# AP 10th Class Maths 5th Chapter Arithmetic Progressions Exercise 5.3 Solutions

Well-designed AP 10th Class Maths Textbook Solutions Chapter 5 Arithmetic Progressions Exercise 5.3 offers step-by-step explanations to help students understand problem-solving strategies.

## Arithmetic Progressions Class 10 Exercise 5.3 Solutions – 10th Class Maths 5.3 Exercise Solutions

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) Given A.P is : 2, 7, 12,…
a = 2, d = a2 – a1 = 7 – 2 = 5, n = 10
We know that,
Sn = $$\frac{\mathrm{n}}{2}$$ [2a + (n – 1)d]
S10 = $$\frac{10}{2}$$ [2 × 2 + (10 – 1) (5)]
= 5 [4 + 9 (5)]
= 5 [4 + 45]
Therefore, S10 = 5 × 49 = 245
So, sum of the first 10 terms of A.P is 245.

ii) Given A.P is -37, -33, -29,…
a = -37, d = a2 – a1
= -33 – (-37)
= -33 + 37 = 4
∴ d = 4, n = 12
We know that,
Sn = $$\frac{\mathrm{n}}{2}$$ [2a + (n – 1) d]

So, sum of first 12 terms of given AP is -180.

iii) Given A.P is 0.6, 1.7, 2,8,…
a = 0.6, d = a2 – a1 = 1.7 – 0.6 = 1.1,
n = 100
We know that,
Sn = $$\frac{\mathrm{n}}{2}$$[2a + (n – 1)d]
S100 = $$\frac{100}{2}$$[2 × (0.6) + (100 – 1) × 1.1]
= 50 [1.2 + 99 × 1.1]
= 50 [1.2 + 108.9]
= 50 [110.1]
= 5505
Therefore, S100 = 5505.
So, sum of first 100 terms of given A.P is 5505.

iv) Given A.P is $$\frac{1}{15}$$, $$\frac{1}{12}$$, $$\frac{1}{10}$$,……..
a = $$\frac{1}{15}$$, d = a2 – a1 = $$\frac{1}{12}$$ – $$\frac{1}{15}$$
d = $$\frac{5-4}{60}$$ = $$\frac{1}{60}$$
We know that,

So, the sum of first 11 terms in given A.P. 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) Given A.P is 7 + 10$$\frac{1}{2}$$ + 14 + ……. + 84
a = 7, last term l = 84,
d = a2 – a1 = 10$$\frac{1}{2}$$ – 7 = 3$$\frac{1}{2}$$
We know that,
an = a + (n – 1) d

So, sum of the terms of given A.P is 1046$$\frac{1}{2}$$.

ii) Given A.P is 34 + 32 + 30 + … + 10
a = 34, d = a2 – a1 = 32 – 34 = -2,
last term l = 10
We know that,
an = a + (n – 1) d
⇒ 34 + (n – 1) (-2) = 10
⇒ (n – 1) (-2) = 10 – 34 = -24

⇒ n = 12 + 1 = 13
Sn = $$\frac{n}{2}$$(a + l)
S13 = $$\frac{13}{2}$$(34 + 10)

Therefore, S13 = 286
So, sum of the terms of given A.P is 286.

iii) Given A.P is -5 + (-8) + (-11) + …+ (-230)
a = -5, d = a2 – a1 = -8 – (-5)
= -8 + 5 = -3,
l = -230
We know that,
an = a + (n – 1) d
⇒ 5 + (n – 1) (-3) = -230
⇒ (n – 1) (-3) = -230 + 5 = -225

⇒ n = +75 + 1 = 76
Sn = $$\frac{\mathrm{n}}{2}$$(a + l)

Therefore, Sn = -8930
So, sum of the terms of given A.P is -8930.

