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NCERT Solutions For Class 10 Maths Chapter 5 Arithmetic Progressions

NCERT Solutions For Class 10 Maths Chapter 5 Arithmetic Progressions

Edited By Apoorva Singh | Updated on Sep 05, 2023 08:12 PM IST | #CBSE Class 10th

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions provided here. These solutions are created by expert team at Careers360 keeping in mind the latest syllabus and pattern of CBSE 2023-24. Arithmetic Progression class 10 solutions are designed covering all the topics and concepts comprehensively with step by step solutions of all exercise. After practicing these NCERT solutions for Class 10 students get command and in-depth understanding of the concepts which ultimately lead to score well in the exams.

This Class 10 Maths Chapter 5 covers the type of questions in which succeeding terms are obtained by adding the same number to the preceding terms. NCERT solutions are not only important for Board exams but also foundation for higher class therefore, students can take help from these solutions if they are facing any problems. here in this page students can find detailed arithmetic progression class 10 notes NCERT solutions also they can download class 10 maths chapter 5 pdf.

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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 - Important Formulae

  • Arithmetic Progressions (AP) involve sequence terms denoted as a1, a2, a3, ..., an representing a series of integers.

  • An AP maintains a constant difference between consecutive terms, known as the common difference.

  • For terms a1, a2, a3, a4, a5, and a6 in an AP, the common difference can be expressed as D = a2 – a1 = a3 – a2 = a4 – a3 = ...

  • The nth term of an AP is given by the formula: an = a + (n – 1) d.

Where

a = First term of the sequence.

n = Term's position in the sequence.

d = Common difference.

  • The sum of the first 'n’ terms in an AP is calculated using the formula:

  • Sn = (n/2) [2a + (n – 1)d]

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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.

Free download NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions PDF for CBSE Exam.

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions (Intext Questions and Exercise)

NCERT Solutions for 5 chapter maths class 10 Arithmetic Progressions Excercise: 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 \small Rs\hspace{1mm}15 for the first km and \small Rs\hspace{1mm}8 for each additional \small km .

Answer:

It is given that
Fare for 1^{st} \ km = Rs. 15
And after that \small Rs\hspace{1mm}8 for each additional \small km
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 n km = 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 \small \frac{1}{4} of the air remaining in the cylinder at a time.

Answer:

It is given that
vacum pump removes \small \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 \small Rs\hspace{1mm}150 for the first metre and rises by \small Rs\hspace{1mm}50 for each subsequent meter.

Answer:

It is given that
Cost of digging of 1st meter = Rs 150
and
rises by \small Rs\hspace{1mm}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 \small Rs\hspace{1mm}10000 is deposited at compound interest at \small 8\hspace{1mm}\% per annum .

Answer:

Amount in the beginning = Rs. 10000

Interest at the end of 1st year at the rate of \small 8\hspace{1mm}\%
is \small 8\hspace{1mm}\% of 10000 = \frac{8\times 10000}{100}= 800

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

= 10000 + 800

= 10800

Now,

Interest at the end of 2nd year at rate of \small 8\hspace{1mm}\%
is \small 8\hspace{1mm}\% 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 \small a=10,d=10

Answer:

It is given that
\small a=10,d=10
Now,
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
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: \small a=-2,d=0
Answer:

It is given that
\small a=-2,d=0
Now,
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
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 \small a=4,d=-3

Answer:

It is given that
\small a=4,d=-3
Now,
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
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 \small a=-1,d=\frac{1}{2}

Answer:

It is given that
\small a=-1,d=\frac{1}{2}
Now,
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}
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 \small a=-1.25,d=-0.25

Answer:

It is given that
\small a=-1.25,d=-0.25
Now,
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
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: \small 3,1,-1,-3,...

Answer:

Given AP series is

\small 3,1,-1,-3,...

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: \small -5,-1,3,7,...

Answer:

Given AP series is

\small -5,-1,3,7,...

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: \small \frac{1}{3},\frac{5}{3},\frac{9}{3},\frac{13}{3},...

Answer:

Given AP series is

\small \frac{1}{3},\frac{5}{3},\frac{9}{3},\frac{13}{3},...
Now, the first term of this AP series is \frac{1}{3}

Therefore,

The 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: \small 0.6,1.7,2.8,3.9,...

Answer:

Given AP series is

\small 0.6,1.7,2.8,3.9,...

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. \small 2,4,8,12,...

Answer:

Given series is
\small 2,4,8,12,...
Now,
the first term to this series is = 2
Now,
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
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. \small 2,\frac{5}{2},3,\frac{7}{2},...

Answer:

Given series is
\small 2,\frac{5}{2},3,\frac{7}{2},...
Now,
first term to this series is = 2
Now,
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}
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
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

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. \small -1.2,-3.2,-5.2,-7.2,...

Answer:

Given series is
\small -1.2,-3.2,-5.2,-7.2,...
Now,
the first term to this series is = -1.2
Now,
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
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
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
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. \small -10,-6,-2,2,...

Answer:

Given series is
\small -10,-6,-2,2,...
Now,
the first term to this series is = -10
Now,
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
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
a_5=a_4+d = 2+4 =6
a_6=a_5+d = 6+4=10
a_7=a_6+d = 10+4=14
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. \small 3,3+\sqrt{2},3+2\sqrt{2},3+3\sqrt{2},...

Answer:

Given series is
\small 3,3+\sqrt{2},3+2\sqrt{2},3+3\sqrt{2},...
Now,
the first term to this series is = 3
Now,
a_1 = 3 \ \ and \ \ a_2 = 3+\sqrt2 \ \ and \ \ a_3 = 3+2\sqrt2 \ \ and \ \ a_4 = 3+3\sqrt2
a_2-a_1 = 3+\sqrt2-3= \sqrt2
a_3-a_2 = 3+2\sqrt2-3-\sqrt2 = \sqrt2
a_4-a_3 = 3+3\sqrt2-3-2\sqrt2 = \sqrt2
We can clearly see that the difference between terms are equal and equal to \sqrt2
Hence, given series is in AP
Now, the next three terms are
a_5=a_4+d = 3+3\sqrt2+\sqrt2=3+4\sqrt2
a_6=a_5+d = 3+4\sqrt2+\sqrt2=3+5\sqrt2
a_7=a_6+d = 3+5\sqrt2+\sqrt2=3+6\sqrt2
Therefore, next three terms of given series are 3+4\sqrt2, 3+5\sqrt2,3+6\sqrt2

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

Answer:

Given series is
\small 0.2,0.22,0.222,0.2222,...
Now,
the first term to this series is = 0.2
Now,
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

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. \small 0,-4,-8,-12,...

Answer:

Given series is
\small 0,-4,-8,-12,...
Now,
first term to this series is = 0
Now,
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
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
a_5=a_4+d = -12-4 =-16
a_6=a_5+d = -16-4=-20
a_7=a_6+d = -20-4=-24
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. \small -\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},...

Answer:

Given series is
\small -\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},...
Now,
the first term to this series is = -\frac{1}{2}
Now,
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}-\left ( -\frac{1}{2} \right ) = -\frac{1}{2}+\frac{1}{2}=0
a_3-a_2 = -\frac{1}{2}-\left ( -\frac{1}{2} \right ) = -\frac{1}{2}+\frac{1}{2}=0
a_4-a_3 = -\frac{1}{2}-\left ( -\frac{1}{2} \right ) = -\frac{1}{2}+\frac{1}{2}=0
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
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}
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. \small 1,3,9,27,...

