These NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.3

Question 1.

Find the sum of the following APs:

(i) 2, 7, 12,…… to 10 terms.

(ii) -37, -33, -29, …… to 12 terms.

(iii) 0.6, 1.7, 2.8, ……, to 100 terms.

(iv) \(\frac { 1 }{ 15 }\), \(\frac { 1 }{ 12 }\), \(\frac { 1 }{ 10 }\), …….., to 11 terms.

Solution:

(i) We have

2, 7, 12, … to 10 terms

Here, a = 2, d = 7 – 2 = 5 and n = 10

Therefore, the sum of 10 terms of the AP 2, 7, 12, ….. is 245

(ii) We have,

– 37, – 33, – 29, … to 12 terms

Here, a = – 37, d = – 33 – (- 37) = 4 and n = 12

Therefore, the sum of 12 terms of the AP – 37, – 33, – 29,… is – 180.

(iii) We have,

0.6,1.7, 2.8,… to 100 terms

Here, a = 0.6, d = 1.7 – 0.6 = 1.1 and n = 100

Therefore, the sum of 100 terms of the AP 0.6, 1.7,2.8,. .. is 5505.

(iv) We have,

Therefore, the sum of 11 terms of the AP \(\frac { 1 }{ 15 }\), \(\frac { 1 }{ 12 }\), \(\frac { 1 }{ 10 }\), ….. is \(\frac { 33 }{ 20 }\).

Question 2.

Find the sums given below:

(i) 7 + 10\(\frac { 1 }{ 2 }\) + 14 + … + 84

(ii) 34 + 32 + 30 + … + 10

(iii) -5 + (-8) + (-11) + ….. + (-230)

Solution:

(i) We have

7 + 10\(\frac { 1 }{ 2 }\) + 14 + … + 84

Therefore 7 + 10\(\frac { 1 }{ 2 }\) + 14 + … + 84 = 1046\(\frac { 1 }{ 2 }\)

(ii) We have

34 + 32 + 30 + … + 10

Therefore, 34 + 32 + 30 + … + 10 = 286

(iii) We have

-5 + (-8) + (-11) + ….. + (-230)

Therefore, -5 + (-8) + (-11) + ….. + (-230) is – 8390.

Question 3.

In an AP:

(i) given a = 5, d = 3, a_{n} = 50, find n and S_{n}.

(ii) given a = 7, a_{13} = 35, find d and S_{13}.

(iii) given a_{12} = 37, d = 3, find a and S_{12}.

(iv) given a_{3} = -15, S_{10} = 125, find d and a_{10}.

(v) given d = 5, S_{9} = 75, find a and a_{9}.

(vi) given a = 2, d = 8, S_{n} = 90, find n and a_{n}.

(vii) given a = 8, a_{n} = 62, S_{n} = 210, find n and d.

(viii) given a_{n} = 4, d = 2, S_{n} = -14, find n and a.

(ix) given a = 3, n = 8, S = 192, find d.

(x) given l = 28, S = 144, and there are total 9 terms. Find a.

Solution:

(i) We have

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

But, We know that

Therefore, the value of n = 16 and S_{n} = 440.

(ii) We have a = 7, a_{13} = 35

But, We know that

a_{13} = a + 12d

or 35 = 7 + 12 d

or d = \(\frac { 28 }{ 12 }\) = \(\frac { 7 }{ 3 }\)

Therefore, the value of d = \(\frac { 7 }{ 3 }\) and S_{13} = 273.

(iii) We have a_{12} = 37, d = 3

But, We know that

a_{12} = a + 11d

or 37 = a + 11 x 3

or 37 = a ÷ 33

or a = 37 – 33 = 4

Again we know that

Therefore, the value of a = 4 and S_{12} = 246.

(iv) We have a_{3} = -15, S_{10} = 125

But, We know that

Multiplying equation (i) by 10 then subtracting from equation (ii), we get,

Putting the value of d = – 1 in equation, (i) we get,

a + 2(-1) = 15

a – 2 = 15 ⇒ a = 17

Now, a = a+ (n – 1)d

a_{10} = 17 + (10 – 1) (- 1)

a_{10} = 17 – 9 = 8

Therefore, d = – 1 and a_{10} = 8

(v) We have d = 5, S_{9} = 75

But, We know that

Therefore, the value of a = \(\frac { -35 }{ 3 }\) and a_{9} = \(\frac { 85 }{ 3 }\).

(vi) We have a = 2, d = 8, S_{n} = 90

But, We know that

But number of terms cannot be in negative

∴ n = \(\frac { -9 }{ 2 }\) is neglected.