Question 3.
In an AP:
i) given a = 5, d = 3, an = 50, find n and Sn.
ii) given a = 7, a13 = 35, find d and S13.
iii) given a12 = 37, d = 3, find a and S12.
iv) given a3 = 15, S10 = 125, find d and a10.
v) given d = 5, S9 = 75, find a and a9.
vi) given a = 2, d = 8, Sn = 90, find n and an.
vii) given a = 8, an = 62, Sn = 210, find n and d.
viii) given an = 4, d = 2, Sn = -14, find n and a.
ix) given a = 3, n = 8, Sn = 192, find d.
x) given l = 28, S = 144, and there are total 9 terms. Find a.
Solution:
i) We know that,
an = a + (n – 1) d
⇒ 5 + (n – 1) 3 = 50
⇒ (n- 1)3 = 50 – 5 = 45

⇒ n = 15 + 1 = 16
n = 15 + 1 = 16
Sn = $$\frac{\mathrm{n}}{2}$$ [2a + (n – 1)d]

S16 = 8 [10 + 45] = 8 × 55
Therefore, S16 = 440
So, sum of the 16 terms is 440.

ii) We know that,
an = a + (n – 1) d
a13 = 7 + (13 – 1) d = 35
⇒ 12d = 35 – 7 = 28

iii) We know that,
an = a + (n – 1) d
a12 = a + (12 – 1)3 = 37
⇒ a + 11 × 3 = 37
⇒ a = 37 – 33 = 4
Sn = $$\frac{\mathrm{n}}{2}$$ [2a + (n – 1) d]

So, sum of the 12 terms of given A.P is 246.

iv) We know that,
an = 3 + (n – 1) d
a3 = a + (3 – 1) d = 15
⇒ a + 2d = 15
⇒ a = 15 – 2d → (1)
Sn = $$\frac{\mathrm{n}}{2}$$ [2a + (n – 1) d]

⇒ [2a + 9d] = $$\frac{125}{5}$$ = 25
⇒ 2a + 9d = 25 → (2)
Put (1) in (2)
⇒ 2(15 – 2d) + 9d = 25
⇒ 30 – 4d + 9d = 25
⇒ 5d = 25 – 30 = – 5
∴ d = $$\frac{-5}{5}$$ = -1
Put d = -1 in (1)
a = 15 – 2(-1)
a = 15 + 2 = 17
an = a + (n – 1) d
a10 = 17 + (10 – 1) (-1)
= 17 + 9(-1) = 17 – 9 = 8
Therefore, a10 = 8.

v) We know that,
an = a + (n – 1) d
a9 = a + (9 – 1)5 = a + (8 × 5)
a9 = a + 40 → (1)

vi) We know that,
an = a + (n – 1) d
= 2 + (n – 1)8
an = 2 + 8n – 8 = 8n – 6
Sn = [2a + (n – 1)d]
⇒ $$\frac{n}{2}$$[2 × 2 + (n – 1)8] = 90
⇒ n(4 + 8n – 8) = 90 × 2
⇒ n(8n – 4) = 180
⇒ 8n2 – 4n – 180 = 0
⇒ 4(2n2 – n – 45) = 0
⇒ 2n2 – n – 45 = 0
⇒ 2n2 – 10n + 9n – 45 = 0
⇒ 2n(n – 5) + 9(n – 5) = 0
⇒ (2n + 9) (n – 5) = 0
∴ n = $$\frac{-9}{2}$$ (or) 5
Number of terms cannot be rational.
So, n = 5.
Put n = 5 in an = 8n – 6
⇒ an = 8(5) – 6 = 40 – 6
⇒ an = 34
Therefore, an = 34

vii) We know that,
an = a + (n – 1)d
⇒ 8 + (n – 1)d = 62
⇒ (n – 1)d = 62 – 8 = 54 → (1)
Sn = $$\frac{\mathrm{n}}{2}$$[2a + (n – 1)d]
⇒ $$\frac{\mathrm{n}}{2}$$[2 × 8 + (n – 1)d] = 210
⇒ n[16 + (n – 1)d] = 210 × 2
⇒ n[16 + (n – 1)d] = 420 → (2)
Put (1) in (2)
⇒ n[16 + 54] = 420
⇒ n × 70 = 420
∴ n = $$\frac{420}{70}$$ = 6
Therefore, n = 6
(6 – 1)d = 54 ⇒ 5d = 54
Therefore, d = $$\frac{54}{5}$$