Answer:

Given series is
\small 1,3,9,27,...
Now,
the first term to this series is = 1
Now,
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


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. \small a,2a,3a,4a,...

Answer:

Given series is
\small a,2a,3a,4a,...
Now,
the first term to this series is = a
Now,
a_1 = a \ \ and \ \ a_2 = 2a \ \ and \ \ a_3 = 3a \ \ and \ \ a_4 = 4a
a_2-a_1 = 2a-a =a
a_3-a_2 = 3a-2a =a
a_4-a_3=4a-3a=a
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
a_5=a_4+d =4a+a=5a
a_6=a_5+d =5a+a=6a
a_7=a_6+d =6a+a=7a
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. \small a,a^2,a^3,a^4,...
Answer:

Given series is
\small a,a^2,a^3,a^4,...
Now,
the first term to this series is = a
Now,
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)

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. \small \sqrt{2},\sqrt{8},\sqrt{18},\sqrt{32},...

Answer:

Given series is
\small \sqrt{2},\sqrt{8},\sqrt{18},\sqrt{32},...
We can rewrite it as
\sqrt2,2\sqrt2,3\sqrt2,4\sqrt2,....
Now,
first term to this series is = a
Now,
a_1 = \sqrt2 \ \ and \ \ a_2 = 2\sqrt2 \ \ and \ \ a_3 = 3\sqrt2 \ \ and \ \ a_4 = 4\sqrt2
a_2-a_1 = 2\sqrt2-\sqrt2 =\sqrt2
a_3-a_2 = 3\sqrt2-2\sqrt2 =\sqrt2
a_4-a_3=4\sqrt2-3\sqrt2=\sqrt2
We can clearly see that difference between terms are equal and equal to \sqrt2
Hence, given series is in AP
Now, the next three terms are
a_5=a_4+d =4\sqrt2+\sqrt2=5\sqrt2
a_6=a_5+d =5\sqrt2+\sqrt2=6\sqrt2
a_7=a_6+d =6\sqrt2+\sqrt2=7\sqrt2
Therefore, next three terms of given series are 5\sqrt2,6\sqrt2,7\sqrt2

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. \small \sqrt{3},\sqrt{6},\sqrt{9},\sqrt{12},...

Answer:

Given series is
\small \sqrt{3},\sqrt{6},\sqrt{9},\sqrt{12},...
Now,
the first term to this series is = \sqrt3
Now,
a_1 = \sqrt3 \ \ and \ \ a_2 = \sqrt6 \ \ and \ \ a_3 = \sqrt9 \ \ and \ \ a_4 = \sqrt{12}
a_2-a_1 = \sqrt6-\sqrt3 =\sqrt3(\sqrt2-1)
a_3-a_2 = 3-\sqrt3 =\sqrt3(\sqrt3-1)
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. \small 1^2,3^2,5^2,7^2,...

Answer:

Given series is
\small 1^2,3^2,5^2,7^2,...
we can rewrite it as
1,9,25,49,....
Now,
the first term to this series is = 1
Now,
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
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. \small 1^2,5^2,7^2,73,...

Answer:

Given series is
\small 1^2,5^2,7^2,73,...
we can rewrite it as
1,25,49,73....
Now,
the first term to this series is = 1
Now,
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
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
a_5=a_4+d = 73+24=97
a_6=a_5+d = 97+24=121
a_7=a_6+d = 121+24=145
Therefore, the next three terms of given series are 97,121,145


NCERT Solutions Class 10 Chapter 5 Arithmetic Progressions Excercise: 5.2

Q1 Fill in the blanks in the following table, given that a is the first term, d the common difference and \small a_n the \small n th term of the AP:


a

d

n

\small a_n

(i)

(ii)

(iii)

(iv)

(v)

7

\small -18

\small ...

\small -18.9

\small 3.5

3

\small ...

\small -3

\small 2.5

0

8

10

18

\small ...

105

\small ...

0

\small -5

\small 3.6

\small ...

Answer:

(i)
It is given that
a=7, d = 3 , n = 8
Now, we know that
a_n = a+(n-1)d
a_8 = 7+(8-1)3= 7+7\times 3 = 7+21 = 28
Therefore,
a_8 = 28

(ii) It is given that
a=-18, n = 10, a_{10} = 0
Now, we know that
a_n = a+(n-1)d
a_{10} = -18+(10-1)d
0 +18=9d
d = \frac{18}{9}=2
(iii) It is given that
d=-3, n = 18, a_{18} = -5
Now, we know that
a_n = a+(n-1)d
a_{18} = a+(18-1)(-3)
-5=a+17\times (-3)
a = 51-5 = 46
Therefore,
a = 46

(iv) It is given that
a=-18.9, d = 2.5, a_{n} = 3.6
Now, we know that
a_n = a+(n-1)d
a_{n} = -18.9+(n-1)2.5
3.6+18.9= 2.5n-2.5
n = \frac{22.5+2.5}{2.5}= \frac{25}{2.5}= 10
Therefore,
n = 10

(v) It is given that
a=3.5, d = 0, n = 105
Now, we know that
a_n = a+(n-1)d
a_{105} = 3.5+(105-1)0
a_{105} = 3.5
Therefore,
a_{105} = 3.5

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

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

Answer:

Given series is
\small 10,7,4,...,
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,
a_{30} = 10+(30-1)(-3)
a_{30} = 10+(29)(-3)
a_{30} = 10-87 = -77
Therefore, \small 30 th term of the AP: \small 10,7,4,..., is -77
Hence, Correct answer is (C)

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

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

Answer:

Given series is
\small -3,-\frac{1}{2},2,...,
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,
a_{11} = -3+(11-1)\left ( \frac{5}{2} \right )
a_{11} = -3+(10)\left ( \frac{5}{2} \right )
a_{11} = -3+5\times 5 = -3+25 = 22
Therefore, 11th term of the AP: \small -3,-\frac{1}{2},2,..., is 22
Hence, the Correct answer is (B)


Q3 (i) In the following APs, find the missing terms in the boxes : \small 2,\hspace {1mm}\fbox{ },\hspace {1mm} 26

Answer:

Given AP series is
\small 2,\hspace {1mm}\fbox{ },\hspace {1mm} 26
Here, a = 2 , n = 3 \ and \ a_3 = 26
Now, we know that
a_n = a+(n-1)d
\Rightarrow a_3 =2+(3-1)d
\Rightarrow 26 -2=(2)d
\Rightarrow d = \frac{24}{2}= 12
Now,
a_2= a_1+d
a_2= 2+12 = 14
Therefore, the missing term is 14

Q3 (ii) In the following APs, find the missing terms in the boxes: \small \fbox { },\hspace {1mm}13,\hspace{1mm}\fbox { }, \hspace {1mm} 3

Answer:

Given AP series is
\small \fbox { },\hspace {1mm}13,\hspace{1mm}\fbox { }, \hspace {1mm} 3
Here, a_2 = 13 , n = 4 \ and \ a_4 = 3
Now,
a_2= a_1+d
a_1= a = 13 - d
Now, we know that
a_n = a+(n-1)d
\Rightarrow a_4 =13-d+(4-1)d
\Rightarrow 3-13=-d+3d
\Rightarrow d = -\frac{10}{2}= -5
Now,
a_2= a_1+d
a_1= a = 13 - d= 13-(-5 ) = 18
And
a_3=a_2+d
a_3=13-5 = 8
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 : \small 5,\: \fbox { },\: \fbox { },\: 9\frac{1}{2}

