - 1) d = 13$
$\Rightarrow a + d = 13$

Put $n = 5$ in equation (1):
$a + (5 - 1) d = 25$
$\Rightarrow a + 4 d = 25$

Subtract equation (2) from equation (3):
$(a + 4 d) - (a + d) = 25 - 13$
$\Rightarrow 3 d = 12$
$\Rightarrow d = \frac{12}{3} = 4$

Substitute $d = 4$ in equation (2):
$a + 4 = 13$
$\Rightarrow a = 13 - 4 = 9$

Now find $a_{7}$ :
$a_{7} = a + (7 - 1) d$
$= 9 + 6 \cdot 4$
$= 9 + 24 = 33$

$a_{7} = 33$

Hence, option B is correct.

Question:8

Which term of the AP: 21, 42, 63, 84... is 210?
(A) 9th
(B) 10th
(C) 11th
(D) 12th

Answer:

$a_{n} = 210 AP$ is $21, 42, 63, 84, \dots$

Here $a_{1} = 21 d = a_{2} - a_{1} ($ where $d$ is the common difference $)$

$d = 42 - 21 (a_{1}$ is the first term $)$

$d = 21$

$a_{n} = 210$

$a_{1} + (n - 1) \cdot d = 210$

$21 + (n - 1) \cdot 21 = 210$

$21 + 21 n - 21 = 210$

$21 n = 210$

$n = \frac{210}{21}$

$n = 10$

Hence the $10^{th}$ term of the AP is 210

Question:9

If the common difference of an AP is 5, then what is $a_{18} - a_{13}$ ?
(A) 5
(B) 20
(C) 25
(D) 30

Answer:

Given

$\begin{aligned} d = 5 \\ a_{18} - a_{13} = ? \end{aligned}$

We know that

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

Put $n = 18$ ,

$a_{18} = a + (18 - 1) \cdot d$

Put $n = 13$ ,

$a_{13} = a + (13 - 1) \cdot d$

Now,

$\begin{matrix} a_{18} - a_{13} = (a + (18 - 1) \cdot d) - (a + (13 - 1) \cdot d) \\ = a + 17 d - a - 12 d \\ = 17 d - 12 d \\ = 5 d \\ = 5 \times 5 \\ = 25 \end{matrix}$
Option (C) is correct.

Question:10

What is the common difference of an AP in which $a 18 - a 14 = 32$ ?
(A) 8
(B) – 8
(C) – 4
(D) 4

Answer:

Given,

$a_{18} - a_{14} = 32$
We know that $a_{n} = a + (n - 1) \cdot d$
Put $n = 18$

$\begin{aligned} a_{18} = a + (18 - 1) \cdot d \\ a_{18} = 9 + 17 d \dots \dots \dots \dots \dots \dots (1) \\ n = 14 \\ a_{14} = a + (14 - 1) \cdot d \\ a_{14} = 9 + 13 d \dots \dots \dots \dots \dots (2) \\ subtract eq (2) from eq(1) \\ a_{18} - a_{14} = 32 \\ (a + 17 d) - (a + 13 d) = 32 \\ a + 17 d - a - 13 d = 32 \\ 4 d = 32 \\ d = \frac{32}{4} \\ d = 8 \end{aligned}$
Hence common difference is 8.

Question:11

Two APs have the same common difference. The first term of one of these is 1, and that of the other is 8. Then the difference between their 4th terms is
(A) –1
(B) – 8
(C) 7
(D) –9

Answer:

Let the first AP be $a_{1}, a_{2}, a_{3}, \dots \dots \dots$ .
Let the second AP be $b_{1}, b_{2}, b_{3}, \dots \dots \dots \dots$
Given, $a_{1} = - 1, b_{1} = - 8$
Difference between their $4_{th}$ terms is:

$a_{4} - b_{4} = (a_{1} + 3 d) - (b_{1} + 3 d)$

Given that the common difference of both A.Ps is equal,

$\begin{matrix} a_{4} - b_{4} = - 1 - (- 8) \\ = - 1 + 8 \\ = 7 \end{matrix}$

Hence, option (C) is correct

Question:12

If 7 times the 7th term of an AP is equal to 11 times its 11th term, then its 18th term will be
(A) 7
(B) 11
(C) 18
(D) 0

Answer:

Given $7 \times (a_{7}) = 11 \times (a_{11})$

$\begin{matrix} 7 \times (a + 6 d) = 11 \times (a + 10 d) (Using a_{n} = a + (n - 1) \times d) \\ 7 a + 42 d = 11 a + 110 d \\ 11 a - 7 a + 110 d - 42 d = 0 \\ 4 a + 68 d = 0 \end{matrix}$

Dividing by 4 , we get

$\begin{matrix} a + 17 d = 0 \dots \\ a_{18} = a + (18 - 1) d \\ = a + 17 d \\ = 0 \end{matrix}$

Hence, option D is correct.

Question:13

The 4th term from the end of the AP: –11, –8, –5, ..., 49 is
(A) 37
(B) 40
(C) 43
(D) 58

Answer:

Given

$\begin{matrix} 7 \times (a_{7}) = 11 \times (a_{11}) \\ 7 \times (a + 6 d) = 11 \times (a + 10 d) (Using a_{n} = a + (n - 1) \times d) \\ 7 a + 42 d = 11 a + 110 d \\ 11 a - 7 a + 110 d - 42 d = 0 \\ 4 a + 68 d = 0 \end{matrix}$

Dividing by 4 , we get

$\begin{matrix} a + 17 d = 0 \dots \\ a_{18} = a + (18 - 1) d \\ = a + 17 d \\ = 0 \end{matrix}$

Hence, option B is correct.

Question:14

The famous mathematician associated with finding the sum of the first 100 natural numbers is
(A) Pythagoras
(B) Newton
(C) Gauss
(D) Euclid

Answer:

[C]
Famous mathematician Gauss found the sum of the first 100 natural numbers.
According to this when he added the first and the last number in the series, the second and the second last number in the series, and so on. He gets a sum of 101 and50 pairs exist so the sum of 100 natural numbers is 50×101=5050.

Question:15

If the first term of an AP is i–5 and the common difference is 2, then the sum of the first 6 terms is
(A) 0
(B) 5
(C) 6
(D) 15

Answer:

Given $a = - 5, d = 2$ (where $a$ is the first term and $d$ is the common difference)

$\begin{matrix} S_{6} = \frac{6}{2} (2 a + (6 - 1) d) \\ = 3 (2 \times (- 5) + 5 \times 2) \\ = 3 (- 10 + 10) \\ = 3 (0) = 0 \end{matrix}$
Hence, option A is correct

Question:16

The sum of the first 16 terms of the AP: 10, 6, 2... is
(A) –320
(B) 320
(C) –352
(D) –400

Answer:

Given: $AP = 10, 6, 2 \dots$ here

$\begin{aligned} a_{1} = 10, a_{2} = 6 \\ d = a_{2} - a_{1} \\ d = 6 - 10 \\ d = - 4 \\ S_{16} = \frac{16}{2} (2 a + (16 - 1) d) \\ = 8 (2 \times 10 + 15 \times (- 4)) \\ = 8 (20 - 60) \\ = 8 \times (- 40) = - 320 \end{aligned}$
Hence, option A is the correct answer.

Question:17

$In an AP if a = 1,$ $a_{n} = 20$ and $S_{n} = 399$ \text{, then n is }$
(A) 19
(B) 21
(C) 38
(D) 42

Answer:

Given : $a = 1, a_{n} = 20, S_{n} = 399$

$\begin{aligned} S_{n} = \frac{n}{2} (a + a_{n}) \\ 38 = n \end{aligned}$
He, once option C is correct.

Question:18

The sum of the first five multiples of 3 is
(A) 45
(B) 55
(C) 65
(D) 75

Answer:

Given:- Multiples of 3 are: $3, 6, 9, 12, \dots \dots$ .

$\begin{matrix} a_{1} = 3, d = a_{2} - a_{1} \\ d = 6 - 3 \\ d = 3 \\ S_{5} = \frac{5}{2} [2 a_{1} + (5 - 1) d] \\ = \frac{5}{2} (2 \times 3 + 4 \times 3) \\ = \frac{5}{2} (6 + 12) \\ = \frac{5}{2} \times 18 \end{matrix}$
Option A is correct.

Class 10 Maths Chapter 5 exemplar solutions Exercise: 5.2
Page number: 49-50
Total questions: 8

Question:1

i)Which of the following forms an AP? Justify your answer.
–1, –1, –1, –1, ...
ii)
Which of the following forms an AP? Justify your answer.
0, 2, 0, 2, ...
iii)
Which of the following is an AP? Justify your answer.
1, 1, 2, 2, 3, 3,...
iv)
Which of the following forms an AP? Justify your answer.
11, 22, 33,...
v)
Which of the forms form an AP? Justify your answer.
$\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, . . . . . . . . . . . . . . . . . . . . . .$
vi)
Which of the following forms an AP? Justify your answer.
$2, 2^{2}, 2^{3}, 2^{4}, . . .$
vii)
Which of the following forms an AP? Justify your answer.
$\sqrt{3}, \sqrt{12}, \sqrt{27}, \sqrt{48}, \dots$

Answer:

i)
Answer. [True]
Solution.

$- 1, - 1, - 1, - 1, \dots$ here $a_{1} = - 1, a_{2} = - 1, a_{3} = - 1, a_{4} = - 1$

$\begin{aligned} a_{2} - a_{1} = - 1 - (- 1) = - 1 + 1 = 0 \\ a_{3} - a_{2} = - 1 - (- 1) = - 1 + 1 = 0 \\ a_{4} - a_{3} = - 1 - (- 1) = - 1 + 1 = 0 \end{aligned}$
Here, the difference between the successive terms is the same, hence it is an AP.
ii)
Answer. [No]
Solution.

The given series is $0, 2, 0, 2, \dots$ here $a_{1} = 0, a_{2} = 2, a_{3} = 0, a_{4} = 2$

$\begin{matrix} a_{2} - a_{1} = 2 - 0 = 2 \\ a_{3} - a_{2} = 0 - 2 = - 2 \\ a_{4} - a_{3} = 2 - 0 = 2 \end{matrix}$
Here the difference in successive terms is not the same. He, once it is not an AP.
iii)
Answer. [No]
Solution.

The given series is $1, 1, 2, 2, 3, 3, \dots$ here $a_{1} = 1, a_{2} = 1, a_{3} = 2, a_{4} = 2$

$\begin{aligned} a_{2} - a_{1} = 1 - 1 = 0 \\ a_{3} - a_{2} = 2 - 1 = 1 \\ a_{4} - a_{3} = 2 - 2 = 0 \end{aligned}$
Here the difference in successive terms is not the same. He, it is not an AP.
iv)
Answer. [Yes]
Solution.

The given series is $11, 22, 33, \dots$ here $a_{1} = 11, a_{2} = 22, a_{3} = 33$

$\begin{aligned} a_{2} - a_{1} = 22 - 11 = 11 \\ a_{3} - a_{2} = 33 - 22 = 11 \end{aligned}$
Here, the difference between the successive terms is the same. Hence, it is an AP.
v)
Answer. [No]
Solution.

