# Arithmetic Progressions Exercise 5.3

Arithmetic Progressions Exercise 5.3

Page No: 112

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) 1/15, 1/12, 1/10, ...... , to 11 terms

(i) 2, 7, 12 ,…, to 10 terms
For this A.P.,
a= 2
da2 − a1 = 7 − 2 = 5
n= 10
We know that,
Sn = n/2 [2a + (n - 1) d]
S10 = 10/2 [2(2) + (10 - 1) × 5]
= 5[4 + (9) × (5)]
= 5 × 49 = 245

(ii) −37, −33, −29 ,…, to 12 terms
For this A.P.,
a= −37
da2 − a1 = (−33) − (−37)
= − 33 + 37 = 4
n= 12
We know that,
Sn = n/2 [2a + (n - 1) d]
S12 = 12/2 [2(-37) + (12 - 1) × 4]
= 6[-74 + 11 × 4]
= 6[-74 + 44]
= 6(-30) = -180

(iii) 0.6, 1.7, 2.8 ,…, to 100 terms
For this A.P.,
a= 0.6
da2 − a1 = 1.7 − 0.6 = 1.1
n= 100
We know that,
Sn = n/2 [2a + (n - 1) d]
S12 = 50/2 [1.2 + (99) × 1.1]
= 50[1.2 + 108.9]
= 50[110.1]
= 5505

(iv) 1/15, 1/12, 1/10, ...... , to 11 terms
For this A.P.,

2. Find the sums given below
(i) 7 + + 14 + .................. +84
(ii)+ 14 + ………… + 84
(ii) 34 + 32 + 30 + ……….. + 10
(iii) − 5 + (− 8) + (− 11) + ………… + (− 230)

(i) For this A.P.,
a = 7
l = 84
d = a2 − a1 = - 7 = 21/2 - 7 = 7/2
Let 84 be the nth term of this A.P.
l = a (n - 1)d
84 = 7 + (n - 1) × 7/2
77 = (n - 1) × 7/2
22 = n − 1
n= 23
We know that,
Sn = n/2 (a + l)
Sn = 23/2 (7 + 84)
= (23×91/2) = 2093/2

(ii) 34 + 32 + 30 + ……….. + 10
For this A.P.,
a= 34
da2 − a1 = 32 − 34 = −2
l= 10
Let 10 be the nth term of this A.P.
la + (− 1) d
10 = 34 + (n − 1) (−2)
−24 = (− 1) (−2)
12 = n − 1
n= 13
Sn = n/2 (a + l)
= 13/2 (34 + 10)
= (13×44/2) = 13 × 22
= 286

(iii) (−5) + (−8) + (−11) + ………… + (−230)

For this A.P.,
a= −5
l= −230
da2 − a1 = (−8) − (−5)
= − 8 + 5 = −3
Let −230 be the nth term of this A.P.
la + (− 1)d
−230 = − 5 + (n − 1) (−3)
−225 = (n − 1) (−3)
(n− 1) = 75
n= 76
And,
Sn = n/2 (a + l)
= 76/2 [(-5) + (-230)]
= 38(-235)
= -8930

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 andan.
(vii) Given a = 8, an = 62, Sn = 210, find n and d.
(viii) Given an= 4, d = 2, Sn = − 14, find nand 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.

(i) Given that, a= 5, d = 3, an = 50
As an = a + (n − 1)d,
⇒ 50 = 5 + (n - 1) × 3
⇒ 3(n - 1) = 45
⇒ n - 1 = 15
⇒ n = 16
Now, Sn = n/2 (a + an)
Sn = 16/2 (5 + 50) = 440

(ii) Given that, a= 7, a13 = 35
As an = a + (n − 1)d,

⇒ 35 = 7 + (13 - 1)d
⇒ 12d = 28
⇒ d = 28/12 = 2.33
Now, Sn = n/2 (a + an)
S13 = 13/2 (7 + 35) = 273

(iii)Given that, a12 = 37, d = 3

Asan = a + (n − 1)d,
⇒ a12a + (12 − 1)3
⇒ 37 = a + 33
⇒ a= 4
Sn = n/2 (a + an)
Sn = 12/2 (4 + 37)
= 246

(iv) Given that, a3= 15, S10 = 125
Asan = a + (n − 1)d,
a3a + (3 − 1)d
15 = a + 2d ... (i)
Sn = n/2 [2a + (n - 1)d]
S10 = 10/2 [2a + (10 - 1)d]
125 = 5(2a + 9d)
25 = 2a + 9... (ii)
On multiplying equation (i) by (ii), we get
30 = 2a + 4d ... (iii)
On subtracting equation (iii) from (ii), we get
−5 = 5d
d= −1
From equation (i),
15 = a + 2(−1)
15 = a − 2
a= 17
a10a + (10 − 1)d
a10= 17 + (9) (−1)
a10= 17 − 9 = 8