Answer:

Given AP series is
\small 5,\: \fbox { },\: \fbox { },\: 9\frac{1}{2}
Here, a = 5 , n = 4 \ and \ a_4 = 9\frac{1}{2}= \frac{19}{2}
Now, we know that
a_n = a+(n-1)d
\Rightarrow a_4 =5+(4-1)d
\Rightarrow \frac{19}{2} -5=3d
\Rightarrow d = \frac{19-10}{2\times 3} = \frac{9}{6} = \frac{3}{2}
Now,
a_2= a_1+d
a_2 = 5+\frac{3}{2} = \frac{13}{2}
And
a_3=a_2+d
a_3=\frac{13}{2}+\frac{3}{2} = \frac{16}{2} = 8
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 : \small -4,\: \fbox { },\: \fbox { },\: \fbox { },\: \fbox { },\: 6

Answer:

Given AP series is
\small -4,\: \fbox { },\: \fbox { },\: \fbox { },\: \fbox { },\: 6
Here, a = -4 , n = 6 \ and \ a_6 = 6
Now, we know that
a_n = a+(n-1)d
\Rightarrow a_6 =-4+(6-1)d
\Rightarrow 6+4 = 5d
\Rightarrow d = \frac{10}{5} = 2
Now,
a_2= a_1+d
a_2 = -4+2 = -2
And
a_3=a_2+d
a_3=-2+2 = 0
And
a_4 = a_3+d
a_4 = 0+2 = 2
And
a_5 = a_4 + d
a_5 = 2+2 = 4
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 : \small \fbox { },\: 38,\; \fbox { },\: \fbox { },\: \fbox { },\; -22

Answer:

Given AP series is
\small \fbox { },\: 38,\; \fbox { },\: \fbox { },\: \fbox { },\; -22
Here, a_2 = 38 , n = 6 \ and \ a_6 = -22
Now,
a_2=a_1+d
a_1=a =38-d \ \ \ \ \ \ \ \ \ -(i)
Now, we know that
a_n = a+(n-1)d
\Rightarrow a_6 =38-d+(6-1)d \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))
\Rightarrow -22-38-=-d+5d
\Rightarrow d = -\frac{60}{4} = - 15
Now,
a_2= a_1+d
a_1 = 38-(-15) = 38+15 = 53
And
a_3=a_2+d
a_3=38-15 = 23
And
a_4 = a_3+d
a_4 = 23-15 = 8
And
a_5 = a_4 + d
a_5 =8-15 = -7
Therefore, missing terms are 53,23,8,-7
AP series is 53,38,23,8,-7,-22

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

Given AP is
\small 3,8,13,18,...,
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
a_n = a + (n-1)d
\Rightarrow 78 = 3 + (n-1)5
\Rightarrow 78 -3 = 5n-5
\Rightarrow n = \frac{75 +5}{5}= \frac{80}{5} = 16
Therefore, value of 16th term of given AP is 78

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

Answer:

Given AP series is
\small 7,13,19,...,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
a_n =a + (n-1)d
\Rightarrow 205=7 + (n-1)6
\Rightarrow 205-7 = 6n-6
\Rightarrow n = \frac{198+6}{6} = \frac{204}{6} = 34
Therefore, there are 34 terms in given AP

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

Answer:

Given AP series is
\small 18,15\frac{1}{2},13,...,-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
a_n =a + (n-1)d
\Rightarrow -47=18 + (n-1)\left ( -\frac{5}{2} \right )
\Rightarrow -47-18= -\frac{5n}{2}+\frac{5}{2}
\Rightarrow -\frac{5n}{2}= -65-\frac{5}{2}
\Rightarrow -\frac{5n}{2}= -\frac{135}{2}
\Rightarrow n = 27
Therefore, there are 27 terms in given AP

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

Answer:

Given AP series is
\small 11,8,5,2...
Here, a = 11
And
d = a_2-a_1 = 8-11 = -3
Now,
suppose -150 is nth term of the given AP
Now, we know that
a_n = a+(n-1)d
\Rightarrow -150 = 11+(n-1)(-3)
\Rightarrow -150- 11=-3n+3
\Rightarrow =n = \frac{161+3}{3}= \frac{164}{3} = 54.66
Value of n is not an integer
Therefore, -150 is not a term of AP \small 11,8,5,2...

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

Answer:

It is given that
\small 11 th term of an AP is \small 38 and the \small 16 th term is \small 73
Now,
a_{11} =38= a+ 10d \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)
And
a_{16} =73= a+ 15d \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(ii)
On solving equation (i) and (ii) we will get
a= -32 \ \ \ and \ \ \ d = 7
Now,
a_{31} = a+30d = -32 + 30\times 7 = -32+210 = 178
Therefore, 31st terms of given AP is 178

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

It is given that
AP consists of \small 50 terms of which \small 3 rd term is \small 12 and the last term is \small 106
Now,
a_3 = 12=a+2d \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)
And
a_{50} = 106=a+49d \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(ii)
On solving equation (i) and (ii) we will get
a= 8 \ \ \ and \ \ \ d = 2
Now,
a_{29} = a+28d=8+28\times 2 = 8 +56 = 64
Therefore, 29th term of given AP is 64

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

It is given that
\small 3 rd and the \small 9 th terms of an AP are \small 4 and \small -8 respectively
Now,
a_3 = 4=a+2d \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)
And
a_{9} = -8=a+8d \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(ii)
On solving equation (i) and (ii) we will get
a= 8 \ \ \ and \ \ \ d = -2
Now,
Let nth term of given AP is 0
Then,
a_{n} = a+(n-1)d
0 = 8+(n-1)(-2)
2n = 8+2= 10
n = \frac{10}{2} = 5
Therefore, 5th term of given AP is 0

Q10 The \small 17 th term of an AP exceeds its \small 10 th term by \small 7 . Find the common difference.
Answer:

It is given that
\small 17 th term of an AP exceeds its \small 10 th term by \small 7
i.e.
a_{17}= a_{10}+7
\Rightarrow a+16d = a+9d+7
\Rightarrow a+16d - a-9d=7
\Rightarrow 7d=7
\Rightarrow d = 1
Therefore, the common difference of AP is 1

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

Given AP is
\small 3,15,27,39,...
Here, a= 3
And
d= a_2-a_1 = 15 - 3 = 12
Now, let's suppose nth term of given AP is \small 132 more than its \small 54 th term
Then,
a_n= a_{54}+132
\Rightarrow a+(n-1)d = a+53d+132
\Rightarrow 3+(n-1)12 = 3+53\times 12+132
\Rightarrow 12n = 3+636+132+12
\Rightarrow 12n = 636+132+12
\Rightarrow n = \frac{780}{12}= 65
Therefore, 65th term of given AP is \small 132 more than its \small 54 th term

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

It is given that
Two APs have the same common difference and difference between their \small 100 th terms is \small 100
i.e.
a_{100}-a'_{100}= 100
Let common difference of both the AP's is d
\Rightarrow a+99d-a'-99d=100
\Rightarrow a-a'=100 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ - (i)
Now, difference between 1000th term is
a_{1000}-a'_{1000}
\Rightarrow a+999d -a'-999d
\Rightarrow a-a'
\Rightarrow 100 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i) )
Therefore, difference between 1000th term is 100

Q 13 How many three-digit numbers are divisible by \small 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
a_n = a+ (n-1)d
\Rightarrow 994 = 105 + (n-1)7
\Rightarrow 7n = 896
\Rightarrow n = \frac{896}{7} = 128
Therefore, there are 128 three-digit numbers divisible by 7

Q14 How many multiples of \small 4 lie between \small 10 and \small 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
a_n = a+ (n-1)d
\Rightarrow 248 = 12 + (n-1)4
\Rightarrow 4n = 240
\Rightarrow n = \frac{240}{4} = 60
Therefore, there are 60 numbers between 10 to 250 that are divisible by 4

Q15 For what value of \small n , are the \small n th terms of two APs: \small 63,65,67,... and \small 3,10,17,... equal?