The given series is $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dots$

$\begin{matrix} a_{1} = \frac{1}{2}, a_{2} = \frac{1}{3}, a_{3} = \frac{1}{4} \\ a_{2} - a_{1} = \frac{1}{3} - \frac{1}{2} = \frac{2 - 3}{6} = \frac{- 1}{6} \\ a_{3} - a_{2} = \frac{1}{4} - \frac{1}{3} = \frac{3 - 4}{12} = \frac{- 1}{12} \end{matrix}$
Here the difference between the successive terms, is not the same. Hence, it is not an AP.
vi)
Answer. [No]
Solution.

The given series is $2, 2^{2}, 2^{3}, 2^{4}, \dots$

$\begin{matrix} a_{1} = 2, a_{2} = 2^{2} = 4, a_{3} = 2^{3} = 8, a_{4} = 2^{4} = 16 \\ a_{2} - a_{1} = 4 - 2 = 2 \\ a_{3} - a_{2} = 8 - 4 = 4 \\ a_{4} - a_{3} = 16 - 8 = 8 \end{matrix}$
Here the difference of the successive terms is not the same. Hence, it is not an AP.
vii)
Answer. [Yes]
Solution.

Here the given series is $\sqrt{3}, \sqrt{12}, \sqrt{27}, \sqrt{48}, \dots$ . here $a_{1} = \sqrt{3}, a_{2} = \sqrt{12}, a_{3} = \sqrt{27}$ ,

$\begin{aligned} a_{4} = \sqrt{48} \\ a_{2} - a_{1} = \sqrt{12} - \sqrt{3} = 2 \sqrt{3} - \sqrt{3} = \sqrt{3} = \sqrt{3} (2 - 1) = \sqrt{3} \\ a_{3} - a_{2} = \sqrt{27} - \sqrt{12} = 3 \sqrt{3} - 2 \sqrt{3} = \sqrt{3} \\ a_{4} - a_{3} = \sqrt{48} - \sqrt{27} = 4 \sqrt{3} - 3 \sqrt{3} = \sqrt{3} \end{aligned}$
Here the difference of the successive term is the same. Hence, it is an AP.

Question:2

Justify whether it is true to say that $- 1, - \frac{3}{2}, - 2, \frac{5}{2}, \dots$ forms an A.P.as $a_{2} - a_{1} = a_{3} - a_{2} .$

Answer. [True]

Solution.

$\begin{matrix} - 1, - \frac{3}{2}, - 2, \frac{5}{2}, \dots \dots \\ here a_{1} = - 1, a_{2} = - \frac{3}{2}, a_{3} = - 2, a_{4} = \frac{5}{2} \\ a_{2} - a_{1} = - \frac{3}{2} - (- 1) = \frac{- 3}{2} + 1 = \frac{- 3 + 2}{2} = \frac{- 1}{2} \\ a_{3} - a_{2} = - 2 - (\frac{- 3}{2}) = - 2 + \frac{3}{2} = \frac{- 4 + 3}{2} = \frac{- 1}{2} \end{matrix}$

Here $a_{2} - a_{1} = a_{3} - a_{2}$ , Hence it is an A.P.

Question:3

For the AP: –3, –7, –11, ..., can we find directly $a_{30} - a_{20}$ without actually finding $a_{30}$ and $a_{20}$ ?
Give reasons for your answer.

Answer. [True]
Solution. Here, the given series is
$- 3, - 7, - 11 \dots$
$a_{1} = - 3, a_{2} = - 7$
$\begin{matrix} d = a_{2} - a_{1} \\ = - 7 - (3) \\ = - 7 + 3 \\ = - 4 \\ a_{30} - a_{20} = a + (30 - 1) d - (a + (20 - 1) d) \\ = a + 29 d - a - 19 d \\ = 10 d \\ = 10 \times (- 4) \\ = - 40 \end{matrix}$
$Yes we can find$ $a_{30} - a_{20}$ $without actually calculating$ $a_{30} a n d a_{20} .$

Question:4

Two APs have the same common difference. The first term of one AP is 2 and that of the other is 7. The difference between their 10th terms is the same as the difference between their 21st terms, which is the same as the difference between any two corresponding terms. Why?

Answer. [True]
Solution.
It is given that the first term of an AP is 2 and the other is 7.

Let

$a_{1} = 2, b_{1} = 7$

Common difference of both A.P is same $a_{n}$ is nth term for first A.P, $b_{n}$ is $n$ th term of second A.P

As per the question

$\begin{matrix} a_{10} - b_{10} = a_{1} + 9 d - b_{1} - 9 d \\ = a_{1} - b_{1} = 2 - 7 = - 5 \\ a_{21} - b_{21} = a_{1} + 20 d - b_{1} - 20 d \\ = a_{1} - b_{1} = - 5 \end{matrix}$

Difference between first term $= a_{1} - b_{1} = - 5$

It is because when we find the difference between the 10th and 21st terms, it is equal to the a₁-b_1, which is the difference between the first terms.

Question:5

Is 0 a term of the AP: 31, 28, 25, ...? Justify your answer.

Answer. [False]
Solution.
Here the given series is $31, 28, 25, \dots \dots .$ .

$\begin{aligned} a = 31 \\ d = a_{2} - a_{1} = 28 - 31 = - 3 \end{aligned}$
If 0 is a term of this A.P., then the value of $n$ must be a positive integer.

$\begin{aligned} a_{n} = a + (n - 1) d \\ 0 = a + (n - 1) d \\ 0 = 31 + (n - 1) (- 3) \\ 0 = 31 - 3 n + 3 \\ 3 n = 34 \\ n = \frac{34}{3} = 11.333 \end{aligned}$
Which is not an integer.
Hence 0 is not a term of this AP.

Question:6

The taxi fare after each km, when the fare is Rs 15 for the first km and s. 8 for each additional km, does not form an AP as the total fare (in Rs) after each km is
15, 8, 8, 8, ...
Is the statement true? Give reasons.

Answer: [True]
Solution.

Given series is $15, 8, 8, 8, \dots$ .

Here $a_{1} = 15, a_{2} = 8, a_{3} = 8, a_{4} = 8$

$\begin{matrix} a_{2} - a_{1} = 8 - 15 = - 7 \\ a_{3} - a_{2} = 8 - 8 = 0 \\ a_{4} - a_{3} = 8 - 8 = 0 \end{matrix}$

The difference of the successive terms is not the same, hence it does not form an AP.

Therefore given statement is true.

Question:7

i)In which of the following situations do the lists of numbers involved form an AP? Give reasons for your answers.
The fee is charged to a student every month by a school for the whole session, when the monthly fee is Rs 400.
ii)
In which of the following situations do the lists of numbers involved form an AP? Give reasons for your answers.
The fee is charged every month by a school from Classes I to XI, when the monthly fee for Class I is Rs 250, and it increases by Rs 50 for the next higher class.
iii)
In which of the following situations do the lists of numbers involved form an AP? Give reasons for your answers.
The amount of money in the account of Varun at the end of every year when Rs 1000 with a simple interest of 10% per annum.
iv)
In which of the following situations do the lists of numbers involved form an AP? Give reasons for your answers.
The number of bacteria in a certain food item after each second, when they double every second.

Answer. [Yes]
Solution.
According to the question:
The monthly fee charged by students is $= 400$ Rs.
Hence series is: 400, 400, 400, 400
The difference of the successive terms is the same, so it forms an AP with common difference $d = 400 - 400 = 0$
ii)

Answer. [Yes]
Solution.
According to the question:
Fees started at Rs. 250 and increased by 50 Rs. from I to XII are

Here first term $(a_{1}) = 250$

common difference

$\begin{aligned} (d) = 50, a_{1} = 250 \\ a_{2} = 250 + 50 = 300 \\ a_{3} = 300 + 50 = 350 \\ a_{4} = 350 + 50 = 400 \end{aligned}$

So the A.P. is:

$250, 300, 350, 400 \dots$

It forms an AP with a common difference of 50.

iii)

Answer. [Yes]
Solution. We know that: Simple interest $= (P \times R \times T) / 100$

This is definite from an AP with a common difference of $SI = 100$ Rs.
iv)

Answer. [No]
Solution.
Let the number of bacteria $= x$
According to the question:
They double in every second.
Here $a_{1} = x, a_{2} = 2 x, a_{3} = 4 x, a_{4} = 8 x, \dots x, 2 x, 4 x, 8 x, \dots$
Since the difference between the successive terms is not the same therefore it does not form an AP.

Question:8

i)
Justify whether it is true to say that the following are the nth terms of an AP.
2n–3
ii)
Justify whether it is true to say that the following are the nth terms of an AP.
$3 n^{2} + 5$
iii)
Justify whether it is true to say that the following are the nth terms of an AP.
$1 + n + n^{2}$

Answer: [Yes]
Solution.

Hence, $2 n - 3$ is the $n$ th term of an AP because the common difference is the same.
ii)

Answer. [ No ]
Solution.

Here, $a_{n} = 3 n^{2} + 5$
Put $n = 1, a_{1} = 3 (1)^{2} + 5 = 3 + 5 = 8$
Put $n = 2, a_{2} = 3 (2)^{2} + 5 = 12 + 5 = 17$
Put $n = 3, a_{3} = 3 (3)^{2} + 5 = 27 + 5 = 32$
Numbers are:

$\begin{matrix} 8, 17, 32 \dots \\ a_{2} - a_{1} = 17 - 8 = 9 \\ a_{3} - a_{2} = 32 - 17 = 15 \end{matrix}$
Here, the common difference is not the same
Therefore series $3 n^{2} + 5$ is not the nth term of an AP.
iii)

Answer. [No]
Solution.

$a_{n} = 1 + n + n^{2} P u t n = 1, a_{1} = 1 + 1 + (1)^{2} = 3 P u t n = 2, a_{2} = 1 + 2 + (2)^{2} = 7 P u t n = 3, a_{3} = 1 + 3 + (3)^{2} = 13 Thenumbersar c 3, 7, 13 \dots a_{2} - a_{1} = 7 - 3 = 4 a_{3} - a_{2} = 13 - 7 = 6$

Here, the common difference is not the same.

Class 10 MathsChapterr 5 exemplar solutions Exercise: 5.3
Page number: 51-54
Total questions:35

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Question:1

Match the APs given in column A with the suitable common differences given in column B.