(v) Given that,d = 5, S9 = 75
As Sn = n/2 [2a + (n - 1)d]
S9 = 9/2 [2a + (9 - 1)5]
25 = 3(a + 20)
25 = 3a + 60
3a= 25 − 60
a = -35/3
ana + (n − 1)d
a9a + (9 − 1) (5)
= -35/3 + 8(5)
= -35/3 + 40
= (35+120/3) = 85/3

(vi) Given that, a= 2, d = 8, Sn = 90
As Sn = n/2 [2a + (n - 1)d]
90 = n/2 [2a + (n - 1)d]
⇒ 180 = n(4 + 8n - 8) = n(8n - 4) = 8n2 - 4n
⇒ 8n2 - 4n - 180 = 0
⇒ 2n2 - n - 45 = 0
⇒ 2n2 - 10n + 9n - 45 = 0
⇒ 2n(n -5) + 9(n - 5) = 0
⇒ (2n - 9)(2n + 9) = 0
So, n = 5 (as it is positive integer)
∴ a5 = 8 + 5 × 4 = 34

(vii) Given that, a= 8, an = 62, Sn = 210
As Sn = n/2 (a + an)
210 = n/2 (8 + 62)
⇒ 35n = 210
⇒ n = 210/35 = 6
Now, 62 = 8 + 5d
⇒ 5d = 62 - 8 = 54
⇒ d = 54/5 = 10.8

(viii) Given that, an = 4, d = 2, Sn = −14
an = a + (n − 1)d
4 = a + (− 1)2
4 = a + 2n − 2
a + 2n = 6
= 6 − 2n ... (i)
Sn = n/2 (a + an)
-14 = n/2 (a + 4)
−28 = (a + 4)
−28 = (6 − 2n + 4)  {From equation (i)}
−28 = (− 2n + 10)
−28 = − 2n2 + 10n
2n2− 10n − 28 = 0
n2− 5−14 = 0
n2− 7n + 2n − 14 = 0
n(n − 7) + 2(n − 7) = 0
(n− 7) (n + 2) = 0
Eithern − 7 = 0 or n + 2 = 0
n= 7 or n = −2
However,n can neither be negative nor fractional.
Therefore,n = 7
From equation (i), we get
a= 6 − 2n
a= 6 − 2(7)
= 6 − 14
= −8

(ix) Given that,a = 3, n = 8, S = 192
As Sn = n/2 [2a + (n - 1)d]
192 = 8/2 [2 × 3 + (8 - 1)d]
192 = 4 [6 + 7d]
48 = 6 + 7d
42 = 7d
d = 6

(x) Given that,l = 28, S = 144 and there are total of 9 terms.
Sn = n/2 (a + l)
144 = 9/2 (a + 28)
(16) × (2) = a + 28
32 = a + 28
a= 4

Page No: 113

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

Let there be nterms of this A.P.
For this A.P., a= 9
d = a2− a1 = 17 − 9 = 8
As Sn = n/2 [2a + (n - 1)d]
636 = n/2 [2 × a + (8 - 1) × 8]
636 = n/2 [18 + (n- 1) × 8]
636 = [9 + 4n− 4]
636 = (4n+ 5)
4n2 + 5n − 636 = 0
4n2 + 53n − 48n − 636 = 0
(4n + 53) − 12 (4n + 53) = 0
(4n + 53) (n− 12) = 0
Either 4+ 53 = 0 or n − 12 = 0
n = (-53/4) or n = 12
cannot be (-53/4). As the number of terms can neither be negative nor fractional, therefore, n = 12 only.

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.

Given that,
a = 5
l = 45
Sn= 400
Sn = n/2 (a + l)
400 = n/2 (5 + 45)
400 = n/2 (50)
n = 16
l = a + (n− 1) d
45 = 5 + (16 − 1)d
40 = 15d
d = 40/15 = 8/3

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

Given that,

a = 17
l = 350
d = 9
Let there be n terms in the A.P.
l = a + (n − 1) d
350 = 17 + (n − 1)9
333 = (n − 1)9
(n − 1) = 37
n = 38
Sn = n/2 (a + l)
S38 = 13/2 (17 + 350)
= 19 × 367
= 6973
Thus, this A.P. contains 38 terms and the sum of the terms of this A.P. is 6973.

