Class 10 Chapter 5 - Arithmetic Progressions

Exerise 5.1
1. 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 ` 15 for
the first km and ` 8 for each additional km.
(ii) The amount of air present in a cylinder when a vacuum

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 Class 10 Chapter 5 - Arithmetic Progressions

1
pump removes of the air remaining in the cylinder
4
at a time.
(iii) The cost of digging a well after every metre of digging,
when it costs ` 150 for the first metre and rises by
` 50 for each subsequent metre.
(iv) The amount of money in the account every year, when
` 10000 is deposited at compound interest at 8% per
annum.
Sol. (i) Yes. ` 15, ` 23, ` 31, ...... is an A.P. as common
difference is ` 8.

3 9 27
(ii) No. 1, , , , ...... is not an A.P. as common
4 16 64
difference is not constant.
(iii) Yes. ` 150, ` 200, ` 250, ...... is an A.P. as common
difference is ` 50.
2
108  108 
(iv) No. ` 10000 × , ` 10000 ×   ,
100  100 
3
 108 
` 10000   ,........ is not an A.P. as common
 100 
difference is not constant.
2. Write first four terms of the AP, when the first term a and
the common difference d are given as follows:
(i) a = 10, d = 10 (ii) a = – 2, d = 0
1
(iii) a = 4, d = – 3 (iv) a = – 1, d =
2
(v) a = – 1.25, d = – 0.25
Sol. We will use: A.P. is a, a + d, a + 2d, ........
(i) terms are 10, 10 + 10, 10 + 20, 10 + 30, i.e., 10, 20, 30, 40.
(ii) terms are – 2, – 2 + 0, – 2 + 0, – 2 + 0, i.e., – 2, – 2,
– 2, – 2.
(iii) terms are 4, 4 – 3, 4 – 6, 4 – 9, i.e., 4, 1, – 2, – 5.
1 3
(iv) terms are – 1, – 1 + , – 1 + 1, – 1 + ,
2 2
1 1
i.e., – 1, – , 0, .
2 2

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 Class 10 Chapter 5 - Arithmetic Progressions

(v) terms are – 1.25, – 1.25 – 0.25, – 1.25 – 0.50, – 1.25

– 0.75, i.e., – 1.25, – 1.50, – 1.75, – 2.00.
3. For the following APs, write the first term and the common
difference:
(i) 3, 1, – 1, – 3, ...... (ii) – 5, – 1, 3, 7, .......

1 5 9 13
(iii) , , , , ...... (iv) 0.6, 1.7, 2.8, 3.9, .......
3 3 3 3
Sol. (i) a = 3, d = 1 – 3 = –2
(ii) a = – 5, d = – 1 + 5 = 4

1 5 1 4
(iii) a = , d = – =
3 3 3 3
(iv) a = 0.6, d = 1.7 – 0.6 = 1.1.
4. Which of the following are APs? If they form an AP, find
the common difference d and write three more terms.
5 7
(i) 2, 4, 8, 16, ....... (ii) 2, , 3, , .......
2 2
(iii) – 1.2, – 3.2, – 5.2, – 7.2, ......
(iv) – 10, – 6, – 2, 2, .......
(v) 3, 3 + 2 , 3 + 2 2 , 3 + 3 2 , ......
(vi) 0.2, 0.22, 0.222, 0.2222, .......
(vii) 0, – 4, – 8, – 12, ......
1 1 1 1
(viii) – , – , – , – , ....... (ix) 1, 3, 9, 27, ......
2 2 2 2
(x) a, 2a, 3a, 4a, ....... (xi) a, a2, a3, a4, ......
(xii) 2, 8, 18 , 32 , .......
(xiii) 3 , 6 , 9 , 12 , ......
(xiv) 12, 32, 52, 72, ....... (xv) 12, 52, 72, 73, ......
Sol. (i) 4 – 2 = 2; 8 – 4 = 4; 16 – 8 = 8.
Difference is not constant. Hence, it is not an A.P.
5 1 5 1 7 1
(ii) – 2 = ; 3 – = ; – 3 = .
2 2 2 2 2 2
1
Difference is constant, i.e., d = . Hence, it is an A.P.
2
Next three terms are:

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Class 10 Chapter 5 - Arithmetic Progressions

7 1 1 9 9 1
+ = 4, 4 + = and + = 5.
2 2 2 2 2 2
(iii) – 3.2 + 1.2 = – 2; – 5.2 + 3.2 = – 2; – 7.2 + 5.2 = – 2.
Difference is constant, i.e., d = – 2. Hence, an A.P.
Next three terms are:
– 7.2 – 2 = – 9.2, – 9.2 – 2 = – 11.2
and – 11.2 – 2 = – 13.2.
(iv) – 10, – 6, – 2, 2, .........
Here, – 6 + 10 = 4, – 2 + 6 = 4, 2 + 2 = 4 .......
Difference is constant, i.e., d = 4.
Hence, it is an A.P.
Three more terms are:
2 + 4 = 6, 6 + 4 = 10 and 10 + 4 = 14.
(v) 3, 3 + 2 , 3 + 2 2 , 3 + 3 2 , ........

Here, 3 + 2 – 3
= 2, 3 + 2 2 – 3 – 2 = 2,

3 + 3 2 – 3 – 2 2 = 2 .....
Difference is constant, i.e.,

d = 2 . Hence, it is an A.P.
Next three terms are:

3 + 3 2 + 2 = 3 + 4 2,
3 + 4 2 + 2 = 3 + 5 2 and
3 + 5 2 + 2 = 3 + 6 2.
(vi) 0.22 – 0.2 = 0.02; 0.222 – 0.22 = 0.002.
Difference is not constant. Hence, it is not an A.P.
(vii) 0, – 4, – 8, – 12, ......
Here, – 4 – 0 = – 4, – 8 + 4 = – 4,
– 12 + 8 = – 4, ......
Difference is constant, i.e.,
d = – 4. Hence, it is an A.P.

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Next three terms are:

– 12 – 4 = – 16, – 16 – 4 = – 20
and – 20 – 4 = – 24.
1 1 1 1
(viii) – , – , – , – , ......
2 2 2 2
1 1 1 1 1 1
Here, – + = 0, – + = 0, – + = 0.
2 2 2 2 2 2
Difference is constant, i.e., d = 0. Hence, it is an A.P.
Next three terms are:

1 1 1 1 1 1
– + 0 = – , – + 0 = – and – +0=– .
2 2 2 2 2 2
(ix) 1, 3, 9, 27, .......
Here, 3 – 1 = 2, 9 – 3 = 6, 27 – 9 = 18, .......
Difference is not constant. Hence, it is not an A.P.
(x) 2a – a = a; 3a – 2a = a; 4a – 3a = a.
Difference is constant, i.e., d = a. Hence, it is an A.P.
Next three terms are: 4a + a = 5a, 5a + a = 6a and
6a + a = 7a.
(xi) a2 – a = a(a – 1); a3 – a2 = a2(a – 1).
Difference is not constant. Hence, it is not an A.P.

(xii) 2 , 2 2 , 3 2 , ........

2 2 – 2 = 2; 3 2 – 2 2 = 2;

4 2 – 3 2 = 2.

Difference is constant, i.e., d = 2 . Hence,

it is an A.P.
Next three terms are: 4 2 + 2 = 5 2 = 50 ,

5 2 + 2 = 6 2 = 72

and 6 2 + 2 = 7 2 = 98 .

(xiii) 3, 6, 9, 12 , .......

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Here, 6 – 3 ≠ 9 – 6 ≠ 12 – 6.
Difference is not constant. Hence, it is not an A.P.
(xiv) 12, 32, 52, 72, ......., i.e., 1, 9, 25, 49, .......
Here, 9 – 1 = 8, 25 – 9 = 16, 49 – 25 = 24 ........
Difference is not constant. Hence, it is not an A.P.
(xv) 12, 52, 72, 73 ......., i.e., 1, 25, 49, 73, .......
Here, 25 – 1 = 24, 49 – 25 = 24, 73 – 49 = 24, ......
Difference is constant, i.e., d = 24.
Hence, it is an A.P.
Next three terms are:
73 + 24 = 97, 97 + 24 = 121, 121 + 24 = 145.