viii) We know that,
an = a + (n – 1)d
⇒ a + (n – 1)2 = 4
⇒ a + 2n – 2 = 4
⇒ a + 2n = 4 + 2 = 6
⇒ a + 2n = 6
⇒ a = 6 – 2n → (1)
Sn = $$\frac{\mathrm{n}}{2}$$[2a + (n – 1)d]
⇒ $$\frac{\mathrm{n}}{2}$$[2a + (n – 1)2] = -14
⇒ n[2a + 2n – 2] = -14 × 2
⇒ n[2a + 2n – 2] = -28 → (2)
Put (1) in (2)
⇒ n [2(6 – 2n) + 2n – 2] = -28
⇒ n[12 – 4n + 2n – 2] = -28
⇒ n[10 – 2n] = -28
⇒ 2n[5 – n] = -28
⇒ 5n – n2 = -14
⇒ n2 – 7n + 2n – 14 = 0
⇒ n(n – 7) + 2(n – 7) = 0
⇒ (n + 2)(n – 7) = 0
⇒ n = -2 (or) 7
Number of terms cannot be negative.
So, n = 7
Put n = 7 in (1)
a = 6 – 2n
a = 6 – 2(7) = 6 – 14 = -8
Therefore, a = -8

ix) We know that,
Sn = $$\frac{\mathrm{n}}{2}$$[2a + (n – 1)d]

x) Give l = 28, n = 9, S = 144
We know that,
S = $$\frac{\mathrm{n}}{2}$$(a + l)
⇒ $$\frac{9}{2}$$(a + 28) = 144

⇒ a + 28 = 32
∴ a = 32 – 28
Therefore, a = 4

Question 4.
How many terms of the AP :9, 17,25,… must be taken to give a sum of 636?
Solution:
Given A.P is 9, 17, 25,. . . and Sn = 636
a = 9, d = a2 – a1 = 17 – 9 = 8
We know that,
Sn = $$\frac{\mathrm{n}}{2}$$[2a + (n – 1)d]
⇒ $$\frac{\mathrm{n}}{2}$$[2 × 9 + (n – 1)8] = 636
⇒ n [18 + 8n – 8] = 63 × 2
⇒ n [10 + 8n] = 1272
⇒ 8n2 + 10n – 1272 = 0
⇒ 2(4n2 + 5n – 636) = 0
⇒ 4n2 + 5n – 636 = 0
⇒ 4n2 + 53n – 48n – 636 = 0
⇒ n(4n + 53) -12(4n + 53) = 0
⇒ (n – 12)(4n + 53) = 0
∴ n = 12 (or) $$\frac{-53}{4}$$
Number of terms cannot be rational number.
Therefore, number of terms n = 12

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:
Given first term a = 5, last term l = 45
Sum of terms Sn = 400
We know that,

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:
Given, first term a = 17, last term l = 350
Common difference d = 9
We know that,
an = a + (n – 1) d
⇒ 17 + (n – 1) 9 = 350
⇒ (n- 1) 9 = 350 – 17

∴ n = 37 + 1 = 38
Therefore, number of terms n = 38
We know that,

Question 7.
Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
Solution:
Given, d = 7, a22 = 149, n = 22
We know that,
an = a + (n – 1)d
a22 = a + (22 – 1)7 = 149
⇒ a + 21 × 7 = 149 ⇒ a = 149 – 147
First term a = 2
Sn = $$\frac{n}{2}$$[2a + (n – 1)d]

= 11[4 + 21 × 7]
= 11(4 + 147) = 11(151)
Therefore, sum of 22 terms is 1661.

Question 8.
Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Solution:
Given, a + d = 14 and a + 2d = 18, n = 51
d = a3 – a2 = 18 – 14 = 4
a + 4 = 14
a = 14 – 4 = 10
We know that,
Sn = $$\frac{\mathrm{n}}{2}$$[2a + (n – 1)d]

Therefore sum of first 51 terms is 5610.