Column A	Column B
(A₁) 2, – 2, – 6, – 10,...	(B₁) $\frac{2}{3}$
(A₂) a = –18, n = 10, an = 0	(B₂) –5
(A₃) a = 0, $a_{10}$ = 6	(B₃) 4
(A₄) $a_{2} = 13, a_{4} = 3$	(B₄) –4
	(B₅) 2
	(B₆) $\frac{1}{2}$
	(B₇) 5

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Answer:

$\begin{aligned} (A_{1}) \\ 2, - 2, - 6, - 10, \dots \\ a_{2} = - 2, a_{1} = 2 \\ d = a_{2} - a_{1} \\ - 2 - 2 = - 4 (B_{4}) - 4 \\ (A_{2}) \\ a = - 18, n = 10, a_{n} = 0 \\ a_{n} = 0 \\ ∵ a + (n - 1) d = 0 \\ - 18 + (10 - 1) d = 0 \\ 9 d = 18 \\ d = 2 (B_{5}) 2 \\ \begin{matrix} (A_{3}) \\ a = 0, a_{10} = 6 \\ a + 9 d = 6 \\ 0 + 9 d = 6 \\ d = \frac{6}{9} = \frac{2}{3} (B_{1}) 2 / 3 \\ (A_{4}) \\ a_{2} = 13, a_{4} = 3 \\ ∵ a + (n - 1) d = 0 \\ a + d = 13 \\ a + 3 d = 3 \\ - - - - - - - - \\ - 2 d = 10 \\ d = \frac{10}{- 2} \\ d = - 5 (B_{2}) - 5 \end{matrix} \end{aligned}$

Question:2

Verify that each of the following is an AP, and then write its next three terms.
$0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, \dots$
ii)
Verify that each of the following is an AP, and then write its next three terms.
$5, \frac{14}{3}, \frac{13}{3}, 4, \dots \dots$
iii)
Verify that each of the following is an AP, and then write its next three terms.
$\sqrt{3}, 2 \sqrt{3}, 3 \sqrt{3}, \dots$
iv)
Verify that each of the following is an AP, and then write its next three terms.
a + b, (a + 1) + b, (a + 1) + (b + 1), ...
v)
Verify that each of the following is an AP, and then write its next three terms.
a, 2a + 1, 3a + 2, 4a + 3,...

Answer:

Here is the given series,

$\begin{aligned} 0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4} \\ a_{1} = 0, a_{2} = \frac{1}{4} a_{3} = \frac{3}{4} \\ a_{2} - a_{1} = \frac{1}{4} - 0 = \frac{1}{4} \\ a_{3} - a_{2} = \frac{1}{4} \\ a_{4} - a_{3} = \frac{1}{4} \end{aligned}$
Here, the common difference is the same. Hence given series is an AP

$\begin{aligned} a_{5} = a_{1} + 4 d (u sing a_{n} = a_{1} + (n - 1) \cdot d) \\ a_{6} = a_{1} + 5 d \\ a_{7} = a_{1} + 6 d \\ = 0 + 6 (\frac{1}{4}) \\ = \frac{3}{2} \end{aligned}$
The next three terms are

$1, \frac{5}{4}, \frac{3}{2}$

ii)

Here the given series is, $5, \frac{14}{3}, \frac{13}{3}, 4, \dots \dots$
Here, the common difference is the same.
Hence, the given series is an AP.

$\begin{aligned} a_{5} = a_{1} + 4 d (u sing a_{n} = a_{1} + (n - 1) \cdot d) \\ a_{6} = a_{1} + 5 d \\ a_{7} = a_{1} + 6 d \end{aligned}$
Next terms are $\frac{11}{3}, \frac{10}{3}, 3$

iii)

Here the given series is $\sqrt{3}, 2 \sqrt{3}, 3 \sqrt{3}, \dots$
Here the common difference is the same. Hence, the given series is an AP.

$\begin{matrix} a_{4} = a_{1} + 3 d (Using a_{n} = a_{1} + (n - 1) \cdot d) \\ = \sqrt{3} + 3 \sqrt{3} = 4 \sqrt{3} \\ a_{5} = a_{1} + 4 d \\ = \sqrt{3} + 4 \sqrt{3} = 5 \sqrt{3} \\ a_{6} = a_{1} + 5 d \\ = \\ \sqrt{3} + 5 \sqrt{3} = 6 \sqrt{3} \end{matrix}$

Next terms are:

$4 \sqrt{3}, 5 \sqrt{3}, 6 \sqrt{3}$
iv)

Here the given series is $a + b, (a + 1) + b, (a + 1) + (b + 1), \dots$

$\begin{matrix} a_{1} = a + b, a_{2} = (a + 1) + b, a_{3} = (a + 1) + (b + 1) \\ a_{2} - a_{1} = (a + 1) + b - (a + b) = 1 \\ a_{3} - a_{2} = (a + 1) + (b + 1) - [(a + 1) + b] = 1 \end{matrix}$

Here the common difference is the same, hence the given series is an AP.

$\begin{matrix} a_{4} = a_{1} + 3 d (Using a_{n} = a_{1} + (n - 1) \cdot d) \\ = (a + b) + 3 (1) = (a + 1) + (b + 1) + 1 \\ a_{5} = a_{1} + 4 d \\ = (a + b) + 4 (1) = (a + 1) + (b + 1) + 1 + 1 \\ a_{6} = a_{1} + 5 d \\ = (a + b) + 5 (1) = (a + 1) + (b + 1) + 1 + 1 + 1 \end{matrix}$

Next three terms are:

$(a + 1) + (b + 1) + 1, (a + 1) + (b + 1) + 1 + 1, (a + 1) + (b + 1) + 1 + 1 + 1$

Here the given series is $a, 2 a + 1, 3 a + 2, 4 a + 3, \dots$

$\begin{matrix} a_{1} = a, a_{2} = 2 a + 1, a_{3} = 3 a + 2 \\ a_{2} - a_{1} = (2 a + 1) - a = a + 1 \\ a_{3} - a_{2} = (3 a + 2) - (2 a + 1) = a + 1 \end{matrix}$

Here the common difference is the same, hence the given series is an AP.

$\begin{aligned} a_{5} = a_{1} + 4 d (Using a_{n} = a_{1} + (n - 1) \cdot d) \\ = a + 4 (a + 1) = a + 4 a + 4 = 5 a + 4 \\ = a + 5 (a + 1) = a + 5 a + 5 = 6 a + 5 \\ a_{6} = a_{1} + 5 d \\ a_{7} = a_{1} + 6 d \\ = a + 6 (a + 1) = a + 6 a + 6 = 7 a + 6 \end{aligned}$

Next three terms are:

$5 a + 4, 6 a + 5, 7 a + 6$

Question:3

i)
Write the first three terms of the APs when a and d are as given below:
$a = \frac{1}{2}, d = - \frac{1}{6}$
ii)
Write the first three terms of the APs when a and d are as given below:
a = – 5, d = – 3
iii)
Write the first three terms of the APs when a and d are as given below:
$a = \sqrt{2}, d = \frac{1}{\sqrt{2}}$

Answer:

$\begin{matrix} Here a = \frac{1}{2}, d = - \frac{1}{6} \\ a_{1} = a = \frac{1}{2} \\ a_{2} = a + d (Using a_{n} = a + (n - 1) \cdot d) \\ a_{3} = a + 2 d \\ = \frac{1}{2} + 2 (\frac{- 1}{6}) \\ = \frac{1}{2} - \frac{2}{6} \\ = \frac{3 - 2}{6} = \frac{1}{6} \end{matrix}$

Hence, the first three terms are:

$\frac{1}{2}, \frac{1}{3}, \frac{1}{6}, \dots$
ii)

Here $a = - 5, d = - 3$

$\begin{matrix} a_{1} = a = - 5 \\ a_{2} = a + d (Using a_{n} = a + (n - 1) \cdot d) \\ = - 5 - 3 = - 8 \\ a_{3} = a + 2 d \\ = - 5 + 2 (- 3) \\ = \end{matrix}$

Here, the first three terms are:

$- 5, - 8, - 11 .$
iii)

Here $a = \sqrt{2}, d = \frac{1}{\sqrt{2}}$

$\begin{aligned} a_{2} = a + d & (Using a_{n} = a + (n - 1) d) \\ = \sqrt{2} + \frac{1}{\sqrt{2}} \\ = & \frac{2 + 1}{\sqrt{2}} = \frac{3}{\sqrt{2}} \\ a_{3} = a + 2 d \\ = & \sqrt{2} + 2 (\frac{1}{\sqrt{2}}) \\ = \sqrt{2} + \frac{2}{\sqrt{2}} \\ = & \frac{2 + 2}{\sqrt{2}} = \frac{4}{\sqrt{2}} \end{aligned}$

Here, the first three terms are:

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

Question:4

Find a, b and c such that the following numbers are in AP: a, 7, b, 23, c.

Answer:

Given AP is a, $7, b, 23, c$
We know that an f series is in AP, then

$a_{1} = a, a_{2} = 7, a_{3} = b, a_{4} = 23, a_{5} = c$

(We know that the common difference of an A.P. is Equal)

$\begin{aligned} a_{2} - a_{1} = a_{3} - a_{2} = a_{4} - a_{3} = a_{5} - a_{4} \\ ∴ 7 - a = b - 7 = 23 - b = c - 23 \end{aligned}$
Taking $2^{nd}$ and $3^{rd}$ term

$\begin{aligned} b - 7 = 23 - b \\ 2 b = 30 \\ b = \frac{30}{2} = 15 \end{aligned}$
Taking $1^{st}$ and $2^{nd}$ term

$\begin{aligned} 7 - a = b - 7 \\ 7 - a = 15 - 7 (∵ b = 15) \\ 7 - 8 = a \\ - 1 = a \end{aligned}$
Taking $1^{st}$ and fourth term

$\begin{aligned} 7 - a = c - 23 \\ 7 - (- 1) = c - 23 (∵ a = - 1) \\ 7 + 1 = c - 23 \\ 8 + 23 = 6 \\ 31 = c \end{aligned}$
Hence $a = - 1, b = 15, c = 31$

Question:5

Determine the AP whose fifth term is 19 and the difference between the eighth term from the thirteenth term is 20.

Answer:

Given:
The fifth term is 19 :

$a + 4 d = 19$

The difference between the eighth and thirteenth terms is 20 :

$(a + 12 d) - (a + 7 d) = 20$

This simplifies to $5 d = 20$ , so $d = 4$ .

Substitute $d = 4$ into the first equation:

$a + 16 = 19, so a = 3$

Therefore, the first term is $a = 3$ and the common difference is $d = 4$ .

Question:6

The 26th, 11thand the last term of an AP are 0, 3 and -1/5, respectively. Find the common difference and the number of terms.

Answer:

$Given a_{26} = 0, a_{11} = 3, a_{n} = - \frac{1}{5}$
$\begin{aligned} a + 25 d = 0 \dots (1) (u sing a_{n} = a + (n - 1) d) \\ a + 10 d + 3 \dots (2) (u sing a_{n} = a + (n - 1) d) \\ 15 d = - 3 \end{aligned}$
Put $d = - \frac{1}{5}$ in (1) we get

$\begin{aligned} a - 5 = 0 \\ a = 5 \\ a_{n} = - \frac{1}{5} \end{aligned}$
We know that $a_{n} = a + (n - 1) d$

$\begin{aligned} \Rightarrow - \frac{1}{5} - 5 = (n - 1) (- \frac{1}{5}) \\ \Rightarrow \frac{- 1 - 25}{5} = (n - 1) (- \frac{1}{5}) \\ \Rightarrow \frac{- 26}{5} = (n - 1) (- \frac{1}{5}) \\ \Rightarrow \frac{- 26}{5} \times \frac{- 5}{1} = n - 1 \\ ∴ 26 = n - 1 \\ \Rightarrow n = 26 + 1 \\ \Rightarrow n = 27 \end{aligned}$
ANS: -

$d = - \frac{1}{5}, n = 27$

Question:7

The sum of the 5th and the 7th terms of an AP is 52, and the 10th term is 46. Find the AP.