Answer:

Given two AP's are
\small 63,65,67,... and \small 3,10,17,...
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
a_n = a'_n
\Rightarrow a+(n-1)d=a'+(n-1)d'
\Rightarrow 63+(n-1)2=3+(n-1)7
\Rightarrow 5n=65
\Rightarrow n=\frac{65}{5} = 13
Therefore, the 13th term of both the AP's are equal

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

It is given that
3rd term of AP is \small 16 and the \small 7 th term exceeds the \small 5 th term by \small 12
i.e.
a_3=a+2d = 16 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)
And
a_7=a_5+12
a+6d=a+4d+12
2d = 12
d = 6
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,.....

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

Answer:

Given AP is
\small 3,8,13,...,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
a_n = a+(n-1)d
253= 3+(n-1)5
5n = 255
n = 51
So, there are 51 terms in the given AP and 20th term from the last will be 32th term from the starting
Therefore,
a_{32} = a+31d
a_{32} = 3+31\times 5 = 3+155 = 158
Therefore, 20th term from the of given AP is 158

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

It is given that
sum of the \small 4 th and \small 8 th terms of an AP is \small 24 and the sum of the \small 6 th and \small 10 th terms is \small 44
i.e.
a_4+a_8=24
\Rightarrow a+3d+a+7d=24
\Rightarrow 2a+10d=24
\Rightarrow a+5d=12 \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)
And
a_6+a_{10}=44
\Rightarrow a+5d+a+9d=44
\Rightarrow 2a+14d=44
\Rightarrow a+7d=22 \ \ \ \ \ \ \ \ \ \ \ \ \ -(ii)
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
a_n = a+(n-1)d
\Rightarrow 7000=5000+(n-1)200
\Rightarrow 2000=200n-200
\Rightarrow 200n=2200
\Rightarrow n = 11
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 \small 1.75 . If in the \small n th week, her weekly savings become Rs \small 20.75 , find \small 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 \small 1.75
Therefore, a = 5 \ and \ d = 1.75
after \small n th week, her weekly savings become Rs \small 20.75
Now, we know that
a_n = a +(n-1)d
\Rightarrow 20.75= 5+(n-1)1.75
\Rightarrow 15.75= 1.75n-1.75
\Rightarrow 1.75n=17.5
\Rightarrow n=10
Therefore, after 10 weeks her saving will become Rs 20.75


NCERT Arithmetic Progression Class 10 Solutions Excercise: 5.3

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

Answer:

Given AP is
\small 2,7,12,..., to \small 10 terms
Here, a = 2 \ and \ n = 10
And
d = a_2-a_1=7-2=5
Now, we know that
S = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S = \frac{10}{2}\left \{ 2\times 2 +(10-1)5\right \}
\Rightarrow S = 5\left \{ 4 +45\right \}
\Rightarrow S = 5\left \{ 49\right \}
\Rightarrow S =245
Therefore, the sum of AP \small 2,7,12,..., to \small 10 terms is 245

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

Answer:

Given AP is
\small -37,-33,-29,..., to \small 12 terms.
Here, a = -37 \ and \ n = 12
And
d = a_2-a_1=-33-(-37)=4
Now, we know that
S = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S = \frac{12}{2}\left \{ 2\times (-37) +(12-1)4\right \}
\Rightarrow S = 6\left \{ -74 +44\right \}
\Rightarrow S = 5\left \{ -30\right \}
\Rightarrow S =-180
Therefore, the sum of AP \small -37,-33,-29,..., to \small 12 terms. is -180

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

Answer:

Given AP is
\small 0.6,1.7,2.8,..., to \small 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
S = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S = \frac{100}{2}\left \{ 2\times (0.6) +(100-1)(1.1)\right \}
\Rightarrow S = 50\left \{ 1.2 +108.9\right \}
\Rightarrow S = 50\left \{ 110.1\right \}
\Rightarrow S =5505
Therefore, the sum of AP \small 0.6,1.7,2.8,..., to \small 100 terms. is 5505

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

Answer:

Given AP is
\small \frac{1}{15},\frac{1}{12},\frac{1}{10},..., to \small 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
S = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S = \frac{11}{2}\left \{ 2\times \frac{1}{15} +(11-1)(\frac{1}{60})\right \}
\Rightarrow S = \frac{11}{2}\left \{ \frac{2}{15} +\frac{1}{6}\right \}
\Rightarrow S = \frac{11}{2}\left \{ \frac{9}{30}\right \}
\Rightarrow S =\frac{99}{60}= \frac{33}{20}
Therefore, the sum of AP \small \frac{1}{15},\frac{1}{12},\frac{1}{10},..., to \small 11 terms. is \frac{33}{20}


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

Answer:

Given AP is
\small 7+10\frac{1}{2}+14+...+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
a_n = a+(n-1)d
\Rightarrow 84 = 7 + (n-1)\frac{7}{2}
\Rightarrow \frac{7n}{2}= 77+\frac{7}{2}
\Rightarrow n = 23
Now, we know that
S = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S = \frac{23}{2}\left \{ 2\times7 +(23-1)(\frac{7}{2})\right \}
\Rightarrow S = \frac{23}{2}\left \{ 14 +77\right \}
\Rightarrow S = \frac{23}{2}\left \{ 91\right \}
\Rightarrow S =\frac{2093}{2}=1046\frac{1}{2}
Therefore, the sum of AP \small 7+10\frac{1}{2}+14+...+84 is 1046\frac{1}{2}

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

Answer:

Given AP is
\small 34+32+30+...+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
a_n = a+(n-1)d
\Rightarrow 10 = 34 + (n-1)(-2)
\Rightarrow -26 = -2n
\Rightarrow n = 13
Now, we know that
S = \frac{n}{2}\left \{ a+a_n \right \}

\Rightarrow S = \frac{13}{2}\left \{ 44\right \}
\Rightarrow S =13\times 22 = 286
Therefore, the sum of AP \small 34+32+30+...+10 is 286

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

Answer:

Given AP is
\small -5+(-8)+(-11)+...+(-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
a_n = a+(n-1)d
\Rightarrow -230 = -5 + (n-1)(-3)
\Rightarrow -228 = -3n
\Rightarrow n = 76
Now, we know that
S = \frac{n}{2}\left \{ a+a_n \right \}
\Rightarrow S = \frac{76}{2}\left \{ (-5-230 )\right \}

\Rightarrow S = 38\left \{ -235\right \}
\Rightarrow S = -8930
Therefore, the sum of AP \small -5+(-8)+(-11)+...+(-230) is -8930

Q3 (i) In an AP: given \small a=5 , \small d=3 , \small a_n=50 , find \small n and \small 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
a_n = a+(n-1)d
\Rightarrow 50 = 5 + (n-1)3
\Rightarrow 48 = 3n
\Rightarrow n = 16
Now, we know that
S = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S = \frac{16}{2}\left \{ 2\times(5) +(16-1)(3)\right \}
\Rightarrow S = 8\left \{ 10+45\right \}
\Rightarrow S = 8\left \{ 55\right \}
\Rightarrow S =440
Therefore, the sum of the given AP is 440

Q3 (ii) In an AP: given \small a=7 , \small a_1_3=35 , find \small d and \small S_1_3 .