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 Class 10 Chapter 5 - Arithmetic Progressions

Exerise 5.2
1. Fill in the blanks in the following table, given that a is the
first term, d the common difference and an the nth term of
the AP:
a d n an

(i) 7 3 8 ....
(ii) – 18 .... 10 0
(iii) .... –3 18 –5
(iv) – 18.9 2.5 .... 3.6
(v) 3.5 0 105 ....

Sol. (i) an = a + (n – 1)d = 7 + (8 – 1) 3 = 7 + 21 = 28.

(ii) an = a + (n – 1)d ⇒ 0 = – 18 + (10 – 1)d
⇒ 9d = 18 ⇒ d = 2.
(iii) an = a + (n – 1)d ⇒ – 5 = a + (18 – 1)(– 3)
⇒ a = – 5 + 51 = 46.
(iv) 3.6 = – 18.9 + (n – 1)(2.5)
⇒ 3.6 + 18.9 = (n – 1)(2.5)
⇒ 22.5 = 2.5n – 2.5 ⇒ 2.5n = 25
⇒ n = 10.
(v) an = a + (n – 1)d = 3.5 + (105 – 1) 0 = 3.5.

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2. Choose the correct choice in the following and justify:

(i) 30th term of the AP: 10, 7, 4, ....., is
(A) 97 (B) 77 (C) – 77 (D) – 87
1
(ii) 11th term of the AP: – 3, – , 2, ......, is
2
1
(A) 28 (B) 22 (C) – 38 (D) – 48
2
Sol. (i) a = 10, d = 7 – 10 = – 3,
a30 = 10 + (30 – 1)(– 3) = 10 – 87 = – 77.
Hence, option (C) is correct.
1 5
(ii) a = – 3, d = – + 3 = ;
2 2
5
a11 = – 3 + (11 – 1) = – 3 + 25 = 22.
2
Hence, option (B) is correct.
3. In the following APs, find the missing terms in the boxes:

(i) 2, , 26 (ii) , 13, , 3

1
(iii) 5, , , 9 (iv) – 4, , , , ,6
2

(v) , 38, , , , – 22
Sol. (i) a1 = 2, a3 = 26; 26 = 2 + (3 – 1) d
⇒ 24 = 2d ⇒ d = 12.
∴ a2 = 2 + 12 = 14.

Hence, 14 .
(ii) a2 = 13 ⇒ a + d = 13 and a4 = 3 ⇒ a + 3d = 3
Solving for a and d, we get a = 18, d = – 5
∴ a3 = 18 + 2 × (– 5) = 8.

Hence, 18 , 8 .

19 19
(iii) a1 = 5, a4 = ; = 5 + (4 – 1) d
2 2
9 3
⇒ 3d = ⇒ d =
2 2

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3 13 1 13 3 16
∴ a2 = 5 + = = 6 ; a3 = + = = 8.
2 2 2 2 2 2
1
Hence, 6 , 8
2
(iv) a = – 4 and a6 = 6 ⇒ – 4 + 5d = 6
⇒ d = 2
∴ a2 = – 2, a3 = 0, a4 = 2, a5 = 4.

Hence, −2 , 0 , 2 , 4 .

(v) Hint: a + d = 38; a + 5d = – 22.

On solving, we get a = 53, d = – 15. Find a3, a4, a5.
4. Which term of the AP: 3, 8, 13, 18, ....., is 78?
Sol. Let an = 78, then 78 = 3 + (n – 1)5 ⇒ n = 16.
5. Find the number of terms in each of the following APs:

1
(i) 7, 13, 19, ......, 205 (ii) 18, 15 , 13, ......, – 47
2
Sol. (i) Let an = 205, a = 7, d = 6
Then, 205 = 7 + (n – 1)6 ⇒ 198 = (n – 1)6
⇒ n – 1 = 33 ⇒ n = 34.
5
(ii) Let an = – 47, a = 18, d = – .
2
 5  5
Then, – 47 = 18 + (n –1)  −  ⇒ – 65 = (n – 1)  − 
 2  2
⇒ n – 1 = 26 ⇒ n = 27.
6. Check whether – 150 is a term of the AP: 11, 8, 5, 2, ..... .
Sol. Let an = – 150, then – 150 = 11 + (n – 1)(– 3)
⇒ – 161 = – 3n + 3 ⇒ 3n = 164
164
⇒ n = = 54.67 ∉ N.
3
Hence, – 150 is not a term of the given AP.
7. Find the 31st term of an AP whose 11th term is 38 and the
16th term is 73.
Sol. a + 10d = 38; a + 15d = 73 ⇒ a = – 32, d = 7.
∴ a31 = – 32 + 30 × 7 = – 32 + 210 = 178.