Given $a_{5} + a_{7} = 52$

$a_{5} = a + 4 d a_{7} = a + 6 d (u \sin g a_{n} = a + (n - 1) \cdot d)$
Given $a_{10} = 46$

$\begin{aligned} ∴ a + 9 d = 46 \dots (1) \\ a + 4 d + a + 6 d = 52 \\ 2 a + 10 d = 52 \end{aligned}$
Dividing both sides by 2

$\begin{aligned} a + 5 d = 26 \dots (2) \\ a + 9 d = - 46 \\ - 4 d = - 20 \\ d = \frac{- 20}{- 4} \\ d = 5 \end{aligned}$
Put $d = 5$ in equation (1) we get

$\begin{aligned} a + 5 \times (5) = 26 \\ a + 25 = 26 \\ a = 26 - 25 \\ a = 1 \\ a_{1} = a = 1 \\ a_{2} = a + d \\ = 1 + 5 \\ = 6 \\ a_{3} = a + 2 d \\ = 1 + 2 \times (5) \\ = 1 + 10 \\ = 11 \end{aligned}$
Hence the AP is $1, 6, 11 \dots$ .

Question:8

Find the 20th term of the AP whose 7th term is 24 less than the 11th term, the first term being 12.

Given $a = 12$

$\begin{aligned} a_{7} = a_{11} - 24 \\ a + 6 d = a + 10 d - 24 (using a_{n} = a + (n - 1) \cdot d) \\ a - a + 24 = 10 d - 6 d \\ 24 = 4 d \\ d = 6 \end{aligned}$
We know that

$a_{n} = a + (n - 1) d$
$\begin{aligned} Put n = 20, a_{20} = a + (20 - 1) d \\ = 12 + (19) \times (6) \\ = 12 + 114 \\ = 126 \end{aligned}$
Hence

$20^{th} term is 126$

Question:9

If the 9th term of an AP is zero, prove that its 29th term is twice its 19th term.

Answer:

Let the first term = a
common difference = d
Given:- $a_{9} = 0$

$\begin{aligned} \Rightarrow∴ a + 8 d & = 0 (using a_{n} = a + (n - 1) d) \\ \Rightarrow∴ a = - 8 d \dots (1) \end{aligned}$

To prove:

$\begin{matrix} 2 a_{19} = a_{29} \\ a_{19} = a + 18 d (using a_{n} = a + (n - 1) d) \\ = - 8 d + 18 d {a = - 8 d from equation (1)} \\ = 10 d \dots (2) \\ a_{29} = a + (29 - 1) d \\ = a + 28 d \\ = - 8 d + 28 d {a = - 8 d from equation (1)} \\ = 20 d \\ = 2 \times 10 d \end{matrix}$

By equation (2),

$a_{29} = 2 a_{19}$
Hence proved

Question:10

Find whether 55 is a term of the AP: 7, 10, 13, ______or not. If yes, find which term it is.

Answer:

Let 55 be the nth term of an AP
$∴ a_{n} = 55$ \ldots $(1)$
Given AP is 7, 10, 13........
Here a = 7, d = 10 - 7
d = 3
By equation (1) is $a_{n} = 55$
$\begin{matrix} a + (n - 1) \times d = 55 \dots (2) (using a_{n} = a + (n - 1) d) \\ 7 + (n - 1) \times 3 = 55 \\ (n - 1) \times 3 = 55 - 7 \\ (n - 1) \times 3 = 48 \\ n - 1 = \frac{48}{3} \\ n - 1 = 16 \\ n = 17 \end{matrix}$
Here n is a positive integer.
Hence, 55 is the 17^th term of an AP

Question:11

Determine k so that $k^{2} + 4 k + 8, 2 k^{2} + 3 k + 6, 3 k^{2} + 4 k + 4$ are three consecutive terms of an AP.

Answer:

Here, the given AP is

$\begin{matrix} k_{2} + 4 k + 8, 2 k_{2} + 3 k + 6, 3 k_{2} + 4 k + 4 \\ a_{1} = k_{2} + 4 k + 8, a_{2} = 2 k_{2} + 3 k + 6, a_{3} = 3 k_{2} + 4 k + 4 \\ a_{2} - a_{1} = 2 k_{2} + 3 k + 6 - (k_{2} + 4 k + 8) \\ = 2 k_{2} + 3 k + 6 - k_{2} - 4 k - 8 \\ = k_{2} - k - 2 \\ a_{3} - a_{2} = 3 k_{2} + 4 k + 4 - (2 k_{2} + 3 k + 6) \\ = 3 k_{2} + 4 k + 4 - 2 k_{2} - 3 k - 6 \\ = k_{2} + k - 2 \end{matrix}$

Hence, it is given that these terms are an AP
So the common difference between the A.P

$\begin{matrix} a_{2} - a_{1} = a_{3} - a_{2} \\ k_{2} - k - 2 = k_{2} + k - 2 \\ k_{2} - k - 2 - k_{2} - k + 2 = 0 \\ - 2 k = 0 \\ k = 0 \end{matrix}$

Question:12

Split 207 into three parts such that these are in AP and the product of the two smaller parts is 4623.

Answer:

Let the three parts be (a-d), a, (a+d)
As per the question -

$\begin{matrix} a - d + a + a + d = 207 \\ 3 a = 207 \\ a = \frac{207}{3} = 69 \\ a = 69 \end{matrix}$

It is given that the product of the two similar parts is 4623

$\begin{matrix} (a - d) \times a = 4623 \\ (69 - d) \times 69 = 4623 \\ 69 - 67 = d \\ d = 2 \end{matrix}$

So the required terms are

$\begin{matrix} a - d = 69 - 2 = 67 \\ a = 69 \\ a + d = 69 + 2 = 71 \end{matrix}$
The required numbers are 67, 69, and 71

Question:13

The angles of a triangle are in AP. The greatest angle is twice the least. Find all the angles of the triangle.

Answer:

Let the three angles $be (a - d), a, (a + d)$ in AP.
According to the question, the greatest is twice the least:

$\begin{matrix} (a + d) = 2 (a - d) \\ a + d = 2 a - 2 d \\ 2 a - a - 2 d - d = 0 \\ a - 3 d = 0 \\ a = 3 d \end{matrix}$

We know that the sum of the angles of a triangle is $180^{\circ}$ :

$\begin{matrix} a + a - d + a + d = 180^{\circ} \\ 3 a = 180^{\circ} \\ a = \frac{180^{\circ}}{3} = 60^{\circ} \end{matrix}$

Substituting $a = 60^{\circ}$ in $a = 3 d$ :

$\begin{matrix} 60^{\circ} = 3 d \\ d = 20^{\circ} \end{matrix}$

So the angles are:

$\begin{matrix} a - d = 60^{\circ} - 20^{\circ} = 40^{\circ} \\ a = 60^{\circ} \\ a + d = 60^{\circ} + 20^{\circ} = 80^{\circ} \end{matrix}$

Required angles: $40^{\circ}, 60^{\circ}, 80^{\circ}$

Question:14

If the nth terms of the two APs: 9, 7, 5, ... and 24, 21, 18,... are the same, find the value of n. Also, find that term.

Answer:

The given APs are

$9, 7, 5, \dots and 24, 21, 18, \dots$
In AP

$9, 7, 5, \dots \dots$

$\begin{aligned} a_{1} = a = 9, a_{2} = 7 \\ d_{1} = a_{2} - a_{1} = 7 - 9 = - 2 \\ a_{n} = a + (n - 1) d_{1} \end{aligned}$

(1) In the AP $24, 21, 18, \dots$

$\begin{aligned} b_{1} = b = 24, b_{2} = 21 \\ d_{2} = b_{2} - b_{1} = 21 - 24 = - 3 \\ b_{n} = b + (n - 1) d_{2} \end{aligned}$

Here, the number of terms $n$ is equal in both cases
According to the question

$a_{n} = b_{n}$

Question:15

If the sum of the 3rd and the 8th terms of an AP is 7 and the sum of the 7th and the 14th terms is 3, find the 10th term.

Answer:

It is given that the sum of $3^{rd}$ and $8^{th}$ term is $7 \cdot$

$\begin{aligned} Solve (1) and (2) 2 a + 19 d = - 3 \\ 2 a + 9 d = 7 \\ 10 d = - 10 \\ d = - 1 \\ Putd = - 1 in (1) \\ \Rightarrow 2 a + 9 (- 1) = 7 \\ \Rightarrow 2 a = 7 + 9 \\ \Rightarrow a = \frac{16}{2} = 8 \\ a_{10} = (a + (10 - 1) d) \\ \Rightarrow a_{10} = 8 + 9 (- 1) \\ \Rightarrow a_{10} = 8 - 9 \\ \Rightarrow a_{10} = - 1 \end{aligned}$

Question:16

Find the 12th term from the end of the AP: -2, -4, -6,..., -100.

Answer:

The given AP : is $- 2, - 4, - 6, \dots \dots ., - 100$
AP from end is $- 100, - 98 \dots \dots . . . .4, - 6, - 2$

$\begin{aligned} a_{1} = a = - 100 \\ d = a_{2} - a_{1} = - 98 - (- 100) = - 98 + 100 = 2 \\ a_{12} = a + (12 - 1) \cdot d (using a_{n} = a + (n - 1) \times d) \\ a_{12} = - 100 + 11 \times (2) \\ a_{12} = - 100 + 22 \\ a_{12} = - 78 \end{aligned}$

Question:17

Which term of the AP: 53, 48, 43,... is the first negative term?
Answer:

The given arithmetic progression (AP) is:
$53, 48, 43, \dots$
Here, the first term $a = 53$ and the common difference $d = 48 - 53 = - 5$ .
To find the first negative term, we use the formula for the $n$ -th term of an AP:

$T_{n} = a + (n - 1) \cdot d$

We want to find $n$ such that $T_{n} < 0$ .
Substitute $a = 53$ and $d = - 5$ :

$\begin{matrix} T_{n} = 53 + (n - 1) (- 5) \\ T_{n} = 53 - 5 (n - 1) \\ T_{n} = 53 - 5 n + 5 \\ T_{n} = 58 - 5 n \end{matrix}$

Set $T_{n} < 0$ to find the first negative term:

$\begin{matrix} 58 - 5 n < 0 \\ 58 < 5 n \\ n > \frac{58}{5} = 11.6 \end{matrix}$

Since $n$ must be a whole number, the first negative term occurs at $n = 12$ . Thus, the first negative term is the 12th term of the AP.

Question:18

How many numbers lie between 10 and 300, which when divided by 4 leave a remainder of 3?