So, n = 5

Again we know that

a_{n} = a + (n – 1)d

∴ a_{5} = 2 + (5 – 1) x 8

= 2 + 4 x 8

or a_{5} = 2 + 32 = 34

Therefore, the value of n = 5 and a_{5} = 34.

(vii) We have a = 8, a_{n} = 62, S_{n} = 210

But, We know that

a_{n} = a + (n – 1)d

62 = 8 + (n – 1) d

∴ (n – 1)d = 62 – 8 = 54

or a_{5} = 2 + 32 = 34 … (i)

Again

Therefore, the value of n = 6 and d = \(\frac { 54 }{ 5 }\).

(viii) We have a_{n} = 4, d = 2, S_{n} = – 14

But, We know that

a_{n} = a + (n – 1)d

or 4 = a+ (n – 1)2

or 4 = a + 2n – 2

∴ a + 2n = 6

∴ a = 6 – 2n … (i)

Again, we know that

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

or – 14 = \(\frac { n }{ 2 }\)[2 x a + (n – 1) x 2]

or – 14 x 2 = n (2a + 2n – 2)

or – 28 = n[2(6 – 2n) + 2n – 2]

[Putting the value of a from equation (i)]

or – 28 = n (12 – 4n + 2n – 2)

or – 28 = n (- 2n +10)

or – 28 = – 2n² + 10n

or 2n² – 10n – 28 = 0

or n² – 5n – 14 = 0

or n² – 7n + 2n – 14 = 0

or n (n – 7) + 2 (n – 7) = 0

or (n + 2) (n – 7) = 0

∴ n = 7 and n= -2

But, number of terms cannot be in negative

∴ n = – 2 is neglected.

So, n = 7

Putting the value of n in equation (i), we get,

a = 6 – 2n = 6 – 2 x 7

∴ a = 6 – 14 = – 8

Therefore, the value of a = – 8 and n = 7.

(ix) We have a = 3, n = 8, S = 192

But, We know that

Therefore, the value of d = 6.

(x) We have l = 28, S = 144, and n = 9

But, We know that

a_{n} = l

or a + (n – 1)d = 28

or a + (9 – 1)d = 28

or a + 8d = 28 … (i)

Again we know that

Multiply equation (i) by 2 then subtract equation (ii) from equation (i) we get,

Putting the value of d in equation (i)

a + 8 d = 28

or a + 8 x 3 = 28

∴ a = 28 – 24 = 4

Therefore, the value of a = 4.

Question 4.

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

Solution:

We have, 9,17,25 Here,

a = 9 and d = 17 – 9 = 8

Let n terms of this AP must be taken to give a sum of 636.

We know that

Question 5.

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

Solution:

Question 6.

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

Solution:

Question 7.

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

Solution:

Question 8.

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

Solution:

Question 9.

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

Solution:

We have

S_{7} and S_{17} = 289

But, we know that

Again, we know that

Subtracting equation (ii) from equation (i), we get,

3d – 8d = 7 – 17

or – 5d = – 10

Putting the value of d in equation (i), we get,

a + 3d = 7

or a + 3 x 2 = 7

∴ a = 7 – 6 = 1

Therefore, sum of n terms of this AP is

Question 10.

Show that a_{1}, a_{2}, ……. a_{n},…… form an AP where an is defined as below:

(i) a_{n} = 3 + 4n

(ii) a_{n} = 9 – 5n

Also find the sum of the first 15 terms in each case.

Solution:

(i) We have

a = 3 + 4n

Put n = 1 in this equation

∴ a_{1} = 3 + 4 x 1 = 7

Now, put n = 2 in this equation

a_{2} = 3 + 4 x 2 = 11

Again, put n = 3 in this equation

a_{3} = 3 + 4 x 3 = 15

Therefore, the required AP is 7, 11, 15, ….

Here, a = 7, d = 11 – 7 = 4

We know that

Therefore, sum of 15 terms of the sequence a_{n} = 3 + 4n is 525.

(ii) We have,

∴ a_{n} = 9 – 5n

a_{1} = 9 – 5 x 1 = 4

a_{2} = 9 – 5 x 2 = – 1

and a_{3}= 9 – 5 x 3 = – 6

∴ AP is 4,-1,-6,…

∴ Common difference (d) = – 1 – 4 = – 5

Therefore, sum of 15 terms of the sequence a_{n} = 9 – 5n is – 465.

Question 11.