Answer:

It is given that
a = 7 \ and \ a_{13} = 35
a_{13}= a+12d = 35
= 12d = 35-7 = 28
d = \frac{28}{12}= \frac{7}{3}
Now, we know that
S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S_{13} = \frac{13}{2}\left \{ 2\times(7) +(13-1)(\frac{7}{3})\right \}
\Rightarrow S_{13} = \frac{13}{2}\left \{14 +28\right \}
\Rightarrow S_{13} = \frac{13}{2}\left \{42\right \}
\Rightarrow S_{13} = 13 \times 21 = 273
Therefore, the sum of given AP is 273

Q3 (iii) In an AP: given \small a_1_2=37,d=3, find \small a and \small S_1_2 .

Answer:

It is given that
d = 3 \ and \ a_{12} = 37
a_{12}= a+11d = 37
= a= 37-11\times 3 = 37-33=4
Now, we know that
S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S_{12} = \frac{12}{2}\left \{ 2\times(4) +(12-1)3\right \}
\Rightarrow S_{12} = 6\left \{ 8+33\right \}
\Rightarrow S_{12} = 6\left \{41\right \}
\Rightarrow S_{12} =246
Therefore, the sum of given AP is 246

Q3 (iv) In an AP: given \small a_3=15, S_1_0=125, find \small d and \small S_1_0
Answer:

It is given that
\small a_3=15, S_1_0=125
a_{3}= a+2d = 15 \ \ \ \ \ \ \ \ -(i)
Now, we know that
S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S_{10} = \frac{10}{2}\left \{ 2\times(a) +(10-1)d\right \}
\Rightarrow 125 = 5\left \{ 2a+9d\right \}
\Rightarrow 2a+9d = 25 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(ii)
On solving equation (i) and (ii) we will get
a= 17 \ and \ d = -1
Now,
a_{10} = a+ 9d = 17 + 9(-1)= 17-9 = 8
Therefore, the value of d and 10th terms is -1 and 8 respectively

Q3 (v) In an AP: given \small d=5, S_9=75 , find \small a and \small a_9 .

Answer:

It is given that
\small d=5, S_9=75
Now, we know that
S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S_{9} = \frac{9}{2}\left \{ 2\times(a) +(9-1)5\right \}
\Rightarrow 75= \frac{9}{2}\left \{ 2a +40\right \}
\Rightarrow 150= 18a+360
\Rightarrow a = -\frac{210}{18}=-\frac{35}{3}
Now,
a_{9} = a+ 8d = -\frac{35}{3} + 8(5)= -\frac{35}{3}+40 = \frac{-35+120}{3}= \frac{85}{3}

Q3 (vi) In an AP: given \small a=2,d=8,S_n=90, find \small n and \small a_n .

Answer:

It is given that
\small a=2,d=8,S_n=90,
Now, we know that
S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow 90 = \frac{n}{2}\left \{ 2\times(2) +(n-1)8\right \}
\Rightarrow 180 = n\left \{ 4+8n-8\right \}
\Rightarrow 8n^2-4n-180=0
\Rightarrow 4(2n^2-n-45)=0
\Rightarrow 2n^2-n-45=0
\Rightarrow 2n^2-10n+9n-45=0
\Rightarrow (n-5)(2n+9)=0
\Rightarrow n = 5 \ \ and \ \ n = - \frac{9}{2}
n can not be negative so the only the value of n is 5
Now,
a_{5} = a+ 4d = 2+4\times 8 = 2+32 = 34
Therefore, value of n and nth term is 5 and 34 respectively

Q3 (vii) In an AP: given \small a=8,a_n=62,S_n=210, find \small n and \small d .

Answer:

It is given that
\small a=8,a_n=62,S_n=210,
Now, we know that
a_n = a+(n-1)d
62 = 8+(n-1)d
(n-1)d= 54 \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)

Now, we know that
S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow 210 = \frac{n}{2}\left \{ 2\times(8) +(n-1)d\right \}
\Rightarrow 420 = n\left \{ 16+54 \right \} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))
\Rightarrow 420 = n\left \{ 70 \right \}
\Rightarrow n = 6
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 \small a_n=4,d=2,S_n=-14, find \small n and \small a .

Answer:

It is given that
\small a_n=4,d=2,S_n=-14,
Now, we know that
a_n = a+(n-1)d
4 = a+(n-1)2
a+2n = 6\Rightarrow a = 6-2n \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)

Now, we know that
S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow -14 = \frac{n}{2}\left \{ 2\times(a) +(n-1)2\right \}
\Rightarrow -28 = n\left \{ 2(6-2n)+2n-2 \right \} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))
\Rightarrow -28 = n\left \{ 10-2n \right \}
\Rightarrow 2n^2-10n-28=0
\Rightarrow 2(n^2-5n-14)=0
\Rightarrow n^2-7n+2n-14=0
\Rightarrow(n+2)(n-7)=0
\Rightarrow n = -2 \ \ and \ \ n = 7
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 \small a=3,n=8,S=192, find \small d .

Answer:

It is given that
\small a=3,n=8,S=192,
Now, we know that
S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow 192 = \frac{8}{2}\left \{ 2\times(3) +(8-1)d\right \}
\Rightarrow 192=4\left \{6 +7d\right \}
\Rightarrow 7d = 48-6
\Rightarrow d = \frac{42}{7} = 6
Therefore, the value of d is 6

Q3 (x) In an AP: given \small l=28,S=144 \ and \ n = 9 and there are total \small 9 terms. Find \small a .

Answer:

It is given that
\small l=28,S=144 \ and \ n = 9
Now, we know that
l = a_n = a+(n-1)d
28 = a_n = a+(n-1)d \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)
Now, we know that
S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow 144 = \frac{9}{2}\left \{ a + a +(n-1)d\right \}
\Rightarrow 288 =9\left \{ a+ 28\right \} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(using \ (i))
\Rightarrow a+28= 32
\Rightarrow a=4
Therefore, the value of a is 4

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

Answer:

Given AP is
\small 9,17,25,...
Here, a =9 \ and \ d = 8
And S_n = 636
Now , we know that
S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow 636 = \frac{n}{2}\left \{ 18+(n-1)8 \right \}
\Rightarrow 1272 = n\left \{ 10+8n \right \}
\Rightarrow 8n^2+10n-1272=0
\Rightarrow 2(4n^2+5n-636)=0
\Rightarrow 4n^2+53n-48n-636=0
\Rightarrow (n-12)(4n+53)=0
\Rightarrow n = 12 \ \ and \ \ n = - \frac{53}{4}
Value of n can not be negative so the only the value of n is 12
Therefore, the sum of 12 terms of AP \small 9,17,25,... must be taken to give a sum of \small 636 .

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

Answer:

It is given that
\small a=5,a_n=45,S_n=400,
Now, we know that
a_n = a+(n-1)d
45 = 5+(n-1)d
(n-1)d= 40 \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)

Now, we know that
S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow 400 = \frac{n}{2}\left \{ 2\times(5) +(n-1)d\right \}
\Rightarrow 800 = n\left \{ 10+40 \right \} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))
\Rightarrow 800 = n\left \{ 50 \right \}
\Rightarrow n = 16
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 \small 17 and \small 350 respectively. If the common difference is \small 9 , how many terms are there and what is their sum?