Answer:

The numbers between 10 and 300 which when divided by 4 leave remainder 3 are: $11, 15, 19, 23, \dots, 299$

$\begin{matrix} a_{1} = 11, a_{2} = 15, a_{3} = 19 \\ a_{2} - a_{1} = 15 - 11 = 4 \\ a_{3} - a_{2} = 19 - 15 = 4 \end{matrix}$

It is an AP with $a_{1} = a = 11, d = 4$
Let there be $n$ terms in this AP

$\begin{matrix} a_{n} = 299 \\ a + (n - 1) \times d = 299 (using a_{n} = a + (n - 1) d) \\ 11 + (n - 1) \times 4 = 299 \\ (n - 1) \times 4 = 299 - 11 \\ (n - 1) \times 4 = 288 \\ (n - 1) = \frac{288}{4} \\ (n - 1) = 72 \\ n = 72 + 1 \\ n = 73 \end{matrix}$

Question:19

Find the sum of the two middlemost terms of the AP:
$- \frac{4}{3}, - 1, - \frac{2}{3}, \dots, 4 \frac{1}{3}$

Answer:

Here, the given AP is $- \frac{4}{3}, - 1, - \frac{2}{3}, \dots, 4 \frac{1}{3}$
In this AP: $a_{1} = a = - \frac{4}{3}$

$d = a_{2} - a_{1} = - 1 - (- \frac{4}{3}) = - 1 + \frac{4}{3} = \frac{- 3 + 4}{3} = \frac{1}{3}$

Let there be $n$ terms in the AP
The nth term of AP is $4 \frac{1}{3}$

$\begin{matrix} a_{n} = 4 \frac{1}{3} \\ a + (n - 1) \cdot d = \frac{13}{3} (Using a_{n} = a + (n - 1) d) \\ - \frac{4}{3} + (n - 1) \times (\frac{1}{3}) = \frac{13}{3} \\ (n - 1) \times (\frac{1}{3}) = \frac{13 + 4}{3} \\ n - 1 = 17 \\ n = 18 \end{matrix}$

The total terms in this AP are 18, and the middlemost terms are $a_{9}$ and $a_{10}$

$\begin{matrix} a_{9} + a_{10} = a + (9 - 1) d + a + (10 - 1) d (using a_{n} = a + (n - 1) d) \\ = 2 a + 17 d \\ = 2 (- \frac{4}{3}) + 17 (\frac{1}{3}) \\ = \frac{- 8}{3} + \frac{17}{3} \\ a_{9} + a_{10} = \frac{- 8 + 17}{3} = \frac{9}{3} = 3 \\ a_{9} + a_{10} = 3 \end{matrix}$

Question:20

The first term of an AP is -5 and the last term is 45. If the sum of the terms of the AP is 120, then find the number of terms and the common difference.

Answer:

Given that $a = - 5, a_{n} = 45, S_{n} = 120$

$a_{n} = 45$

$\begin{matrix} a + (n - 1) \times d = 45 (Using a_{n} = a + (n - 1) d) \\ - 5 + (n - 1) \times d = 45 \\ (n - 1) \times d = 50 \dots (1) \\ S_{n} = 120 \\ S_{n} = \frac{n}{2} [2 a + (n - 1) d] = 120 \\ \frac{n}{2} [- 10 + 50] = 120 (Using (1)) \\ \frac{n}{2} \times 40 = 120 \\ n \times 20 = 120 \\ n = \frac{120}{20} = 6 \end{matrix}$

Number of terms $n = 6$
Put $n = 6$ in (1)

$\begin{matrix} (6 - 1) \times d = 50 \\ 5 d = 50 \\ d = 10 \end{matrix}$

Common difference $d = 10$

Question:21

i)Find the sum:1 + (-2) + (-5) + (-8) + ... + (-236)
ii)Find the sum: $4 - \frac{1}{n} + 4 - \frac{2}{n} + 4 - \frac{3}{n} + \dots$ up to n terms
iii) Find the sum: $\frac{a - b}{a + b} + \frac{3 a - 2 b}{a + b} + \frac{5 a - 3 b}{a + b} + .$ to 11 terms

Answer:

The given series is $1 + (- 2) + (- 5) + (- 8) + \dots + (- 236)$
Here, $a_{1} = a = 1, a_{2} = - 2$

$\begin{matrix} d = a_{2} - a_{1} = - 2 - 1 = - 3 \\ a_{n} = - 236 \end{matrix}$

$\begin{matrix} a + (n - 1) \times d = - 236 (Using a_{n} = a + (n - 1) d) \\ 1 + (n - 1) \times (- 3) = - 236 \\ (n - 1) \times (- 3) = - 237 \\ n - 1 = 79 \\ n = 80 \end{matrix}$

The formula for the sum is:

$\begin{matrix} S_{n} = \frac{n}{2} [2 a + (n - 1) d] \\ S_{n} = \frac{80}{2} [2 \times 1 + (80 - 1) \times (- 3)] \\ = 40 [2 + 79 \times (- 3)] \\ = 40 [2 - 237] \\ = 40 \times (- 235) \\ S_{n} = - 9400 \end{matrix}$
ii)

The given series is $[4 - \frac{1}{n}] + [4 - \frac{2}{n}] + [4 - \frac{3}{n}] + \dots$

$\begin{matrix} First term (a) = [4 - \frac{1}{n}] \\ \begin{matrix} d = [4 - \frac{2}{n}] - [4 - \frac{1}{n}] \\ = \frac{- 2}{n} + \frac{1}{n} \\ = \frac{- 1}{n} \\ S_{n} = \frac{n}{2} [2 (4 - \frac{1}{n}) + (n - 1) \times (\frac{- 1}{n})] \\ = \frac{n}{2} [8 - \frac{2}{n} - \frac{n - 1}{n}] \\ = \\ \frac{n}{2} [8 - \frac{2 + n - 1}{n}] \\ = \\ \frac{n}{2} [8 - \frac{n + 1}{n}] \\ = \\ \frac{n}{2} \times \frac{8 n - (n + 1)}{n} \\ = \end{matrix} \\ = \frac{n}{2} \times \frac{7 n - 1}{n} \\ = \frac{7 n - 1}{2} \\ S_{n} = \frac{7 n - 1}{2} \end{matrix}$
iii)

The given series is $\frac{a - b}{a + b} + \frac{3 a - 2 b}{a + b} + \frac{5 a - 3 b}{a + b} + \dots$

$\begin{matrix} a_{1} = \frac{a - b}{a + b}, a_{2} = \frac{3 a - 2 b}{a + b} \\ d = a_{2} - a_{1} \end{matrix}$

Given that $n = 11$

$\begin{matrix} S_{11} = \frac{11}{2} [2 a_{1} + (11 - 1) d] \\ = \frac{11}{2} [\frac{2 (a - b)}{a + b} + \frac{10 (2 a - b)}{a + b}] \\ = \frac{11}{2} [\frac{2 a - 2 b + 20 a - 10 b}{a + b}] \\ = \frac{11}{2} [\frac{22 a - 12 b}{a + b}] \\ = \frac{11}{2} \times \frac{2 (11 a - 6 b)}{a + b} \\ S_{11} = \frac{11 (11 a - 6 b)}{a + b} \end{matrix}$

Question:22

Which term of the AP: -2, -7, -12,... will be -77? Find the sum of this AP up to the term -77.

Answer:

The given AP is $- 2, - 7, - 12, \dots$
First term $(a) = - 2$
Common difference $(d) = - 7 - (- 2) = - 7 + 2 = - 5$
Let the $n^{th}$ term be -77

$\begin{matrix} a_{n} = - 77 \\ a + (n - 1) \cdot d = - 77 \\ - 2 + (n - 1) \cdot (- 5) = - 77 \\ (n - 1) \cdot (- 5) = - 77 + 2 \\ (n - 1) = \frac{- 75}{- 5} = 15 \\ n = 15 + 1 = 16 \end{matrix}$

Hence, the $16^{th}$ term is -77
Sum of 16 terms

$\begin{matrix} S_{16} = \frac{16}{2} [2 (- 2) + (16 - 1) (- 5)] \\ = 8 [- 4 + 15 (- 5)] \\ = 8 [- 4 - 75] \\ S_{16} = 8 \times (- 79) \\ S_{16} = - 632 \end{matrix}$

Question:23

If $a_{n} = 3 - 4 n,$ $show that a_{1}, a_{2}, a_{3}, . . .$ form an AP. Also find $S_{20}$

Answer:

Here it is given that

$a_{n} = 3 - 4 n$

Put n =1 then

$a_{1} = 3 - 4 (1)$

$a_{1} = 3 - 4 = - 1$

Put n =2 then

$a_{2} = 3 - 4 (2)$

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

Put n =3 then

$a_{3} = 3 - 4 (3)$

$a_{3} = 3 - 12 = - 9$

$\text{Then A.P is -1, -5, -9} \ldots \ldots $

$a_{2} - a_{1} = - 5 - (- 1) = - 5 + 1 = - 4 a_{3} - a_{2} = - 9 - (- 5) = - 9 + 5 = - 4$

$ a_{\text{2}} - a_{\text{1}} = a_{\text{3}} - a_{\text{2}}\ \text{ Hence}, a_{\text{1}}, a_{\text{2}}, a_{\text{3}} \ldots \text{ forms an AP with } a_{1 }= a = - 1 , d = -4$

$Here d= common difference and first term is a = a_{1} = - 1$

$S_{2}=\frac{20}{2}\left[ 2\left( -1 \right)+\left( 20-1 \right)\left( -4 \right) \right] \left( Using \ S_{n}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right] \right) \ = 10 [- 2 + 19(-4)]$

$= 10 [- 2 - 76] S_{20} = - 780$

Question:24

In an Ap if $S_{n} = n (4 n + 1)$ find the A.P.

Answer:

We are given that the sum of the first $n$ terms of the AP is:

$S_{n} = n (4 n + 1)$
The first term $a$ is $S_{1} = 5$ .
For the second term, $S_{2} = 18$ , so the second term $T_{2} = S_{2} - S_{1} = 18 - 5 = 13$ .
The common difference $d$ is:

$d = T_{2} - T_{1} = 13 - 5 = 8$
The general term of the AP is:

$T_{n} = a + (n - 1) d = 5 + (n - 1) \cdot 8 = 8 n - 3$
Thus, the arithmetic progression is:

$5, 13, 21, 29, \dots$
And the general term is:

$T_{n} = 8 n - 3$

Question:25

In an AP, if $S_{n} = 3 n_{2} + 5 n$ and $a_{k} = 164,$ find the value of k

Answer:

Here it is given that $S_{n} = 3 n^{2} + 5 n \dots \dots$ (1)

$\begin{aligned} a_{k} = 164 \\ Weknowthat S_{1} = a, S_{2} = a + (a + d) \\ Putn = 1 inequation (1) \\ S_{1} = 3 (1)^{2} + 5 (1) = 8 \\ Putn = 2 inequation (1) \\ S_{2} = 3 (2)^{2} + 5 (2) \\ S_{2} = 12 + 10 = 22 \\ a = 8 \\ a + a + d = 22 \\ 2 a + d = 22 \\ 2 (8) + d = 22 \\ d = 22 - 16 \\ d = 6 \\ T h e A P i s a, a + d, a + 2 d \dots \\ a = 8 \\ a + d = 8 + 6 = 14 \\ a + 2 d = 8 + 6 (2) = 8 + 12 = 20 \\ 8, 14, 20 \dots \\ a_{k} = 164 \\ a + (k - 1) \cdot d = 164 \\ 8 + (k - 1) \cdot 6 = 164 \\ (k - 1) \cdot 6 = 164 - 8 \\ k = 26 + 1 \\ k = 27 \end{aligned}$

Question:26

If $S_{n}$ denotes the sum of the first n terms of an AP, prove that $S_{12} = 3 (S_{8} - S_{4})$

Answer:

To prove : $- S_{12} = 3 (S_{8} - S_{4})$
If $S_{n}$ is the sum of $n$ terms, then the first term is $a$ and the common difference is $d$

$\begin{matrix} S_{n} = \frac{n}{2} [2 a + (n - 1) d] \dots (1) \\ Put n = 12 in equation (1) \\ S_{12} = \frac{12}{2} [2 a + 11 d] = 6 [2 a + 11 d] \\ S_{12} = 12 a + 66 d \dots (2) \\ S_{8} = \frac{8}{2} [2 a + 7 d] = 4 [2 a + 7 d] \\ S_{8} = 8 a + 28 d \dots (3) \\ S_{4} = \frac{4}{2} [2 a + 3 d] = 2 [2 a + 3 d] \\ S_{4} = 4 a + 6 d \dots (4) \\ RHS: 3 (S_{8} - S_{4}) \\ = 3 (8 a + 28 d - 4 a - 6 d) (By Equation (3) and (4)) \\ = 3 (4 a + 22 d) \\ = 12 a + 66 d (By Equation (2)) \\ = S_{12} \end{matrix}$

Question:27

Find the sum of the first 17 terms of an AP whose 4th and 9th terms are -15 and -30, respectively.