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8. An AP consists of 50 terms of which 3rd term is 12 and

the last term is 106. Find the 29th term.
Sol. a3 = 12 and a50 = 106 ⇒ a + 2d = 12
and a + 49d = 106
Solving for a and d, we get, a = 8, d = 2
∴ a29 = 8 + (29 – 1) × 2 = 8 + 56 = 64.
9. If the 3rd and the 9th terms of an AP are 4 and – 8
respectively, which term of this AP is zero?
Sol. a + 2d = 4; a + 8d = – 8 ⇒ a = 8, d = – 2.
Let an = 0 ⇒ 8 + (n – 1)(– 2) = 0 ⇒ n = 5.
10. The 17th term of an AP exceeds its 10th term by 7. Find
the common difference.
Sol. a17 = a10 + 7 ⇒ a + 16d = a + 9d + 7 ⇒ d = 1.
11. Which term of the AP: 3, 15, 27, 39, ..... will be 132 more
than its 54th term?
Sol. Let nth term of given AP is 132 more than its 54th term.
Here, a = 3, d = 12
∴ an = a54 + 132
∴ 3 + (n – 1)12 = 3 + (54 – 1)12 + 132
⇒ (n – 1)12 = 768 ⇒ n – 1 = 64 ⇒ n = 65
Hence, 65th term of the AP is 132 more than its 54th term.
12. Two APs have the same common difference. The difference
between their 100th terms is 100, what is the difference
between their 1000th terms?
Sol. Let common difference of two APs be d. {an} and {an′ } are
two APs:
′ = 100
a100 − a100 ⇒ (a1 + 99d) – ( a1′ + 99d) = 100

⇒ a1 – a1′ = 100. ...(i)

′
Now, a1000 − a1000 = (a1 + 999d) – ( a1′ + 999d)

= a1 – a1′ = 100... [From (i)]

13. How many three-digit numbers are divisible by 7?
Sol. Three-digit numbers divisible by 7 are 105, 112, ....., 994.

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 Class 10 Chapter 5 - Arithmetic Progressions

Let the required number of numbers be n

∴ 994 = 105 + (n – 1)7 ⇒ 7n = 896 ⇒ n = 128.
14. How many multiples of 4 lie between 10 and 250?
Sol. Multiples of 4 between 10 and 250 are 12, 16, ....., 248.
Here, a = 12, d = 4, let an = 248
∴ 248 = 12 + (n – 1)4 ⇒ (n – 1)4 = 236
⇒ n – 1 = 59 ⇒ n = 1 + 59 ⇒ n = 60.
15. For what value of n, are the nth terms of two APs: 63, 65,
67, ..... and 3, 10, 17, ...... equal?
Sol. 63 + (n – 1)2 = 3 + (n – 1)7
⇒ 60 = 5(n – 1) ⇒ n = 13.
16. Determine the AP whose third term is 16 and 7th term
exceeds the 5th term by 12.
Sol. a3 = 16 and a7 = a5 + 12
⇒ a + 2d = 16 and a + 6d = a + 4d + 12
⇒ d = 6 and a = 4.
Hence, A.P. is 4, 10, 16, 22, .......
17. Find the 20th term from the last term of the AP:
3, 8, 13, ....., 253.
Sol. From the end, a = 253, d = – 5
∴ 20th term from the end = 253 + (20 – 1)(– 5)
= 253 – 95 = 158.
18. The sum of the 4th and 8th terms of an AP is 24 and the
sum of the 6th and 10th terms is 44. Find the first three
terms of the AP.
Sol. (a + 3d) + (a + 7d) = 24 ⇒ 2a + 10d = 24
⇒ a + 5d = 12 ...(i)
and (a + 5d) + (a + 9d) = 44 ⇒ 2a + 14d = 44
⇒ a + 7d = 22 ...(ii)
Solving (i) and (ii), we get
a = – 13, d = 5
Thus, first three terms are – 13, – 8 and – 3.
19. Subba Rao started work in 1995 at an annual salary of
` 5000 and received an increment of ` 200 each year. In
which year did his income reach ` 7000?