If the sum of the first n terms of an AP is 4n – n^{2}, what is the first term (that is S_{1})? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.

Solution:

S_{n} = 4n – n²

Put n = 1, 2, 3, …

S_{1} = 4(1) – 1

S_{1} = 3,

S_{2}=4,

S_{3}=3,

The second term is

a_{2} = S_{2} – S_{1}

a_{2} = 4 – 3 ⇒ a_{2} = 1;

Third term,

a_{2}= S_{3} – S_{2}

a_{3} = 3 – 4 ⇒ a_{2} = – 1

Third term

a_{2} = S_{3} – S_{2}

a_{3} = 3 – 4 ⇒ a_{3} = – 1;

⇒ – 60 + 45 = – 15

and nth term is

a_{n} = (4n – n^{2}) – [(4n – 4) – (n – 1)²]

= 4n – n^{2} – 4n + 4 + n^{2} + 1 – 2n

= 5 – 2n

nth term is (5 – 2n).

Question 12.

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

Solution:

The positive integers divisible by 6 are

6,12,18, …….

Here, a = 6 and d = 12 – 6 = 6

∴ S_{n} = \(\frac { 40 }{ 2 }\)[2 x 6 + (40 – 1) x 6]

= 20 (12 + 39 x 6) = 20 (12 + 234)

∴ S_{40} = 20 x 246 = 4920

Question 13.

Find the sum of the first 15 multiples of 8.

Solution:

The first 15 multiples of 8 are 8,16, 24…

Here, a = 8,d = 16-8 = 8

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

⇒ S_{15} = \(\frac {15 }{ 2 }\)[2 x 8 + (15 – 1)8]

⇒ S_{15} = \(\frac { 15 }{ 2 }\)[16 + 112]

⇒ S_{15} = \(\frac { 15 }{ 2 }\) x 128

⇒ S_{15} = 15 x 64 = 960

∴ The sum of first 15 multiples of 8 is 960.

Question 14.

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

Solution:

Let odd numbers between 0 and 50 be 1, 3, 5, 7,…….., 49.

Question 15.

A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows:

₹ 200 for the first day, ₹ 250 for the second day, ₹ 300 for the third day, etc. the penalty for each succeeding day being ₹ 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?

Solution:

We have given that penalty for delay of completion beyond a certain date are 200, 250, 300, ….

It is the form of an AP, where a = 200 d = 50

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

S_{30}= \(\frac { 30 }{ 2 }\) [2 x 200 + (30 – 1)50]

= 15 [400 + 1450]

= (15 x 1850) = 27750

The contractor has to pay ₹ 27750 as penalty.

Question 16.

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

Solution:

Let 1st prize be of ₹ a

2nd prize be ₹ (a – 20) and

3rd prize be ₹ (a – 20 – 20) = ₹ (a – 40)

Question 17.

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, example a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?

Solution:

Let the trees be planted 1, 2, 3, 4, 5 , …… 12

Here, a = 1, d = 1, n = 12

Total number of trees planted by each section

S_{12} = \(\frac { 12 }{ 2 }\) [2a + (n – 1) d] = 6 [2 x 1 + (12 – 1) x 1]

= 6 [2 + 11] = 6 x 13 = 78

Total number of trees planted by 3 sections = 78 x 3 = 234

Question 18.

A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm,… as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles?

(Take π = \(\frac { 22 }{ 7 }\))

[Hint:Length of successive semicircles is l_{1}, l_{2}, l_{3}, l_{4}, … with centres at A, B, respectively.]

Solution:

Length of successive semi circles is l_{1}, l_{2}, l_{3}, l_{4}, …

Series is 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm… we have to found l_{n}(total length)

Here, l_{1} = 0.5 π

d = 1.0 – 0.5 = 0.5π

n = 13 (consective semi circles)

Sum of l_{n} terms

∴ Total length of spiral of 143 cm.

Question 19.

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

Solution:

Question 20.

In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see Fig.) A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?

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

Solution:

Distance covered by the girl pick up the first potato = 10m

The distance covered by the girl to pick up the second potato = 8 x 2 = 16m

and the distance covered by the girl to pick up the third potato = 11 x 2 = 22m

So, we get the following series ;

10, 16, 22, 28, … upto 10^{th} term.

We know that, [where a = 10, d = 6]

Total distance We know that

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

= \(\frac { 10 }{ 2 }\)[2 x 10 + (10 – 1)6]

= 5 [20 + 54]

= 5 [74] = 370 m

Hence, total distance covered by the competitor is 370.