Answer:

$\begin{matrix} a_{4} = - 15, a_{9} = - 30 \\ a + 3 d = - 15 \dots (1) \\ a + 8 d = - 30 \dots (2) \\ (Using a_{n} = a + (n - 1) \cdot d) \\ From equations (1) and (2): \\ (a + 8 d) - (a + 3 d) = - 30 + 15 \\ 5 d = - 15 \\ d = - 3 \end{matrix}$

Put $d = - 3$ in equation (1):

$\begin{matrix} a + 3 (- 3) = - 15 \\ a - 9 = - 15 \\ a = - 6 \\ S_{17} = \frac{17}{2} [2 (- 6) + (17 - 1) (- 3)] \\ (U sing S_{n} = \frac{n}{2} [2 a + (n - 1) d]) \\ = \frac{17}{2} [- 12 - 48] \\ = \frac{17}{2} \times (- 60) \\ S_{17} = - 510 \end{matrix}$

Question:28

If the sum of the first 6 terms of an AP is 36 and that of the first 16 terms is 256, find the sum of the first 10 terms.

Answer:

$\begin{aligned} S_{6} = 36, S_{16} = 256 \\ [U sing S_{n} = \frac{n}{2} [2 a + (n - 1) d]] \\ S_{6} = 36 S_{16} = 256 \\ = \frac{6}{2} [2 a + (6 - 1) d] = 36 \frac{16}{2} [2 a + (16 - 1) d] = 256 \\ = 3 [2 a + 5 d] = 368 [2 a + 15 d] = 256 \\ = 2 a + 5 d = 12 \dots (1) \\ 2 a + 15 d = 32 \dots (2) \\ from (1) and (2) \\ - = - 2 \\ 10 d = 20 \\ d = 2 \\ P u t d = 2 in \\ 2 a + 5 (2) = 12 \\ 2 a = 12 - 10 \\ a = 1 \\ ([U sing S_{n} = \frac{n}{2} [2 a + (n - 1) d]] \\ Putn = 1 \\ = 5 [2 + 18] \\ 5 [20] \\ S_{10} = 100 \end{aligned}$

Question:29

Find the sum of all the 11 terms of an AP whose middlemost term is 30.

Answer:

The middle term of an $AP = \frac{n + 1}{2}$
Here $n = 11$
Hence $a_{6} = 30$

$\begin{aligned} a + 5 d = 30 \dots (1) (U sing a_{n} = a + (n - 1) d) \\ S_{n} = \frac{n}{2} [2 a + (n - 1) d] putn = 11 \\ = \frac{11}{2} [2 a + 10 d] \\ = \frac{n}{2} [2 (a + 5 d)] \\ = 11 [30] (U sing (1)) \\ S_{11} = 330 \end{aligned}$

Question:30

Find the sum of the last ten terms of the AP: 8, 10, 12, __________ , 126.

Answer:

The given AP is $8, 10, \dots, 126$
AP from last is $126, 124, \dots, 10, 8$

$a = 126, d = 124 - 126 = - 2$

The sum of 10 terms from the last:

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

Put $n = 10$ :

$\begin{matrix} S_{10} = \frac{10}{2} [2 (126) + (10 - 1) (- 2)] \\ = 5 [252 + 9 (- 2)] \\ = 5 [252 - 18] \\ = 5 \times 234 \\ S_{10} = 1170 \end{matrix}$

Question:31

Find the sum of the first seven numbers which are multiples of 2 as well as of 9.

Answer:

$LCM of 2 and 9 is 18.\ So the terms are 18, 36, \dots . ., 126 First term (a) = 18$
$Common difference (d) = 36 - 18 = 18$
$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$
$P u t n = 7$
$= \frac{7}{2} [36 + 108] = \frac{7}{2} [144]$
$S_{7} = 7 \times 72 S_{7} = 504$

Question:32

How many terms of the AP: -15, -13, -11, _______ are needed to make the sum -55? Explain the reason for the double answer.

Answer:

The given AP is $- 15, - 13, \dots$
First term $a = - 15$
Common difference $d = - 13 - (- 15) = - 13 + 15 = 2$
Let $n$ terms have a sum of -55 , then

$\begin{matrix} S_{n} = - 55 \\ S_{n} = \frac{n}{2} [2 a + (n - 1) d] \\ n [- 30 + 2 n - 2] = - 55 \times 2 \\ - 32 n + 2 n^{2} + 110 = 0 \\ 2 n^{2} - 32 n + 110 = 0 \\ n^{2} - 16 n + 55 = 0 \\ n^{2} - 5 n - 11 n + 55 = 0 \\ n (n - 5) - 11 (n - 5) = 0 \\ (n - 5) (n - 11) = 0 \\ n = 5, 11 \end{matrix}$

There are two possible values for $n$ :
- When $n = 5$ , all terms are negative.
- When $n = 11$ , the AP will contain both positive and negative terms, leading to a sum of -55.

Question:33

The sum of the first n terms of an AP whose first term is 8 and the common difference is 20 is equal to the sum of the first 2n terms of another AP whose first term is 30 and the common difference is 8. Find n.

Answer:

The given first AP has:

$a = 8, d_{1} = 20$

The given second $A P$ has:

$b = - 30, d_{2} = 8$

It is given that $S_{n} = S_{2 n}$ , so

$\frac{n}{2} [2 a + (n - 1) d_{1}] = \frac{2 n}{2} [2 b + (2 n - 1) d_{2}]$

Substituting values:

$\begin{aligned} \frac{n}{2} [2 (8) + (n - 1) 20] & = n [2 (- 30) + (2 n - 1) (8)] \\ \frac{n}{2} [16 + 20 n - 20] & = n [- 60 + 16 n - 8] \\ \frac{n}{2} [20 n - 4] & = n [- 68 + 16 n] \\ \frac{2 (10 n - 2)}{2} & = - 68 + 16 n \\ 10 n - 16 n & = - 68 + 2 \\ - 6 n & = - 66 \\ n = \frac{66}{6} & = 11 \end{aligned}$

Thus, $n = 11$ .

Question:34

Kanika was given her pocket money on Jan 1st, 2008. She puts Re 1 on Day 1, Rs 2 on Day 2, Rs 3 on Day 3, and continues doing so till the end of the month, from this money into her piggy bank. She also spent Rs 204 of her pocket money and found that at the end of the month, she still had Rs 100 with her. How much was her pocket money for the month?

Answer:

According to the question, she deposits Rs. 1 on Day 1, Rs. 2 on Day 2, and so on. Thus, the sequence follows: $1, 2, 3, \dots, 31$ , where $a = 1, d = 1, a_{n} = 31$ , and $n = 31$ .

The sum of an arithmetic series is given by:

$S_{n} = \frac{n}{2} (a + a_{n})$

Substituting the values:

$S_{31} = \frac{31}{2} (1 + 31) = \frac{31}{2} \times 32 = 31 \times 16 = 496$

Total money calculation:

$\begin{matrix} Total money = Total deposited + Spent + Left \\ = 496 + 204 + 100 = 800 \end{matrix}$

Thus, the total amount is Rs. 800.

Question:35

Yasmeen saves Rs 32 during the first month, Rs 36 in the second month and Rs 40 in the third month. If she continues to save in this manner, in how many months will she save Rs 2000?

Answer:

As per the question, she saves Rs. 32 in the first month, Rs. 36 in the second month, and so on.
Thus, the sequence follows: $32, 36, 40, \dots$
Here, the first term $a = 32$ , and the common difference $d = 36 - 32 = 4$ .
Let her savings reach Rs. 2000 in $n$ months.
The sum of an arithmetic series is given by:

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

Substituting the values:

$\begin{matrix} 2000 = \frac{n}{2} [2 \times 32 + (n - 1) \times 4] \\ 2000 = \frac{n}{2} [64 + 4 n - 4] \\ 2000 = \frac{n}{2} [60 + 4 n] \\ 4000 = n [60 + 4 n] \\ 4 n^{2} + 60 n - 4000 = 0 \end{matrix}$

Dividing by 4 :

$n^{2} + 15 n - 1000 = 0$

Solving by factorization:

$\begin{matrix} n^{2} + 40 n - 25 n - 1000 = 0 \\ n (n + 40) - 25 (n + 40) = 0 \\ (n + 40) (n - 25) = 0 \end{matrix}$

Since $n = - 40$ is not possible, we take $n = 25$ .
Hence, it takes 25 months to save Rs. 2000.

Class 10 Maths Chapter 5 exemplar solutions Exercise: 5.4
Page number: 56-58
Total questions:10

Question:1

The sum of the first five terms of an AP and the sum of the first seven terms of the same AP is 167. If the sum of the first ten terms of this AP is 235, find the sum of its first twenty terms.

Answer:

$S_{n} = \frac{n}{2} (2 a + (n - 1) d)$
Using this formula for $S_{5}$ :

$\begin{matrix} S_{5} = \frac{5}{2} (2 a + 4 d) = 167 \\ 5 (2 a + 4 d) = 334 ⟹ 2 a + 4 d = 66 \end{matrix}$
Using this formula for $S_{7}$ :

$\begin{matrix} S_{7} = \frac{7}{2} (2 a + 6 d) = 167 \\ 7 (2 a + 6 d) = 334 ⟹ 2 a + 6 d = 48 \end{matrix}$
Subtract equation (1) from equation (2):

$\begin{matrix} (2 a + 6 d) - (2 a + 4 d) = 48 - 66 \\ 2 d = - 18 ⟹ d = - 9 \end{matrix}$

Substitute $d = - 9$ into equation (1):

$\begin{matrix} 2 a + 4 (- 9) = 66 \\ 2 a - 36 = 66 ⟹ 2 a = 102 ⟹ a = 51 \end{matrix}$
Now, to find the sum of the first 20 terms:

$\begin{matrix} S_{20} = \frac{20}{2} (2 a + 19 d) \\ S_{20} = 10 (2 (51) + 19 (- 9)) \\ S_{20} = 10 (102 - 171) = 10 (- 69) = - 690 \end{matrix}$
Thus, the sum of the first 20th terms is -690.

Question:2

i)Find the sum of those integers between 1 and 500 which are multiples of 2 as well as of 5.
ii)Find the sum of those integers from 1 to 500 which are multiples of 2 as well as of 5.
iii)Find the sum of those integers from 1 to 500 which are multiples of 2 or 5.