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 Class 10 Chapter 5 - Arithmetic Progressions

Sol. Let an = 7000. Then 7000 = 5000 + (n – 1)200

⇒ 2000 = (n – 1) · 200 ⇒ n = 11.
In the year 2006, i.e., 11th year, his income reaches ` 7000.
20. Ramkali saved ` 5 in the first week of a year and then
increased her weekly savings by ` 1.75. If in the nth week,
her weekly savings become ` 20.75, find n.
Sol. an = 20.75. Then 20.75 = 5 + (n – 1)(1.75)
⇒ 15.75 = (n – 1)(1.75) ⇒ n – 1 = 9 ⇒ n = 10
In 10th week, her savings will be ` 19.75.

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 Class 10 Chapter 5 - Arithmetic Progressions

Exerise 5.3
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.
1 1 1
(iv) , , , ......, to 11 terms.
15 12 10
Sol. (i) a = 2, d = 5, n = 10.
10
∴ S10 = [4 + (10 – 1)5] = 5[49] = 245.
2
(ii) a = – 37, d = 4, n = 12
12
∴ S12 = [– 74 + 11 × 4] = 12 × (– 15) = – 180.
2
(iii) a = 0.6, d = 1.1, n = 100
100
∴ S100 = [1.2 + (100 – 1)(1.1)]
2
= 50 × 110.1 = 5505.
1 1 1 1
(iv) a= , d = – = , n = 11
15 12 15 60
11  2 1 11  2 1 
∴ S11 = + (11 − 1)  = +

2 15 60  2  15 6 
11 3 33
= × = .
2 10 20

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2. Find the sums given below:

1
(i) 7 + 10 + 14 + .... + 84
2
(ii) 34 + 32 + 30 + .... + 10
(iii) – 5 + (– 8) + (– 11) + .... + (– 230).
21 7
Sol. (i) a = 7, d = – 7 = , l = 84.
2 2
7 7
∴ 84 = 7 + (n – 1) ⇒ 77 = (n – 1)
2 2
⇒ n = 23.
23 23 2093 1
∴ S23 = (7 + 84) = × 91 = = 1046 .
2 2 2 2
(ii) a = 34, d = – 2, an = 10
⇒ 10 = 34 + (n – 1)(– 2)
⇒ – 24 = (n – 1)(– 2) ⇒ n = 13

13 13
∴ S13 = [34 + 10] = × 44 = 13 × 22 = 286.
2 2
(iii) a = – 5, d = – 3, l = – 230
– 230 = – 5 + (n – 1)(– 3) ⇒ – 225 = (n – 1)(– 3)
⇒ n = 76
76
∴ S76 = (– 5 – 230) = 38 × (– 235) = – 8930.
2
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, S = 192, find d.
(x) given l = 28, S = 144, and there are total 9 terms.
Find a.

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Sol. (i) an = 50 ⇒ 5 + (n – 1)3 = 50

⇒ (n – 1)3 = 45 ⇒ n – 1 = 15
⇒ n = 16
16
Sn = S16 = (10 + 15 × 3) = 8 × 55 = 440.
2
7
(ii) 35 = 7 + 12d ⇒ d =
3


13 7 13
∴ S13 = 14 + 12 × 3  = 2 × 42

2 
= 13 × 21 = 273.
(iii) a12 = 37 ⇒ a + 33 = 37
⇒ a=4

12
S12 = [8 + (12 – 1)3] = 6 × 41 = 246.
2

10
(iv) a + 2d = 15 and 125 = [2a + 9d]
2
⇒ a + 2d = 15 and 25 = 2a + 9d
Solving, we get a = 17, d = – 1
∴ a10 = a + 9d = 17 – 9 = 8.
9
(v) S9 = 75 ⇒ 75 = [2a + (9 – 1)5]
2
35 35
⇒ a = – , a9 = – + (9 – 1)5
3 3
35 85
= – + 40 = .
3 3
n
(vi) Sn = 90 ⇒ 90 = [4 + (n – 1)8].
2
n
⇒ 90 = [8n – 4] ⇒ 90 = 4n2 – 2n
2
⇒ 2n2 – n – 45 = 0 ⇒ 2n2 – 10n + 9n – 45 = 0
⇒ 2n(n – 5) + 9(n – 5) = 0
⇒ (2n + 9)(n – 5) = 0

