**NCERT Solution For Class 10, Maths, Chapter 5 Arithmetic Progressions, Exercise 5.3 solutions** given below which is very important. Students of Class 10 maths first should learn all the basic formula and then practice all the exercise given. Class 10 maths chapter 5 ex. 5.3 is well explained in easy format.

Table of Contents

Toggle## Class 10, Maths, Chapter 5, Exercise 5.3 Solutions

**Q.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.**

**Ans:**

**Q.2. Find the sums given below:**

**(i) 7 + 10$\frac{1}{4}$ + 14, …… + 84 **

**(ii) 34 + 32 + 30 + . . . + 10**

** (iii) –5 + (–8) + (–11) + . . . + (–230)**

**Ans:**

**Q.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 _{a10.}**

**(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 is total 9 terms. Find a.**

**Ans:**

**(iv)** **Given a _{3} = 15, S_{10} = 125, find d and a_{10.}**

We have, a_{3} = 15, S_{10} = 125

Let ‘a’ be the first term and ‘d’ the common difference of the given AP, Then,

a_{3} = 15 and S_{10} = 125

∴ a_{n }= a + (n-1) d

a_{3 }=15 = a + (3-1) d

⟹ 15 = a + 2d ……………….. (i)

and,

∴ ${{S}_{10}}=125=\frac{10}{2}\left[ 2a+(10-1)d \right]$

125 = 5 (2a + 9d)

⟹ 5 (2a + 9d = 125

⟹ 2a + 9d = 25 …………………. (ii)

Now, multiply eq. (i) by 2 and then subtracting from eq.(ii),

2(a + 2d) – (2a + 9d) = (2×15) – 25

⟹ 4d – 9d = 30 – 25

⟹ – 5d = 5 ⟹ d = $-\frac{5}{5}$ = -1

Putting the value of d in eq. (i)

15 = a + 2d

15 = a + 2(-1)

a = 15 + 2 = 17

Now, we have, a = 17 and d = -1

∴ a_{n} = a + (n -1) d

⟹ a_{10} = 17 + (10-1) (-1)

⟹ a_{10} = 17 – 9 = 8

Hence, d = -1 and a_{10} = 8.

**Q.4. How many terms of the AP : 9, 17, 25, . . . must be taken to give a sum of 636?**

**Ans:**

**Q.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.**

**Ans:**

**Q.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?**

**Ans:** Let ‘a’ be the first term and ‘d’ be the common difference. Let ‘l’ be its last term. Then, a = 17, l = a_{n} = 350, d = 9 and l = a_{n} = 350

∴ a_{n} = a + (n – 1) d

⟹ 17 + (n – 1) 9 = 350 (∵ a_{n} = 350)

⟹ 9(n – 1) = 350 – 17 = 333

⟹ n – 1 =$\frac{333}{9}$ = 37

⟹ n = 37 + 1 = 38

Putting a = 17, l = 350, n = 38 in S_{n }= $\frac{n}{2}$ (a + l ), we get

⟹ S_{38} =$\frac{38}{2}$ (17 + 350) = 19 × 367 = 6973

Hence, there are 38 terms in the AP having their sum as 6973.

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

**Ans:** Let ‘a’ be the first term and ‘d’ the common difference of the given AP. then,

So, we have, d = 7 and a_{22} = 149

∴ a_{n} = a + (n+1) d

⟹ a + (22 – 1) 7 = 149 (∵ a_{n} = 149)

⟹ a + 21 × 7 = 149

⟹ a = 149 – 147

⟹ a = 2

Putting a = 2, n = 22 and d = 7 in S_{n }= $\frac{n}{2}$ (a + l ), we get

S_{22} = $\frac{22}{2}$ [(2×2)+(22-1)(7)]

S_{22 }= 11 [4 +( 21× 7)]

S_{22 }= 11 (4 + 147)

S_{22 }= 11 × 151 = 1661

Hence, the sum of first 22 terms is 1661.

**Q.8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.**

**Ans:** Let ‘a’ be the first term and ‘d’ the common difference of the given AP. then,

a_{2} = 14 and a_{3} = 18 and here we need to find ‘a’.

⟹ a + d = 14 …………….. (i)

And a + 2d = 18 …………….. (ii)

Subtracting (i) from (ii), we get

⟹ (a + 2d) – (a + d) = 18 -14

⟹ a + 2d – a – d) = 4

⟹ d = 4

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

⟹ a + d = 14

⟹ a + 4 = 14

⟹ a = 10

Now, we have, a = 10 and d = 4

∴ S_{n }= $\frac{n}{2}$ (a + l )

⟹ S_{51} = [(2×10)+(51-1)4]

⟹ S_{51} = [20 + 50 × 4 ]

⟹ S_{51} = $\frac{51}{2}$ (20 + 200) = $\frac{51}{2}$x 220 = 51 x 110 = 5610

Hence, the sum of first 51 terms is 5610.