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

d = 7
a22 = 149
S22 = ?
ana + (n − 1)d
a22 =a + (22 − 1)d
149 = a + 21 × 7
149 = a + 147
a = 2
Sn = n/2 (a + an)
= 22/2 (2 + 149)
= 11 × 151
= 1661

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

Given that,
a2 = 14
a3 = 18
d = a3− a2 = 18 − 14 = 4
a2 =a + d
14 = a + 4
a = 10
Sn = n/2 [2a + (n - 1)d]
S51 = 51/2 [2 × 10 + (51 - 1) × 4]
= 51/2 [2 + (20) × 4]
= 51×220/2
= 51 × 110
= 5610

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.

Given that,

S7 = 49
S17 = 289
S7

= 7/2 [2a + (n - 1)d]
S7 = 7/2 [2a + (7 - 1)d]
49 = 7/2 [2a + 16d]
7 = (a + 3d)
a + 3d = 7 ... (i)
Similarly,
S17 = 17/2 [2a + (17 - 1)d]
289 = 17/2 (2a + 16d)
17 = (a + 8d)
a + 8d = 17 ... (ii)
Subtracting equation(i) from equation (ii),
5d = 10
d = 2
From equation (i),
a + 3(2) = 7
a + 6 = 7
a = 1
Sn = n/2 [2a + (n - 1)d]
n/2 [2(1) + (n - 1) × 2]
n/2 (2 + 2n - 2)
n/2 (2n)
n2

10. Show that a1a… , 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.

(i) an= 3 + 4n
a1= 3 + 4(1) = 7
a2= 3 + 4(2) = 3 + 8 = 11
a3= 3 + 4(3) = 3 + 12 = 15
a4= 3 + 4(4) = 3 + 16 = 19
It can be observed that
a2− a1 = 11 − 7 = 4
a3− a2 = 15 − 11 = 4
a4− a3 = 19 − 15 = 4
i.e.,ak + 1 − akis same every time. Therefore, this is an AP with common difference as 4 and first term as 7.
Sn = n/2 [2a + (n - 1)d]
S15 = 15/2 [2(7) + (15 - 1) × 4]
= 15/2 [(14) + 56]
= 15/2 (70)
= 15 × 35
= 525

(ii) an= 9 − 5n
a1= 9 − 5 × 1 = 9 − 5 = 4
a2= 9 − 5 × 2 = 9 − 10 = −1
a3= 9 − 5 × 3 = 9 − 15 = −6
a4= 9 − 5 × 4 = 9 − 20 = −11
It can be observed that
a2− a1 = − 1 − 4 = −5
a3− a2 = − 6 − (−1) = −5
a4− a3 = − 11 − (−6) = −5
i.e.,ak + 1 − akis same every time. Therefore, this is an A.P. with common difference as −5 and first term as 4.
Sn = n/2 [2a + (n - 1)d]
S15 = 15/2 [2(4) + (15 - 1) (-5)]
= 15/2 [8 + 14(-5)]
= 15/2 (8 - 70)
= 15/2 (-62)
= 15(-31)
= -465

11. If the sum of the firstn 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, the10th and the nthterms.

Given that,
Sn= 4n − n2
First term, a =S1 = 4(1) − (1)2 = 4 − 1 = 3
Sum of first two terms = S2
= 4(2) − (2)2= 8 − 4 = 4
Second term, a2S2 − S1 = 4 − 3 = 1
d = a2− a = 1 − 3 = −2
ana + (n − 1)
= 3 + (n − 1) (−2)
= 3 − 2n + 2
= 5 − 2n
Therefore, a3= 5 − 2(3) = 5 − 6 = −1
a10 = 5 − 2(10) = 5 − 20 = −15
Hence, the sum of first two terms is 4. The second term is 1. 3rd, 10th, and nth terms are −1, −15, and 5 − 2n respectively.

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

The positive integers that are divisible by 6 are
6, 12, 18, 24 …
It can be observed that these are making an A.P. whose first term is 6 and common difference is 6.
a = 6
d = 6
S40?
Sn = n/2 [2a + (n - 1)d]
S40 = 40/2 [2(6) + (40 - 1) 6]
= 20[12 + (39) (6)]
= 20(12 + 234)
= 20 × 246
= 4920

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

The multiples of 8 are
8, 16, 24, 32…
These are in an A.P., having first term as 8 and common difference as 8.
Therefore, a = 8
d = 8
S15= ?
Sn = n/2 [2a + (n - 1)d]
S15 = 15/2 [2(8) + (15 - 1)8]
= 15/2[6 + (14) (8)]
= 15/2[16 + 112]
= 15(128)/2
= 15 × 64
= 960