Answer:

The numbers that are multiples of both 2 and 5 are the multiples of 10 :

$10, 20, 30, \dots, 490$

Here, the first term $a = 10$ , the common difference $d = 20 - 10 = 10$ , and the last term $a_{n} = 490$ .
To find $n$ , use the formula for the $n$ th term of an arithmetic sequence:

$\begin{matrix} a + (n - 1) d = 490 \\ 10 + (n - 1) \times 10 = 490 \\ (n - 1) \times 10 = 480 \\ n - 1 = \frac{480}{10} = 48 \\ n = 48 + 1 = 49 \end{matrix}$

Now, using the sum formula for an arithmetic series:

$\begin{aligned} S_{n} & = \frac{n}{2} [a + a_{n}] \\ S_{49} & = \frac{49}{2} [10 + 490] \\ = \frac{49}{2} \times 500 \\ = 49 & \times 250 = 12, 250 \end{aligned}$

Thus, the sum of the series is 12,250 .

ii)

The numbers that are multiples of both 2 and 5 are multiples of 10:

$10, 20, 30, \dots, 500$

Here, the first term $a = 10$ and the common difference $d = 20 - 10 = 10$ .
The last term is $a_{n} = 500$ .
Using the formula for the $n$ th term of an arithmetic sequence:

$\begin{matrix} a_{n} = a + (n - 1) d \\ 10 + (n - 1) \times 10 = 500 \\ (n - 1) \times 10 = 490 \\ n - 1 = \frac{490}{10} = 49 \\ n = 49 + 1 = 50 \end{matrix}$

Now, using the sum formula for an arithmetic series:

$S_{n} = \frac{n}{2} [a + a_{n}]$

Substituting $n = 50$ :

$\begin{aligned} S_{50} = \frac{50}{2} [10 + 500] \\ = 25 \times 510 = 12, 750 \end{aligned}$

Thus, the sum of the series is 12,750 .

iii)

The numbers that are multiples of 2 :
$2, 4, 6, 8, \dots, 500$
Here, the first term $a_{1} = 2$ , common difference $d_{1} = 4 - 2 = 2$ , and last term $a_{n_{1}} = 500$ .
To find $n_{1}$ :

$n_{1} = \frac{500}{2} = 250$

Using the sum formula for an arithmetic series:

$\begin{matrix} S_{n_{1}} = \frac{n_{1}}{2} [a_{1} + a_{n_{1}}] \\ S_{n_{1}} = \frac{250}{2} [2 + 500] = 125 \times 502 = 62, 750 \end{matrix}$

The numbers that are multiples of 5 :

$5, 10, 15, \dots, 500$

Here, the first term $a_{2} = 5$ , common difference $d_{2} = 10 - 5 = 5$ , and last term $a_{n_{2}} = 500$ . To find $n_{2}$ :

$n_{2} = \frac{500}{5} = 100$

Using the sum formula:

$\begin{matrix} S_{n_{2}} = \frac{n_{2}}{2} [a_{2} + a_{n_{2}}] \\ S_{n_{2}} = \frac{100}{2} [5 + 500] = 50 \times 505 = 25, 250 \end{matrix}$

The numbers that are multiples of 10 (to be subtracted as they are counted in both multiples of 2 and 5):

$10, 20, 30, \dots, 500$

Here, the first term $a_{3} = 10$ , common difference $d_{3} = 20 - 10 = 10$ , and last term $a_{n_{3}} = 500$ . To find $n_{3}$ :

$n_{3} = \frac{500}{10} = 50$

Using the sum formula:

$\begin{matrix} S_{n_{3}} = \frac{n_{3}}{2} [a_{3} + a_{n_{3}}] \\ S_{n_{3}} = \frac{50}{2} [10 + 500] = 25 \times 510 = 12, 750 \end{matrix}$

Now, applying the principle of inclusion-exclusion:
Sum of numbers that are multiples of 2 or $5 = S_{n_{1}} + S_{n_{2}} - S_{n_{3}}$

$\begin{aligned} = 62, 750 + 25, 250 - 12, 750 \\ = 88, 000 - 12, 750 = 75, 250 \end{aligned}$

Thus, the sum of integers from 1 to 500 that are multiples of 2 or 5 is 75,250 .

Question:3

The eighth term of an AP is half its second term, and the eleventh term exceeds
one-third of its fourth term by 1. Find the 15th term.

Answer:

Let the first term be $a$ and the common difference be $d$ .
Given conditions:
1. $T_{8} = \frac{1}{2} T_{2}$ implies:

$a + 7 d = \frac{1}{2} (a + d) ⟹ 2 a + 14 d = a + d ⟹ a = - 13 d$

2. $T_{11} = \frac{1}{3} T_{4} + 1$ implies:

$a + 10 d = \frac{1}{3} (a + 3 d) + 1 ⟹ 3 a + 30 d = a + 3 d + 3 ⟹ 2 a = - 27 d + 3$
Substitute $a = - 13 d$ into $2 a = - 27 d + 3$ :

$2 (- 13 d) = - 27 d + 3 ⟹ - 26 d = - 27 d + 3 ⟹ d = 3$
Substitute $d = 3$ into $a = - 13 d$ :

$a = - 39$
Find the 15th term:

$T_{15} = a + 14 d = - 39 + 14 (3) = - 39 + 42 = 3$
Thus, the 15th term is 3.

Question:4

An AP consists of 37 terms. The sum of the three middlemost terms is 225, and the sum of the last three is 429. Find the AP.

Answer:

Total terms $= 37$ , middle term $=$

$\frac{37 + 1}{2} = 19$
Three middle most terms, 18th, 19th and 20th, given

$a_{18} + a_{19} + a_{20} = 225$

$a + 17 d + a + 18 d + a + 19 d = 225 (Using a_{n} = a + (n - 1) d)$

$3 a + 54 d = 225$

Dividing by 3, we get

$\begin{aligned} a + 18 d = 75 \\ a = 75 - 18 d \dots (1) \end{aligned}$

According to the question

$\begin{aligned} a_{35} + a_{36} + a_{37} = 429 (Using a_{n} = a + (n - 1) d) \\ a + 34 d + a + 35 d + a + 36 d = 429 \\ 3 a + 105 d = 429 \end{aligned}$

Dividing by three, we get

$a + 35 d = 143$ Putting value from equation (1), we get

$\begin{aligned} 75 - 18 d + 35 d = 143 \\ 17 d = 143 - 75 \\ d = 68 \\ d = \frac{68}{17} d = 4 \end{aligned}$ Puttingt$ $d = 4$ in (1), we get

$\begin{aligned} a = 75 - 72 = 3 \\ a + d = 3 + 4 = 7 \\ a + 2 d = 3 + 2 (4) = 11 \end{aligned}$
Required AP is $3, 7, 11, \dots \dots$

Question:5

i)Find the sum of the integers between 100 and 200 that are divisible by 9
ii)Find the sum of the integers between 100 and 200 that are not divisible by 9

Answer:

Numbers between 100 and 200 which is divisible by 9 are $108, 117, 126, \dots 198$ .
Here $a = 108, d = 117 - 108 = 9$

$\begin{aligned} a_{n} = 198 \\ a + (n - 1) \times d = 198 \\ 108 + (n - 1) \times 9 = 198 \\ (n - 1) \times 9 = 198 - 108 \\ (n - 1) = \frac{90}{9} \\ n = 10 + 1 = 11 \\ S_{n} = \frac{n}{2} [2 a + (n - 1) d] \end{aligned}$
Put $n = 11$

$\begin{aligned} S_{11} & = \frac{11}{2} [2 (108) + (11 - 1) 9] \\ = \frac{11}{2} [216 + 90] \\ = \frac{11}{2} \times 306 \\ = 11 \times 153 \\ S_{11} & = 1683 \end{aligned}$

ii)

Numbers between 100 and 200 are 101, 102, 103...., 199

$\begin{aligned} Here a = 101, d = 102 \cdot 101 = 1 \\ a_{n} = 199 \\ a + (n - 1) \cdot d = 199 (U sing a_{n} = a + (n - 1) d) \\ 101 + (n - 1) 1 = 199 \\ (n - 1) = 199 - 101 \\ n = 98 + 1 \\ n = 99 \\ S_{99} = \frac{99}{2} [2 \times 101 + (99 - 1) (1)] \\ = \frac{99}{2} [202 + 98] = \frac{99}{2} [300] \\ = 99 [150] \\ = 14850 \end{aligned}$

Numbers between 100 and 200 which are divisible by 9 are 108, 117, 126, ... 198

$\begin{aligned} Here b = 108 \\ d_{1} = 117 - 108 = 9 \\ b_{N} = 198 \\ b + (N - 1) \times d_{1} = 198 \\ 108 + (N - 1) \times 9 = 198 \\ (N - 1) \times 9 = 198 - 108 \\ (N - 1) = \frac{90}{9} \end{aligned}$

$\begin{aligned} N = 10 + 1 = 11 \\ S_{N} = \frac{N}{2} [2 b + (N - 1) d_{1}] \end{aligned}$

Put N $= 11$

$\begin{array}{r} \begin{aligned} S_{11} & = \frac{11}{2} [2 (108) + (11 - 1) 9] \\ = \frac{11}{2} [216 + 90] \\ = \frac{11}{2} \times 306 \\ = 11 \times 153 \\ S_{11} & = 1683 \end{aligned} \end{array}$

The sum of the integers between 100 and 200 that are not divisible by 9
$=$ Sum of all numbers - the sum of numbers which is divisible by 9

$\begin{aligned} = 14850 - 1683 \\ = 13167 \end{aligned}$

Question:6

The ratio of the 11th term to the 18th term of an AP is 2/3. Find the ratio of the 5th term to the 21st term, and also the ratio of the sum of the first five terms to the sum of the first 21 terms.