**Q.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.**

**Ans:**

**Q.10. Show that a _{1}, a_{2}, . . ., a_{n}, . . . form an AP where a_{n} 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.**

**Ans:**

**Q.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.**

**Ans:** According to the question,

S_{n} = 4n – n^{2} …………… (i)

Putting n = 1 in eq.(i), we get

S_{1} = 4(1) – (1)^{2} = 4 -1 = 3

Now, S_{1} = a_{1} = 3 (First term)

Putting n = 2 in eq.(i), we get

S_{2} = 4(2) – (2)^{2} = 8 – 4 = 4

Now, a_{2} = S_{2} – S_{1} = 4 – 3 = 1 (Second term)

Putting n = 3 in eq.(i), we get

S_{3} = 4(3) – (3)^{2} = 12 – 9 = 3

Now, a_{3} = S_{3} – S_{2} = 3 – 4 = – 1 (third term)

Now, to calculate 10^{th} term, we need to know sum of 9^{th} term,

Putting n = 9 in eq.(i), we get

S_{9} = 4(9) – (9)^{2} = 36 – 81 = -45

Similarly, putting n = 10 in eq.(i), we get

S_{10} = 4(10) – (10)^{2} = 40 – 100 = – 60

Now, a_{10} = S_{10} – S_{9} = (- 60) – (- 45) = – 15 (tenth term)

Here, common difference, d = a_{n} – a_{(n-1)}

Let say, d = a_{3} – a_{2} = (-1) – (1) = -2

∴ n^{th }term, a_{n} = a+(n-1) d

a_{n} = 3 +(n-1) (-2)

a_{n} = 3 +(-2n + 2)

a_{n} = 5 – 2n

**Q.12. Find the sum of the first 40 positive integers divisible by 6.**

**Ans:** The first positive integers divisible by 6 are 6, 12, 18, …., Clearly, it is an AP with first term a = 6 and common difference d = 6. We want to find, S_{40}.

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

⟹ S_{40} = $\frac{40}{2}$ [(2 × 6) + (40 –1)6]

⟹ S_{40} = 20[12 + (39 × 6)] = 20 (12 + 234)

⟹ S_{40} = 20 × 246

⟹ S_{40} = 4920

Hence, **sum of the first 40 positive integers divisible by 6 is 4920.**

**Q.13. Find the sum of the first 15 multiples of 8.**

**Ans:** The first 15 multiples of 8 are (8×1), (8×2), (8×3), ……., (8×15) i.e., 8, 16, 24, ……120,

It is an AP in which a = 8 and d = a_{2} – a_{1} = 16 – 8 = 8

We know that,

${{S}_{n}}=\frac{n}{2}(a+l)$

∴ ${{S}_{15}}=\frac{15}{2}(8+120)$ (∵l = a_{n} =120)

${{S}_{15}}=\frac{15}{2}\times 128$

S_{15} = 15 × 64 = 960

Hence, the sum of the first 15 multiples of 8 is 960.

**Q.14. Find the sum of the odd numbers between 0 and 50.**

**Ans:** The odd numbers between 0 and 50 are 1, 3, 5,… 49. They form an AP and there are 25 terms.

Here, a = 1, n = 25 terms and l = a_{n} = 49

∴ ${{S}_{n}}=\frac{n}{2}(a+l)$

${{S}_{25}}=\frac{25}{2}(1+49)$ (∵ l = a_{n} = 49)

${{S}_{25}}=\frac{25}{2}\times 50=\,25\times 25=625$

Hence, the sum of the odd number between 0 and 50 is 625.

**Q.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?**

** ****Ans:** penalty of the first day = ₹ 200

penalty of the second day = ₹ (200 + 50) = ₹ 250

penalty of the third day = ₹ (200 + 50 + 50) = ₹ 300

the list is ₹ 200, ₹ 250, ₹ 300 …..

Here, a = 200, d = a_{2} – a_{1}= 250 – 200 = 50 and n = 30 terms (= Days)

∵ ${{S}_{n}}=\frac{n}{2}\left[ 2a+(n-1)d \right]$

∴ ${{S}_{30}}=\frac{30}{2}\left[ (2\times 200)+(30-1)50 \right]$

S_{30} =15 [400 + (29 ×50)]

S_{30} = 15 (400 +1450) = 15 × 1850 = 27750

Hence, a delay of 30 days costs the contractor Rs 27,750

**Q.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.**

**Ans:**

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

**Ans:** Since there are three sections of each class; so

The number of trees planted by class -I = 1 x 3 = 3

The number of trees planted by class -II = 2 x 3 = 6

The number of trees planted by class – III = 3 x 3 = 9

The number of trees planted by class -IV = 4 x 3 = 12

Similarly, the number of trees planted by class – XII = 12 x 3 = 36

Now, we have the list of trees planted = 3, 6, 9, 12, ……….. 36

it is an AP in which a = 3, d = a_{2} – a_{1}= 6 – 3 = 3, n = 12 and l = 36

we know that,

${{S}_{n}}=\frac{n}{2}(a+l)$

${{S}_{12}}=\frac{12}{2}(3+36)$

S_{12} = 6(39) = 234

Hence, the sum of the number of the trees planted by these classes are 234.

**Q.18. A spiral is made up of successive semicircles, with centers 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 Fig. 5.4. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take $\pi =\frac{22}{7}$)**

**Ans:**

**Q.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?**

** **

**Ans:**

**Q.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 Figure).**

**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?**

**Ans: **To pick up the first potato, second potato, third potato, fourth potato, …

The distances (in meters) run by the competitor to pick –

First potato = 2 x 5 = 10 m

Second potato = 2 x (5 + 3) = 16 m

Third potato = 2 x (5 + 3 +3) = 22 m

Fourth potato = 2 x (5 + 3 + 3 + 3) = 28 m

It is in AP with a = 10, d = 16 -10 = 6 and n = 10 terms (= potatoes)

∴ The sum of first ten terms

S_{10} = $\frac{10}{2}$ [(2×10) + (10 –1) × 6]

S_{10} = 5 (20 + 54) = 5 × 74 = 370

∴ The total distance the competitor has to run is 370 m.

**Download**: NCERT Books