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

The odd numbers between 0 and 50 are
1, 3, 5, 7, 9 … 49
Therefore, it can be observed that these odd numbers are in an A.P.
a = 1
d = 2
l = 49
l = a + (n − 1) d
49 = 1 + (n − 1)2
48 = 2(n − 1)
n − 1 = 24
n = 25
Sn = n/2 (a + l)
S25 = 25/2 (1 + 49)
= 25(50)/2
=(25)(25)
= 625

15. A contract on construction job specifies a penalty for delay of completion beyond a certain dateas follows: Rs. 200 for the first day, Rs. 250 for the second day, Rs. 300 for the third day, etc., the penalty for each succeeding day being Rs. 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.

It can be observed that these penalties are in an A.P. having first term as 200 and common difference as 50.
a = 200
d = 50
Penalty that has to be paid if he has delayed the work by 30 days = S30
= 30/2 [2(200) + (30 - 1) 50]

= 15 [400 + 1450]
= 15 (1850)
= 27750
Therefore, the contractor has to pay Rs 27750 as penalty.

16. A sum of Rs 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 20 less than its preceding prize, find the value of each of the prizes.

Let the cost of 1st prize be P.
Cost of 2ndprize = P − 20
And cost of 3rdprize = P − 40
It can be observed that the cost of these prizes are in an A.P. having common difference as −20 and first term as P.
a = P
d = −20
Given that, S7= 700
7/2 [2a + (7 - 1)d] = 700

a + 3(−20) = 100
a − 60 = 100
a = 160
Therefore, the value of each of the prizes was Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60, and Rs 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 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?

It can be observed that the number of trees planted by the students is in an AP.
1, 2, 3, 4, 5………………..12
First term, a = 1
Common difference, d= 2 − 1 = 1
Sn = n/2 [2a + (n - 1)d]
S12 = 12/2 [2(1) + (12 - 1)(1)]
= 6 (2 + 11)
= 6 (13)
= 78
Therefore, number of trees planted by 1 section of the classes = 78
Number of trees planted by 3 sections of the classes = 3 × 78 = 234
Therefore, 234 trees will be planted by the students.

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, 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 = 22/7)

perimeter of semi-circle = πr
P1 = π(0.5) = π/2 cm
P2 = π(1) = π cm
P3 = π(1.5) = 3π/2 cm
P1P2P3 are the lengths of the semi-circles

π/2, π, 3π/2, 2π, ....
P1 = π/2 cm
P2 = π cm
d = P2- P1 = π - π/2 = π/2
First term = P1 = a = π/2 cm
Sn = n/2 [2a + (n - 1)d]
Therefor, Sum of the length of 13 consecutive circles
S13 = 13/2 [2(π/2) + (13 - 1)π/2]
=  13/2 [π + 6π]
=13/2 (7π)

= 13/2 × 7 × 22/7
= 143 cm

Page No: 114

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. In how many rows are the 200 logs placed and how many logs are in the top row?

It can be observed that the numbers of logs in rows are in an A.P.
20, 19, 18…
For this A.P.,
a = 20
d = a2− a1 = 19 − 20 = −1
Let a total of 200 logs be placed in n rows.
Sn= 200
Sn = n/2 [2a + (n - 1)d]
S12 = 12/2 [2(20) + (n - 1)(-1)]
400 = n (40 −n + 1)
400 = (41 −n)
400 = 41n −n2
n2 − 41+ 400 = 0
n2 − 16n − 25n + 400 = 0
(n − 16) −25 (n − 16) = 0
(− 16) (n − 25) = 0
Either (n − 16) = 0 or n − 25 = 0
n = 16 or n= 25
ana + (n − 1)d
a16 = 20 + (16 − 1) (−1)
a16 = 20 − 15
a16 = 5
Similarly,
a25 = 20 + (25 − 1) (−1)
a25 = 20 − 24
= −4
Clearly, the number of logs in 16th row is 5. However, the number of logs in 25throw is negative, which is not possible.
Therefore, 200 logs can be placed in 16 rows and the number of logs in the 16throw is 5.

20. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato and other potatoes 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)]

The distances of potatoes from the bucket are 5, 8, 11, 14…
Distance run by the competitor for collecting these potatoes are two times of the distance at which the potatoes have been kept. Therefore, distances to be run are
10, 16, 22, 28, 34,……….
a = 10
d = 16 − 10 = 6
S10=?
S10 = 12/2 [2(20) + (n - 1)(-1)]
= 5[20 + 54]
= 5 (74)
= 370
Therefore, the competitor will run a total distance of 370 m.