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:
Given, S7 = 49 and S17 = 289
We know that,

Question 10.
Show that a1, a2,…, an,. .. form an AP where an is defined as below :
i) an = 3 + 4n
ii) an = 9 – 5n
Also find the sum of the first 15 terms in each case.
Solution:
i) Given an = 3 + 4n
If n = 1, then a1 = 3 + 4 (1) = 7
If n = 2, then a2 = 3 + 4 (2) = 3 + 8 = 11
If n = 3, then a3 = 3 + 4 (3) = 3 + 12 = 15
A.P is a1, a2, a3,… = 7, 11, 15,…
a = 7, d = a2 – a1 = 11 – 7 = 4, n = 15
Sn = $$\frac{\mathrm{n}}{2}$$[2a + (n – 1)d]
S15 = $$\frac{15}{2}$$[2 × 7 + (15 – 1)4]

Sum of 15 terms = S15 = 525

ii) Given, an = 9 – 5n
If n = 1, then a1 = 9 – 5(1) = 9 – 5 = 4
If n = 2, then a2 = 9 – 5(2) = 9 – 10 = -1
If n = 3, then a3 = 9 – 5(3) = 9 – 15 = -6
A.P is a1, a2, a3,… = 4, -1, -6,…
a = 4, d = a2 – a1 = -1 – 4 = -5, n = 15
Sn = $$\frac{\mathrm{n}}{2}$$ [2a + (n – 1) d]
S15 = $$\frac{15}{2}$$ [2 × 4 + (15 – 1) (-5)]
= $$\frac{15}{2}$$ [8 + 14 × (-5)]

Sum of 15 terms = S15 = – 465

Question 11.
If the sum of the first n terms of an AP is 4n – n2, what is the first term (that is S1)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.
Solution:
Given, Sn = 4n – n2
If n = 1, then S1 = 4 (1) – (1)2 = 4 – 1 = 3
If n = 2, then S2 = 4(2) – (2)2 = 8 – 4 = 4
S1 = a = 3, a2 = S2 – S1 = 4 – 3 = 1
Common difference d = a2 – a1 = 1 – 3 = – 2
We know that, an = a + (n – 1) d
a3 = 3 + (3 – 1)(-2)
= 3 + 2 × (-2)
a3 = 3 – 4 = -1
a10 = 3 + (10 – 1)(-2)
= 3 + 9 × (-2)
a10 = 3 – 18 = -15
Therefore, a3 = -1 and a10 = -15
an = 3 + (n – 1)(-2)
an = 3 – 2n + 2 = 5 – 2n

Question 12.
Find the sum of the first 40 positive integers divisible by 6.
Solution:
We know that, 6, 12, 18, 24,… are the positive integers which are divisible by 6.
a = 6, d = a2 – a1 = 12 – 6 = 6, n = 40
We know that,
Sn = $$\frac{\mathrm{n}}{2}$$[2a + (n – 1)d]
S40= $$\frac{40}{2}$$ [2 × 6 + (40 – 1) 6]
= 20 [12 + 39 × 6] = 20 [12 + 234]
S40 = 20 × 246 = 4920

Question 13.
Find the sum of the first 15 multiples of 8.
Solution:
We know that 8, 16, 24, 32, . . . are 8 multiples.

Therefore, sum of first 15 multiples of 8 is 960.

Question 14.
Find the sum of the odd numbers be-tween 0 and 50.
Solution:
Odd numbers between 0 and 50 are: 1, 3, 5, 7,…, 49.
a = 1, d = a2 – a1 = 3 – 1 = 2
an = a + (n – 1) d
⇒ 1 + (n – 1) 2 = 49
⇒ (n – 1) 2 = 49 – 1 = 48
⇒ n – 1 = $$\frac{48}{2}$$ = 24
∴ n = 24 + 1 = 25
We know that,

Sum of odd numbers between 0 and 50 is 625.

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:
Penalty for the first day = ₹ 200
second day = ₹ 250
third day = ₹ 300
Therefore A.P is 200, 250, 300,… (30 days)
a = 200, d = a2 – a1 = 250 – 200 = 50, n = 30
We know that
Sn = $$\frac{\mathrm{n}}{2}$$ [2a + (n – 1) d]

= 15[400 + 29 × 50]
= 15 [400 + 1450]
S30 = 15 × 1850 = 27750
Penalty to be paid by the contractor for 30 days is ₹ 27750.