Answer:

According to the question

$\begin{aligned} \frac{a_{11}}{a_{18}} = \frac{2}{3} \Rightarrow \frac{a + 10 d}{a + 17 d} = \frac{2}{3} (U sing a_{n} = a + (n - 1) \cdot d) \\ 3 a + 30 d = 2 a + 34 d \\ 3 a - 2 a + 30 d - 34 d = 0 \\ a - 4 d = 0 a = 4 d \dots (1) \end{aligned}$

Ratio of $5^{th}$ and 21 th term is

$\frac{a_{5}}{a_{21}} = \frac{a + 4 d}{a + 20 d} \dots (2)$

Put $a = 4 d$ in (2) we get $= \frac{4 d + 4 d}{4 d + 20 d} = \frac{8 d}{24 d}$

$\frac{a_{5}}{a_{21}} = \frac{1}{3}$

Ratio of $S_{5}$ to $S_{21}$ is $\frac{S_{5}}{S_{21}} = \frac{\frac{5}{2} [2 a + 4 d]}{2 [2 a + 20 d]}$

$\begin{aligned} [∵ S_{n} & = \frac{n}{2} [2 a + (n - 1) d]] \\ \frac{S_{5}}{S_{21}} & = \frac{5 [2 a + 4 d]}{21 [2 a + 200]} \end{aligned}$

Put $a = 4 d$ in equation (3) we get

$\begin{matrix} \frac{S_{5}}{S_{21}} = \frac{5 [4 d d) + 4 d]}{21 [2 (4 d) + 20 d]} \\ = \frac{5 [[d d + 4 d]}{21 [8 d + 20 d]} \\ = \frac{5 [12 d]}{21 [28 d]} \\ = \frac{60 d}{588 d} \\ \frac{S_{5}}{S_{12}} = \frac{5}{49} \end{matrix}$

Question:7

Show that the sum of an AP whose first term is a, the second term b and the last term c, is equal to
$\frac{(a + c) (b + c - 2 a)}{2 (b - a)}$

Answer:

Here

$\begin{aligned} a_{1} = a, a_{2} = b \\ d = a_{2} - a_{1} \\ = b - a \\ a_{n} = c ∵ a_{n} = a + (n - 1) d \\ a_{1} + (n - 1) \cdot d = c Put a_{1} = a, d = b - a \\ (n - 1) (b - a) = c - a \\ n - 1 = \frac{c - a}{b - a} \\ n = \frac{c - a}{b - a + 1} \\ n = \frac{c - a + b - a}{b - a} \\ n = \frac{c + b - 2 a}{b - a} \\ S_{n} = \frac{n}{2} [2 a_{1} + (n - 1) d] \\ = \frac{c + b - 2 a}{2 (b - a)} [a + a + (n - 1) d] \\ = \frac{(c + b - 2 a)}{2 (b - a)} [a + a_{n}] [∵ a + (n - 1) d = a_{n}] \\ \frac{(a + c) (b + c - 2 a)}{2 (b - a)} [∵ a_{n} = c] \end{aligned}$

Hence proved.

Question:8

Solve the equation- 4 + (-1) + 2 +...+ x = 437

Answer:

Given series:

$- 4 + (- 1) + 2 + \dots + x = 437$
Here $a = - 4$

$\begin{aligned} d = - 1 - (- 4) \\ - 1 + 4 = + 3 \\ S_{n} = 437 (given) \end{aligned}$
We know that $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$

$\begin{aligned} 437 = \frac{n}{2} [2 (- 4) + (n - 1) 3] \\ 874 = n [- 8 + 3 n - 3] \\ 874 = 3 n^{2} - 11 n \\ 3 n^{2} - 11 n - 874 = 0 \\ 3 n^{2} - 57 n + 46 n - 874 = 0 \\ 3 n (n - 19) + 46 (n - 19) = 0 \\ \begin{matrix} (n - 19) (3 n + 46) = 0 \\ n - 19 = 0 2 n + 16 = 0 \\ n = 19 n = - 46 / 3 \end{matrix} \end{aligned}$
Hence $n = 19$

$\begin{aligned} a_{n} = x \\ a + (n - 1) d = x (using a_{n} = a + (n - 1) d) \\ - 4 + (19 - 1) 3 = x \\ - 4 + (18) 3 = x \\ - 4 + 54 = x \\ x = 50 \end{aligned}$

Question:9

Jaspal Singh repays his total loan of Rs 118000 by paying every month, starting with the first instalment of Rs 1000. If he increases the instalment by Rs 100 every month, what amount will be paid by him in the 30th instalment? What amount of loan does he still have to pay after the 30th instalment?

Answer:1

Total loan = 118000
First installment (a) = 1000
Installment increased = 100
Hence d = 100
$30_{t h} installment = a_{30} = a + 29 d$
= 1000 + 29(100)
= 1000 + 2900
= 3900
Amount of loan paid after 30 instalments
= Total loan - 30th installments is total loan

$\begin{matrix} 118000 - \frac{30}{2} [2 \times 1000 + (30 - 1) \times 100] \\ 118000 - 15 [2000 + 2900] \\ 118000 - 15 [4900] \\ 118000 - 73500 = 44500 \end{matrix}$

Question:10

The students of a school decided to beautify the school on the Annual Day by fixing colourful flags on the straight passage of the school. They have 27 flags to be fixed at intervals of 2 m. The flags are stored at the position of the middlemost flag. Ruchi was given the responsibility of placing the flags. Ruchi kept her books where the flags were stored. She could carry only one flag at a time. How much distance did she cover in completing this job and returning to collect her books? What is the maximum distance she travelled carrying a flag?

Answer:

Let her fix 13 flags on one side and 13 on the other side of the middle flag. The distance covered for fixing the first flag and returning is 2m + 2m = 4.

The distance covered for fixing the second flag and returning is 4m + 4m = 8m.

The distance covered for fixing the third flag and returning is 6m + 6m = 12m
Similarly, for thirteen flags is
4m + 8m + 12m + …….
Here, a = 4
d = 8 - 4 = 4
n = 13
$\begin{matrix} S_{n} = \frac{n}{2} [2 a + (n - 1) d] \\ S_{13} = \frac{13}{2} [2 (4) + (13 - 1) 4] \\ = \frac{13}{2} [56] \\ = 364 \end{matrix}$
Total distance covered = 364 × 2 = 728 m (for both side )
distance, only to carry the flag = 364.

NCERT Class 10 Maths Exemplar Solutions for Other Chapters:

NCERT Exemplar for Class 10 Mathematics Chapter 1 Real Numbers

NCERT Exemplar for Class 10 Mathematics Chapter 2 Polynomials

NCERT Exemplar for Class 10 Mathematics Chapter 3 Pair of Linear Equations in Two Variables

NCERT Exemplar for Class 10 Mathematics Chapter 4 Quadratic Equations

NCERT Exemplar for Class 10 Mathematics Chapter 5 Arithmetic Progressions

NCERT Exemplar for Class 10 Mathematics Chapter 6 Triangles

NCERT Exemplar for Class 10 Mathematics Chapter 7 Coordinate Geometry

NCERT Exemplar for Class 10 Mathematics Chapter 8 Introduction to Trigonometry & Its Equations

NCERT Exemplar for Class 10 Mathematics Chapter 9 Circles

NCERT Exemplar for Class 10 Mathematics Chapter 10 Constructions

NCERT Exemplar for Class 10 Mathematics Chapter 12 Surface Areas and Volumes

NCERT Exemplar for Class 10 Mathematics Chapter 13 Statistics and Probability

Importance of Solving NCERT Exemplar Class 10 Maths Solutions Chapter 5

NCERT Exemplar Class 10 Maths Solutions Chapter 5 pdf downloads are available through online tools for the students to access this content in an offline version so that no breaks in continuity are faced while practising NCERT Exemplar Class 10 Maths Chapter 5.

These Class 10 Maths NCERT exemplar Chapter 5 solutions provide a basic knowledge of Arithmetic Progressions, which has great importance in higher classes.
The questions based on Arithmetic Progressions can be practised in a better way, along with these solutions.

NCERT Solutions for Class 10 Maths: Chapter Wise

NCERT Solutions for Class 10 Mathematics Chapter 1: Real numbers

NCERT Solutions for Class 10 Mathematics Chapter 2: Polynomials

NCERT Solutions for Class 10 Mathematics Chapter 3: Pair of Linear Equations in Two Variables

NCERT Solutions for Class 10 Mathematics Chapter 4: Quadratic Equations

NCERT Solutions for Class 10 Mathematics Chapter 5: Arithmetic Progressions

NCERT Solutions for Class 10 Mathematics Chapter 6: Triangles

NCERT Solutions for Class 10 Mathematics Chapter 7: Coordinate Geometry

NCERT Solutions for Class 10 Mathematics Chapter 8: Introduction to Trigonometry

NCERT Solutions for Class 10 Mathematics Chapter 9: Some Applications of Trigonometry

NCERT Solutions for Class 10 Mathematics Chapter 10: Circles

NCERT Solutions for Class 10 Mathematics Chapter 11: Areas related to Circles

NCERT Solutions for Class 10 Mathematics Chapter 12: Surface Areas and Volumes

NCERT Solutions for Class 10 Mathematics Chapter 13: Statistics

NCERT Solutions for Class 10 Mathematics Chapter 14: Probability

NCERT solutions of class 10 - Subject-wise

Here are the subject-wise links for the NCERT solutions of class 10:

NCERT Notes of class 10 - Subject Wise

Given below are the subject-wise NCERT Notes of class 10 :

NCERT Books and NCERT Syllabus

Here are some useful links for NCERT books and the NCERT syllabus for class 10:

NCERT Class 10 Exemplar Solutions Subject Wise

Given below are the subject-wise exemplar solutions of class 10 NCERT:

Frequently Asked Questions (FAQs)

1. Can two different Arithmetic Progressions have the same common difference?

Yes, they can have same common difference but they must have different first term then only these two progressions will be considered as different progression

2. Sum of any arithmetic progression can be zero. Comment.

Yes, the summation of AP can be zero but for that exactly one first term or common difference has to be negative and other has to be positive.

3. Is the knowledge of Arithmetic progression of class 10 sufficient for JEE Advanced exam?

Yes, the knowledge of AP given in class 10 is more than sufficient for solving any questions of AP at a higher level. If any student refers NCERT exemplar Class 10 Maths solutions chapter 5, it would be very useful for mathematics at higher levels

4. Is the chapter Arithmetic Progression important for Board examinations?

This chapter is important for Class 10 board exams as it comprises around 4-5% weightage of the whole paper.

5. What percentage of marks does this chapter of Arithmetic Progression holds in board examinations?

Generally, Arithmetics progression is considered among the important topics for board examinations and accounts for 8-10% marks of the total paper.

Also Read

Some Applications of Trigonometry Class 10th Notes - Free NCERT Class 10 Maths Chapter 9 Notes - Download PDF

Coordinate Geometry Class 10th Notes - Free NCERT Class 10 Maths Chapter 7 Notes - Download PDF

Arithmetic Progressions Class 10th Notes - Free NCERT Class 10 Maths Chapter 5 Notes - Download PDF

Introduction to Trigonometry Class 10th Notes - Free NCERT Class 10 Maths Chapter 8 Notes - Download PDF

Triangles Class 10th Notes - Free NCERT Class 10 Maths Chapter 6 Notes - Download PDF

Free NCERT Class 10 Maths Chapter 2 Notes - Download PDF

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

NCERT Solutions for Exercise 12.2 Class 10 Maths Chapter 12 - Areas related to circles

NCERT Solutions for Exercise 12.1 Class 10 Maths Chapter 12 - Areas related to circles

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

NCERT Exemplar Class 10 Solutions - Subject Wise Problems with Solutions

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

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

NCERT Syllabus for Class 10 Maths 2025-26 (Updated) - Download New Syllabus PDF

NCERT Syllabus for Class 10 - Download All Subjects Syllabus PDF

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Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?

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 .

Read Complete Answer

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.

Read Complete Answer

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.

Read Complete Answer

sadafreen356 03 July,2024

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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 Exemplar Class 10 Maths Solutions Chapter 5 Arithmetic Progressions

NCERT Exemplar Class 10 Maths Solutions Chapter 5

NEET/JEE Coaching Scholarship

JEE Main Important Mathematics Formulas

Question:1

NCERT Class 10 Maths Exemplar Solutions for Other Chapters:

Importance of Solving NCERT Exemplar Class 10 Maths Solutions Chapter 5

NCERT solutions of class 10 - Subject-wise

NCERT Notes of class 10 - Subject Wise

NCERT Books and NCERT Syllabus

NCERT Class 10 Exemplar Solutions Subject Wise

Frequently Asked Questions (FAQs)

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