Answer:

It is given that
\small a=17,l=350,d=9,
Now, we know that
a_n = a+(n-1)d
350 = 17+(n-1)9
(n-1)9 = 333
(n-1)=37
n = 38

Now, we know that
S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S_{38}= \frac{38}{2}\left \{ 2\times(17) +(38-1)9\right \}
\Rightarrow S_{38}= 19\left \{ 34 +333\right \}
\Rightarrow S_{38}= 19\left \{367\right \}
\Rightarrow S_{38}= 6973
Therefore, there are 38 terms and their sun is 6973

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

Answer:

It is given that
\small a_{22}=149,d=7,n = 22
Now, we know that
a_{22} = a+21d
149 = a+21\times 7
a = 149 - 147 = 2
Now, we know that
S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S_{22}= \frac{22}{2}\left \{ 2\times(2) +(22-1)7\right \}
\Rightarrow S_{22}= 11\left \{ 4 +147\right \}
\Rightarrow S_{22}= 11\left \{ 151\right \}
\Rightarrow S_{22}= 1661
Therefore, there are 22 terms and their sum is 1661

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

Answer:

It is given that
\small a_{2}=14,a_3=18,n = 51
And d= a_3-a_2= 18-14=4
Now,
a_2 = a+d
a= 14-4 = 10
Now, we know that
S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S_{51}= \frac{51}{2}\left \{ 2\times(10) +(51-1)4\right \}
\Rightarrow S_{51}= \frac{51}{2}\left \{ 20 +200\right \}
\Rightarrow S_{51}= \frac{51}{2}\left \{ 220\right \}
\Rightarrow S_{51}= 51 \times 110
\Rightarrow S_{51}=5304
Therefore, there are 51 terms and their sum is 5610

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

Answer:

It is given that
S_7 = 49 \ and \ S_{17}= 289
Now, we know that
S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S_{7}= \frac{7}{2}\left \{ 2\times(a) +(7-1)d\right \}
\Rightarrow 98= 7\left \{ 2a +6d\right \}
\Rightarrow a +3d = 7 \ \ \ \ \ \ \ -(i)
Similarly,
\Rightarrow S_{17}= \frac{17}{2}\left \{ 2\times(a) +(17-1)d\right \}
\Rightarrow 578= 17\left \{ 2a +16d\right \}
\Rightarrow a +8d = 17 \ \ \ \ \ \ \ -(ii)
On solving equation (i) and (ii) we will get
a = 1 and d = 2
Now, the sum of first n terms is
S_n = \frac{n}{2}\left \{ 2\times 1 +(n-1)2 \right \}
S_n = \frac{n}{2}\left \{ 2 +2n-2 \right \}
S_n = n^2
Therefore, the sum of n terms is n^2

Q10 (i) Show that \small a_1,a_2,...,a_n,... form an AP where an is defined as below : \small a_n=3+4n Also find the sum of the first \small 15 terms.

Answer:

It is given that
\small a_n=3+4n
We will check values of a_n for different values of n
a_1 = 3+4(1) =3+4= 7
a_2 = 3+4(2) =3+8= 11
a_3 = 3+4(3) =3+12= 15
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
S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S_{15}= \frac{15}{2}\left \{ 2\times(7) +(15-1)4\right \}
\Rightarrow S_{15}= \frac{15}{2}\left \{ 14 +56\right \}
\Rightarrow S_{15}= \frac{15}{2}\left \{ 70\right \}
\Rightarrow S_{15}= 15 \times 35
\Rightarrow S_{15}= 525
Therefore, the sum of 15 terms is 525

Q10 (ii) Show that \small a_1,a _2,...,a_n,... form an AP where an is defined as below : \small a_n=9-5n . Also find the sum of the first \small 15 terms in each case.

Answer:

It is given that
\small a_n=9-5n
We will check values of a_n for different values of n
a_1 = 9-5(1) =9-5= 4
a_2 = 9-5(2) =9-10= -1
a_3 = 9-5(3) =9-15= -6
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
S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S_{15}= \frac{15}{2}\left \{ 2\times(4) +(15-1)(-5)\right \}
\Rightarrow S_{15}= \frac{15}{2}\left \{ 8 -70\right \}
\Rightarrow S_{15}= \frac{15}{2}\left \{ -62\right \}
\Rightarrow S_{15}= 15 \times (-31)
\Rightarrow S_{15}= -465
Therefore, the sum of 15 terms is -465

Q11 If the sum of the first \small n terms of an AP is \small 4n-n^2 , what is the first term (that is \small S_1 )? What is the sum of first two terms? What is the second term? Similarly, find the \small 3 rd, the \small 10 th and the \small n th terms

Answer:

It is given that
the sum of the first \small n terms of an AP is \small 4n-n^2
Now,
\Rightarrow S_n = 4n-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
\Rightarrow S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S_2 = \frac{2}{2}\left \{ 2\times 3+(2-1)d \right \}
\Rightarrow 4 = \left \{6+d \right \}
\Rightarrow d = -2
Now,
a_2= a+d = 3+(-2 )= 1
Similarly,
a_3= a+2d = 3+2(-2 )= 3-4=-1
a_{10}= a+9d = 3+9(-2 )= 3-18=-15
a_{n}= a+(n-1)d = 3+(n-1)(-2 )= 5-2n

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

Answer:

Positive integers divisible by 6 are
6,12,18,...
This is an AP with
here, \ a = 6 \ and \ d = 6
Now, we know that
S_n= \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S_{40}= \frac{40}{2}\left \{ 2\times 6+(40-1)6 \right \}
\Rightarrow S_{40}= 20\left \{12+234 \right \}
\Rightarrow S_{40}= 20\left \{246 \right \}
\Rightarrow S_{40}= 4920
Therefore, sum of the first \small 40 positive integers divisible by \small 6 is 4920

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

Answer:

First 15 multiples of 8 are
8,16,24,...
This is an AP with
here, \ a = 8 \ and \ d = 8
Now, we know that
S_n= \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S_{15}= \frac{15}{2}\left \{ 2\times 8+(15-1)8 \right \}
\Rightarrow S_{15}= \frac{15}{2}\left \{ 16+112 \right \}
\Rightarrow S_{15}= \frac{15}{2}\left \{ 128 \right \}
\Rightarrow S_{15}= 15 \times 64 = 960
Therefore, sum of the first 15 multiple of 8 is 960

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

Answer:

The odd number between 0 and 50 are
1,3,5,...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
S_n= \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S_{25}= \frac{25}{2}\left \{ 2\times 1+(25-1)2 \right \}
\Rightarrow S_{25}= \frac{25}{2}\left \{ 2+48 \right \}
\Rightarrow S_{25}= \frac{25}{2}\times 50
\Rightarrow S_{25}= 25 \times 25 = 625
Therefore, sum of the odd numbers between \small 0 and \small 50 625

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

Answer:

It is given that
Penalty for delay of completion beyond a certain date is Rs \small 200 for the first day, Rs \small 250 for the second day, Rs \small 300 for the third day and penalty for each succeeding day being Rs \small 50 more than for the preceding day
We can clearly see that
200,250,300,..... is an AP with
a = 200 \ and \ d = 50
Now, the penalty for 30 days is given by the expression
S_{30}= \frac{30}{2}\left \{ 2\times 200+(30-1)50 \right \}
S_{30}= 15\left ( 400+1450 \right )
S_{30}= 15 \times 1850
S_{30}= 27750
Therefore, the penalty for 30 days is 27750