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⇒ 2n + 9 = 0 or n – 5 = 0
−9
⇒ n = (rejected) or n = 5.
2
∴ an = a5 = 2 + 4 × 8 = 34.
n n
(vii) 210 = (a + an ) ⇒ 210 = (8 + 62)
2 2
⇒ n=6
a6 = 62 ⇒ 62 = 8 + (6 – 1) × d
54
⇒ 54 = 5d ⇒ d = .
5
(viii) an = 4 ⇒ a + (n – 1)d = 4
⇒ a + (n – 1).2 = 4 ...(i)
n
Sn = – 14 ⇒ – 14 = [2a + (n – 1)d]
2
n
⇒ – 14 = [8 – 4(n – 1) + 2(n – 1)] [Using (i)]
2
n
⇒ – 14 = [10 – 2n] ⇒ – 14 = n(5 – n)
2
⇒ n2 – 5n – 14 = 0 ⇒ (n – 7)(n + 2) = 0
⇒ n – 7 = 0 or n + 2 = 0
⇒ n = 7, – 2 (rejected)
Substituting the value of n = 7 in (i), we get
a + 12 = 4 ⇒ a = – 8.
8
(ix) S8 = 192 ⇒ 192 = [6 + (8 – 1)d]
2
⇒ 192 = 4[6 + 7d] ⇒ 48 = 6 + 7d
⇒ d = 6.
9 9
(x) (a + l) ⇒ 144 =
S9 = (a + 28)
2 2
⇒ a + 28 = 32 ⇒ a = 4.
4. How many terms of the AP: 9, 17, 25, ...... must be taken to
give a sum of 636?
n
Sol. Let Sn = 636 ⇒ 636 = [18 + (n − 1)8]
2
⇒ 636 = n[9 + 4n – 4] ⇒ 4n2 + 5n – 636 = 0
⇒ 4n2 + 53n – 48n – 636 = 0

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⇒ n(4n + 53) – 12(4n + 53) = 0

⇒ (4n + 53)(n – 12) = 0
53
⇒ n = – (rejected) or n = 12.
4
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.
Sol. a1 = 5, an = 45, Sn = 400
n
∴ 400 = [5 + 45] ⇒ n = 16
2
40 8
∴ a16 = 45 ⇒ 5 + 15d = 45 ⇒ d = = .
15 3
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?
Sol. a = 17, an = 350, d = 9
∴ 350 = 17 + (n – 1)9 ⇒ n = 38.
38
∴ S38 = (17 + 350) = 19 × 367 = 6973.
2
7. Find the sum of first 22 terms of an AP in which d = 7 and
22nd term is 149.
Sol. a22 = 149 ⇒ 149 = a + 21 × 7 ⇒ a = 2
22
∴ S22 = [4 + (22 – 1) × 7] = 11 × 151 = 1661.
2
8. Find the sum of first 51 terms of an AP whose second and
third terms are 14 and 18 respectively.
Sol. a2 = 14 and a3 = 18 ⇒ a + d = 14
and a + 2d = 18
⇒ a = 10, d = 4

51
∴ S51 = [20 + (51 – 1) × 4] = 51 × 110 = 5610.
2
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.
7
Sol. [2a + (7 – 1)d] = 49 ⇒ a + 3d = 7 ...(i)
2

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 Class 10 Chapter 5 - Arithmetic Progressions