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 the first prize is ₹ x
second prize is ₹ (x – 20)
third prize is ₹ (x – 40)
Therefore, AY is x, (x – 20),(x – 40), …… (7 prizes)
We know that,

First prize = ₹ 160
160, 160 – 20, 160 – 40,……….
Therefore 7 prizes are 160, 140, 120, 100, 80, 60, 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, 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?
Solution:
Given each section of class plants the same number of trees as the class and 3 sections of each class.
= 1 × 3, 2 × 3, 3 × 3,……, 12 × 3
Total number of trees planted by the students
= 1 × 3 + 2 × 3 + 3 × 3 + … + 12 × 3
= 3(1 + 2 + 3 + . . . + 12)
a = 1, n = 12, d = a2 – a1 = 2 – 1 = 1
We know that’,

Total number of trees = 3 (S12)
= 3 × 78 = 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 given 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 l1, l2, l3, l4, . . . with centres at A, B, A, B, . . ., respectively.]
Solution:
Let p1, p2, p3, … are the circumferences of semicircles of radii r1 = 0.5 cm, r2 = 1.0 cm, r3 = 1.5 cm, r4 = 2.0 cm, r5 = 2.5 cm ……… respectively.
Solution:
p1 = πr1 = π × 0.5 = $$\frac{\pi}{2}$$ cm
p2 = πr2 = π × 1.0 = π cm
p3 = πr3 = π × 1.5 = $$\frac{3}{2}$$π cm
………………………….
………………………….
p13 = πr13 = π × $$\frac{13}{2}$$ = $$\frac{13 \pi}{2}$$ cm
Total length of the spiral

Therefore, total length of the spiral = 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 given figure. In how many rows are the 200 logs placed and how many logs are in the top row?

Solution:
Given the number of logs in the first (bottom) row = 20
logs in the second row = 19
So, A.P. is 20, 19, 18, 17,………. of 200 logs.
a= 20, d = a2 – a1 = 19 – 20 = -1
We know that,
Sn = $$\frac{\mathrm{n}}{2}$$ [2a + (n – 1)d]
⇒$$\frac{\mathrm{n}}{2}$$[2 × 20 + (n – 1)(-1)] = 200
⇒ n[40 – n + 1] = 200 × 2
⇒ n(41 – n) = 400
⇒ 41n – n2 = 400
⇒ n2 – 41n + 400 = 0
⇒ n2 – 25n – 16n + 400 = 0
⇒ n(n – 25) – 16(n – 25) = 0
⇒ (n – 16)(ñ – 25) = 0
Therefore, n = 16 (or) 25
If n = 25, then a25 = a + 24d
= 20 + 24 (-1)
a25 = 20 – 24 = -4
Number of rows cannot be negative.
So, n = 16 is possible.
Therefore, logs are placed in 16 rows.
an = a + (n – 1) d
a16 = 20 + (16 – 1)(-1)
= 20 + 15 (-1)
= 20 – 15
a16 = 5
Therefore, number of logs in the top row are 5.

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 potato are placed 3 m apart in a straight line. There are ten potatoes in the line.

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 × 5 + 2 × (5 + 3)]
Solution:
Given distance run by the competitor to pick up the first potato d1 = 5 + 5 = 10 m.
Distance run by the competitor to pick up the second potato d2 = 2 × 5 + 2 × 3 = 16 m
third potato d3 = 2 × 5 + 2 × 3 + 2 × 3 = 22 m
= 2 [5 + 3 × 3] ………. so on
Distance run by the competitor to pick up the nth potato = 2 [5 + (n – 1) 3] m.
Total distance = d1 + d2 + … + dn.
If n = 10 then, distance = 2 [5 + (10 – 1)3]
= 2 [5 + 27]
= 2 × 32 = 64 m.
So, A.P is 10, 16, 22,.. ., 64.
a = 10, d = 16 – 10 = 6, n = 10
Sn = $$\frac{\mathrm{n}}{2}$$ [2a + (n – 1)d]
S10 = $$\frac{10}{2}$$ [2 × 10 + (10 – 1)6]
= 5 [20+ 54]
S10 = 5 × 74 = 370
Total distance run by the competitor to pick up the potatoes is 370 m.