Q16 A sum of Rs \small 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 \small 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 1st student
Then,
S_7 = \frac{7}{2}\left \{ 2a+(7-1)(-20) \right \}
700 = \frac{7}{2}\left \{ 2a-120 \right \}
2a - 120 = 200
a = \frac{320}{2}= 160
Therefore, the prize given to the first student is Rs 160
Now,
Let a_2,a_2,...,a_7 is the prize money given to the next 6 students
then,
a_2 = a+d = 160+(-20)=160-20=140
a_3 = a+2d = 160+2(-20)=160-40=120
a_4 = a+3d = 160+3(-20)=160-60=100
a_5 = a+4d = 160+4(-20)=160-80=80
a_6 = a+5d = 160+5(-20)=160-100=60
a_7 = a+6d = 160+6(-20)=160-120=40
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 \small 1 tree, a section of Class II will plant \small 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
S_{12}= \frac{12}{2}\left \{ 2\times 3+(12-1)\times 3 \right \}
S_{12}= 6\left ( 6+33 \right )
S_{12}= 6 \times 39 = 234
Therefore, number of trees planted by 12 classes is 234

Q18 A spiral is made up of successive semicircles, with centres alternately at \small A and \small B ??????, starting with centre at \small A , of radii \small 0.5\hspace {1mm}cm,1.0\hspace {1mm}cm,1.5\hspace {1mm}cm,2.0\hspace {1mm}cm,... as shown in Fig. \small 5.4 . What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take \pi =\frac{22}{7} )

1635921529757

[ Hint : Length of successive semicircles is \small l_1,l_2,l_3,l_4,... with centres at \small A,B,A,B,..., respectively.]

Answer:

From the above-given figure

Circumference of 1st semicircle l_1 = \pi r_1 = 0.5\pi

Similarly,

Circumference of 2nd semicircle l_2 = \pi r_2 = \pi

Circumference of 3rd semicircle l_3 = \pi r_3 = 1.5\pi

It is clear that this is an AP with a = 0.5\pi \ and \ d = 0.5\pi

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

S_{13} = \frac{13}{2}\left \{ 2\times 0.5\pi + (13-1)0.5\pi\right \}
S_{13} = \frac{13}{2}\left ( \pi+6\pi \right )
S_{13} = \frac{13}{2}\times 7\pi
S_{13} = \frac{91\pi}{2} = \frac{91}{2}\times \frac{22}{7}=143
Therefore, sum of length of 13 such semicircles is 143 cm

Q19 \small 200 logs are stacked in the following manner: \small 20 logs in the bottom row, \small 19 in the next row, \small 18 in the row next to it and so on (see Fig. \small 5.5 ). In how many rows are the \small 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, ..., is an AP.
and here a = 20 \ and \ d = -1
Let suppose 200 logs are arranged in 'n' rows,
Then,
S_n = \frac{n}{2}\left \{ 2\times 20 +(n-1)(-1) \right \}
200 = \frac{n}{2}\left \{ 41-n \right \}
\Rightarrow n^2-41n +400 = 0
\Rightarrow n^2-16n-25n +400 = 0
\Rightarrow (n-16)(n-25) = 0
\Rightarrow n = 16 \ \ and \ \ n = 25
Now,
case (i) n = 25
a_{25} =a+24d = 20+24\times (-1)= 20-24=-4
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+15d = 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 \small 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. \small 5.6 ).

1635921559600

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 \small 2\times5+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
S_{10}= \frac{10}{2}\left \{ 2\times 10+(10-1)6 \right \}
S_{10}= 5\left ( 20+54 \right )
S_{10}= 5\times 74 = 370

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



NCERT solutions for class 10 maths chapter 5 Arithmetic Progressions Excercise: 5.4

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

Answer:

Given AP is
\small 121,117,113,...,
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 nth term is negative then a_n < 0
\Rightarrow 121+(n-1)(-4) < 0
\Rightarrow 125<4n
\Rightarrow n > \frac{125}{4}=31.25
Therefore, first negative term must be 32nd term

Q2 The sum of the third and the seventh terms of an AP is \small 6 and their product is \small 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 \small 8
a_3= a+ 2d
a_7= a+ 6d
Now,
a_3+a_7= a+ 2d+a+6d= 6
\Rightarrow 2a+8d = 6
\Rightarrow a+4d = 3 \Rightarrow a = 3-4d \ \ \ \ \ \ \ \ \ \ \ \ -(i)
And
a_3.a_7 = (a+2d).(a+6d)=a^2+8ad +12d^2 = 8 \ \ \ \ \ \ \ -(ii)
put value from equation (i) in (ii) we will get
\Rightarrow (3-4d)^2+8(3-4d)d+12d^2= 8
\Rightarrow 9+16d^2-24d+24d-32d^2+12d^2=8
\Rightarrow 4d^2 = 1
\Rightarrow d = \pm \frac{1}{2}
Now,
case (i) d = \frac{1}{2}

a= 3 - 4 \times \frac{1}{2} = 1
Then,
S_{16}=\frac{16}{2}\left \{ 2\times 1+(16-1)\frac{1}{2} \right \}

S_{16}=76

case (ii) d = -\frac{1}{2}
a= 3 - 4 \times \left ( -\frac{1}{2} \right ) = 5

Then,
S_{16}=\frac{16}{2}\left \{ 2\times 1+(16-1)\left ( -\frac{1}{2} \right ) \right \}

S_{16}=20

Q3 A ladder has rungs \small 25 cm apart. (see Fig. \small 5.7 ). The rungs decrease uniformly in length from \small 45 cm at the bottom to \small 25 cm at the top. If the top and the bottom rungs are \small 2\frac{1}{2} m apart, what is the length of the wood required for the rungs? [ Hint: Number of rungs =\frac{250}{25}+1]

1635921645173


Answer:

It is given that
The total distance between the top and bottom rung = 2\frac{1}{2}\ m = 250cm
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+ 10d

\Rightarrow 45=25+10d
\Rightarrow d = 2
Now, total length of the wood required for the rungs is equal to
S_{11} = \frac{11}{2}\left \{ 2\times 25+(11-1)2 \right \}
S_{11} = \frac{11}{2}\left \{ 50+20 \right \}
S_{11} = \frac{11}{2}\times 70
S_{11} =385 \ cm
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 \small 1 to \small 49 . Show that there is a value of \small x such that the sum of the numbers of the houses preceding the house numbered \small x is equal to the sum of the numbers of the houses following it. Find this value of \small x . [ Hint : \small S_{x-1}=S_4_9-S_x ]

Answer:

It is given that the sum of the numbers of the houses preceding the house numbered \small x is equal to the sum of the numbers of the houses following it
And 1,2,3,.....,49 form an AP with a = 1 and d = 1
Now, we know that
S_n= \frac{n}{2}\left \{ 2a+(n-1)d \right \}
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}\left \{ 2a+((n-1)-1)d \right \}=\frac{49}{2}\left \{ 2a+(49-1)d \right \}-\frac{n}{2}\left \{ 2a+(n-1)d \right \}

\frac{n-1}{2}\left \{ 2+(n-2) \right \}=\frac{49}{2}\left \{ 2+48 \right \}-\frac{n}{2}\left \{ 2+(n-1) \right \}

\frac{n-1}{2}\left \{ n\right \}=\frac{49}{2}\left \{ 50 \right \}-\frac{n}{2}\left \{ n+1 \right \}