17
and [2a + (17 – 1)d] = 289 ⇒ a + 8d = 17 ...(ii)
2
Solving (i) and (ii), we get
a = 1, d = 2
n n
∴ Sn = [2 + (n – 1)2] = × 2n = n2.
2 2
10. Show that a1 , a2, ....., a n, ...... form an AP where a n is
defined as below:
(i) an = 3 + 4n (ii) an = 9 – 5n
Also find the sum of the first 15 terms in each case.
Sol. (i) an = 3 + 4n
d = an – an – 1 = (3 + 4n) – {3 + 4(n – 1)}
= 3 + 4n – 3 – 4n + 4 = 4
As d is free from n, i.e., d is constant. Hence {an} is an
A.P.
a1 = 7
15 15
∴ S15 = [14 + (15 – 1)4] = × 70 = 525.
2 2
(ii) an = 9 – 5n,
d = (9 – 5n) – {9 – 5(n – 1)} = – 5
As d is free from n, i.e., d is constant. Hence {an} is
an A.P.
Further a1 = 4.
15 15
∴ S15 = [8 + 14 × (– 5)] = × (– 62) = – 465.
2 2
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.
Sol. Sn = 4n – n2 ...(i)
S1 = 4 – 1 = 3 = a
S2 = 8 – 4 = 4 [From (i)]
2 2
an = Sn – Sn – 1 = (4n – n ) – {4(n – 1) – (n – 1) }
= 4n – n2 – 4n + 4 + n2 – 2n + 1
⇒ an = 5 – 2n ⇒ a2 = 1, a3 = – 1 and a10 = – 15.

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 Class 10 Chapter 5 - Arithmetic Progressions

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

Sol. 6, 12, 18, ..... 40 terms
40
S40 = [12 + (40 – 1)6] = 20 × 246 = 4920.
2
13. Find the sum of the first 15 multiples of 8.
Sol. 8 + 16 + 24 + ........ 15 terms.
15 15
S15 = [16 + (15 – 1)8] = × 128 = 960.
2 2
14. Find the sum of the odd numbers between 0 and 50.
Sol. Numbers between 0 and 50 are 1, 3, 5, ...... 49.
Here, n = 25, a = 1, d = 2
25 25
∴ S25 = [2 + (25 – 1)2] = × 50 = 625.
2 2
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?
Sol. 200 + 250 + 300 + ..... 30 terms
30
S30 = [400 + (30 – 1)50] = 15 × 1850 = ` 27750.
2
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.
Sol. Let prizes be a, a – 20, a – 40; a – 60, ...... 7 times.
The sum of these prizes is ` 700.
7
∴ [2a + (7 – 1) (– 20)] = 700 ⇒ a – 60 = 100
2
⇒ a = 160.
The prizes are of ` 160, ` 140, ` 120, ` 100, ` 80, ` 60 and
` 40.
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

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 Class 10 Chapter 5 - Arithmetic Progressions

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?
Sol. Total number of trees planted by the students.
12
= 3 + 6 + 9 + ...... + 36 = [3 + 36] = 6 × 39 = 234.
2
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
 22 
consecutive semicircles.  Take π =
 7 
[Hint. Length of successive semicircle is l1, l2, l3, l4, ......
with centres at A, B, A, B, ..... respectively.]
l3

l1

A B

l2

l4
Sol. Length of semicircle with radius 0.5 cm = 0.5 π cm
Length of semicircle with radius 1.0 cm = π cm
Length of semicircle with radius 1.5 cm = 1.5 π cm
These form an A.P.; which is 0.5π, π, 1.5 π, ........
13 13
∴ S13 = [π + (13 – 1)0.5π] = × 7π
2 2
13 22
= × 7 × = 143 cm.
2 7
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?

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 Class 10 Chapter 5 - Arithmetic Progressions

Sol. 20 + 19 + 18 + ........ n terms = 200

n
[40 + (n – 1)(– 1)] = 200.
2
n
⇒ [41 – n] = 200 ⇒ 41n – n2 = 400
2
⇒ n2 – 41n + 400 = 0
⇒ (n – 25)(n – 16) = 0
⇒ n = 25, 16
25
S25 = [40 + 24 × (– 1)]
2
25
= × 16 = 200;
2
16
S16 = [40 + 15 × (– 1)] = 8 × 25 = 200
2
n = 25 is ruled out as a25 = – 4 < 0.
∴ n = 16. And a16 = 20 + (16 – 1) · (– 1) = 5.
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.).

5m 3m 3m

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?