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

n^2 = 1225
n = \pm 35

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

Q5 A small terrace at a football ground comprises of \small 15 steps each of which is \small 50 m long and built of solid concrete. Each step has a rise of \small \frac{1}{4}\: m and a tread of \small \frac{1}{2}\: m . (see Fig. \small 5.8 ). Calculate the total volume of concrete required to build the terrace.
[ Hint: Volume of concrete required to build the first step \small =\frac{1}{4}\times \frac{1}{2}\times 50\: m^3 ]

1635921665136


Answer:

It is given that
football ground comprises of \small 15 steps each of which is \small 50 m long and Each step has a rise of \small \frac{1}{4}\: m and a tread of \small \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 = \left ( \frac{1}{4}+\frac{1}{4}\right )\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 3rd step = \left ( \frac{1}{4}+\frac{1}{4}+\frac{1}{4}\right )\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
S_{15} =\frac{15}{2}\left \{ 2 \times 6.25 +(15-1)6.25 \right \}
S_{15} =\frac{15}{2}\left \{ 12.5 +87.5\right \}
S_{15} =\frac{15}{2}\times 100
S_{15} =15\times 50 = 750
Therefore, the total volume of concrete required to build the terrace of 15 such steps is 750 \ m^3

Arithmetic Progression Class 10 Maths chapter 5 - Topics

  • Checking whether the series is an arithmetic progression or not.

  • To find the nth term or last term or an arithmetic progression.

  • Questions based on the sum of an arithmetic progression

  • Use of linear equation concept in an arithmetic progression

  • Application of nth term and sum formula

NCERT Class 10 Arithmetic Progression Exercise-Wise Solutions

Features of NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

  • The NCERT solutions cover all the questions given in the exercise and in between the chapters. This makes it really helpful in homework as well as in the board exam preparations.

  • The Arithmetic Progression Class 10 Maths Chapter 5 NCERT solutions offer a step by step description of the solutions.

  • These NCERT solutions for Class 10 Maths ch 5 Arithmetic Progressions is given in a simple language and are easy to understand.

  • For NCERT Solutions for Class 10 Maths Chapter 5 PDF download, you can save the webpage of this article.

NCERT Books and NCERT Syllabus

NCERT Solutions for Class 10 Maths - Chapters Wise

How to use NCERT solutions for class 10 maths chapter 5 Arithmetic Progressions?

  • Before practising the NCERT chapter 5 Maths Class 10 solutions, go through the conceptual theory given in the NCERT textbook.

  • After going through the conceptual part, jump on to practice exercises available.

  • While solving the exercises if you are facing problems in any specific question, then take the help of NCERT solutions for Class 10 Maths chapter 5 Arithmetic Progressions.

  • Once you have done the practice exercises you can move to previous year questions.

Frequently Asked Question (FAQs)

1. What are the important topics of NCERT class 10 maths chapter 5 arithmetic progressions?

Checking whether the series is an arithmetic progression or not, nth term or last term of arithmetic progression, sum of an arithmetic progression, use of linear equation concept in an arithmetic progression and application of nth term and sum of a.p. formulae are the important topics in this chapter.

2. What kind of questions are there in NCERT solutions for class 10 chapter 5 maths?

The NCERT Solutions for Class 10 Maths Chapter 5 includes a range of question types such as multiple choice, long and short answer, fill in the blank, and real-world examples. These questions aim to enhance students' problem-solving and time management abilities, thus preparing them for success in their CBSE exams.

3. How many exercises are there in maths chapter 5 class 10?

The number of exercises in Maths Chapter 5 for Class 10 are four. These are very important because these include all types of questions on which repeatedly questions are being asked in board exams. Students can practice and find solutions at careers360 official website

4. Do I need to practice all questions given in NCERT class 10 maths chapter 5 solutions?

 It is generally a good idea to practice all of the questions provided in the NCERT solutions for class 10 maths arithmetic progression, as they are designed to cover the key concepts and skills that are important for understanding the chapter and doing well in your exams. By practicing all the questions and arithmetic progression solutions, you will be able to identify your strengths and weaknesses and work on them accordingly. Also, chapter 5 questions and ap solutions class 10 help you build your problem-solving skills, time management skills and confidence, which will be beneficial for you in your exams. Students can also use arithmetic progression class 10 pdf as to study offline.

5. Why should I practice NCERT maths class 10 chapter 5 solutions?

There are several reasons why you should practice the class 10 maths NCERT solution arithmetic progression. Practising the NCERT Solutions for Class 10 Maths Chapter 5 is a valuable step in your studies as it will help you to understand the concepts, improve your problem-solving skills, and prepare for your exams. Students can use NCERT solutions for class 10 maths arithmetic progression pdf to study comfortable this chapter.

6. What is the arithmetic progression in class 10 chapter 5 maths?

Arithmetic Progression (AP) is a sequence of numbers in which the difference between any two consecutive terms is always the same. In other words, it is a sequence of numbers in which the difference between two consecutive terms is a constant. Arithmetic progression class 10 NCERT solutions pdf covers various concepts related to arithmetic progressions such as finding the common difference, the nth term, the sum of an arithmetic series, and solving word problems using arithmetic progressions.

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Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?

Here are some options you can explore to get admission in a good school even though admissions might be closed for many:

  • Waitlist: Many schools maintain waitlists after their initial application rounds close.  If a student who secured a seat decides not to join, the school might reach out to students on the waitlist.  So, even if the application deadline has passed,  it might be worth inquiring with schools you're interested in if they have a waitlist and if they would consider adding you to it.

  • Schools with ongoing admissions: Some schools, due to transfers or other reasons, might still have seats available even after the main admission rush.  Reach out to the schools directly to see if they have any open seats in 10th grade.

  • Consider other good schools: There might be other schools in your area that have a good reputation but weren't on your initial list. Research these schools and see if they have any seats available.

  • Focus on excelling in your current school: If you can't find a new school this year, focus on doing well in your current school. Maintain good grades and get involved in extracurricular activities. This will strengthen your application for next year if you decide to try transferring again.


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Dear aspirant !

Hope you are doing well ! The class 10 Hindi mp board sample paper can be found on the given link below . Set your time and give your best in the sample paper it will lead to great results ,many students are already doing so .

https://www.google.com/url?sa=t&source=web&rct=j&opi=89978449&url=https://school.careers360.com/download/sample-papers/mp-board-10th-hindi-model-paper&ved=2ahUKEwjO3YvJu5KEAxWAR2wGHSLpAiQQFnoECBMQAQ&usg=AOvVaw2qFFjVeuiZZJsx0b35oL1x .

Hope you get it !

Thanking you

Hello aspirant,

The dates of the CBSE Class 10th and Class 12 exams are February 15–March 13, 2024 and February 15–April 2, 2024, respectively. You may obtain the CBSE exam date sheet 2024 PDF from the official CBSE website, cbse.gov.in.

To get the complete datesheet, you can visit our website by clicking on the link given below.

https://school.careers360.com/boards/cbse/cbse-date-sheet

Thank you

Hope this information helps you.

Hello aspirant,

The paper of class 7th is not issued by respective boards so I can not find it on the board's website. You should definitely try to search for it from the website of your school and can also take advise of your seniors for the same.

You don't need to worry. The class 7th paper will be simple and made by your own school teachers.

Thank you

Hope it helps you.

The eligibility age criteria for class 10th CBSE is 14 years of age. Since your son will be 15 years of age in 2024, he will be eligible to give the exam.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

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

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

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 .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

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.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

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

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

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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