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 Class 10 Chapter 5 - Arithmetic Progressions

[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)]
Sol. To pick up first, second, third, ....... potatoes, distances
covered are 10 m, 16 m, 22 m, ........ . These form an A.P.
with a = 10, d = 6.
10
∴ S10 = [20 + (10 – 1)6] = 5 × 74 = 370 m.
2

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 Class 10 Chapter 5 - Arithmetic Progressions

Exerise 5.4
1. Which term of the AP: 121, 117, 113, ....., is its first
negative term?
[Hint. Find n for an < 0]
Sol. AP is 121, 117, 113, .......
Let an < 0 ⇒ 121 + (n – 1)(– 4) < 0
⇒ 125 – 4n < 0 ⇒ 125 < 4n
1
⇒ n > 31
4
⇒ n = 32,
i.e., 32nd term is the first negative term.
2. The sum of the third and the seventh terms of an AP is 6
and their product is 8. Find the sum of first sixteen terms
of the AP.
Sol. a3 + a7 = 6 ⇒ a + 2d + a + 6d = 6
⇒ 2a + 8d = 6 ⇒ a + 4d = 3 ...(i)
Also (a + 2d)(a + 6d) = 8
⇒ (3 – 4d + 2d)(3 – 4d + 6d) = 8 [From (i)]
⇒ (3 – 2d)(3 + 2d) = 8 ⇒ 9 – 4d2 = 8
1
⇒ 4d2 = 1 ⇒ d= ±
2
1 16  1
When d = , a = 1, S16 =  2 + (16 − 1) 2 
2 2  
19
= 8 × = 76
2
1 16   1 
When d = – , a = 5, S16 = 10 + (16 − 1)  − 2  
2 2   

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 Class 10 Chapter 5 - Arithmetic Progressions

5
=8× = 20.
2
3. A ladder has rungs 25 cm apart 25 cm
(see figure). The rungs decrease
uniformly in length from 45 cm at
the bottom to 25 cm at the top. If
the top and the bottom rungs are
1
2–m
1 2
2 m apart, what is the length of
2
the wood required for the rungs?
 250  25 cm
 Hint: Number of rungs = 25 + 1
 
45 cm
1
2 m
Sol. Number of rungs = 2 + 1
25 cm
250 cm
= + 1 = 11
25 cm
a1 = 45 and a11 = 25 ⇒ 45 + 10d = 25
− 20
⇒ d= = – 2
10
11
∴ S11 = [90 + 10 × (– 2)]
2
= 5.5 × 70 cm = 385 cm.
4. The houses of a row are numbered consecutively from 1 to
49. Show that there is a value of x such that the sum of the
numbers of the houses preceding the house numbered x is
equal to the sum of the numbers of the houses following it.
Find this value of x.
[Hint: Sx – 1 = S49 – Sx ]
Sol. House numbers are: 1, 2, 3, ......, x, ......, 49
According to the given conditions,
Sx – 1 = S49 – Sx
x −1 49 x
⇒ [2 + (x – 2)1] = [2 + 48] – [2 + (x – 1)1]
2 2 2
⇒ (x – 1)(x) = 49 × 50 – x(x + 1)

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 Class 10 Chapter 5 - Arithmetic Progressions

⇒ x2 – x = 2450 – x2 – x ⇒ x2 = 1225
⇒ x = 35.
5. A small terrace at a football ground comprises of 15 steps
each of which is 50 m long and built of solid concrete.
1 1
Each step has a rise of m and a tread of m (see figure).
4 2
Calculate the total volume of concrete required to build the
terrace.
[Hint: Volume of concrete required to build the first step
1 1
= × × 50 m3]
4 2

1m
–
2
m

1
50

–m
4
1 1
Sol. Volume for the first step = × × 50 m3
4 2
2 1
Volume for the second step = × × 50 m3
4 2
3 1
Volume for the third step = × × 50 m3
4 2
These form an A.P. which are
1 1 2 1 3 1
× × 50, × × 50, × × 50 ...... upto
4 2 4 2 4 2
15 terms
50 100 50 50
Here, a = and d = – =
8 8 8 8
15  100 50  15
∴ S15 = + 14 ×  = × 100 = 750.
2  8 8 2
Hence, total volume of concrete required is 750 m3.

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