NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progression

NCERT
SOLUTIONS
CLASS - 10th

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Class : 10th
Subject : Maths
Chapter : 5
Chapter Name : Arithmetic Progressions

Exercise 5.1

Q1 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 rst km and ₹ 8 for each additional
km.
(ii) The amount of air present in a cylinder when a vacuum pump removes of the air 1

4

remaining in the cylinder at a time.
(iii) The cost of digging a well after every metre of digging, when it costs ₹ 150 for the rst
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.

Answer. (i) It can be observe that
Taxi fare for 1 km = 15
st

Taxi fare for rst 2 km = 15 + 8 = 23
Taxi fare for rst 3 km = 23 + 8 = 31
Taxi fare for rst 4 km = 31 + 8 = 39
Clearly 15, 23, 31, 39.… forms an A.P. because every term is 8 more than the preceding term
(ii) Let the initial volume of air in a cylinder be V lit. In each stroke, the vacuum pump
removes of air remaining in the cylinder at a time.
1

4

In other words, after every stroke ,only 1 − part of air will remain.
1 3
= th
4 4
2 3

Therefore, volumes will be V (
3V 3 3
) ⋅ ( V) ⋅ ( V) …
4 4 4

Clearly, it can be observed that the adjacent terms of this series do
not have the same difference between them. Therefore, this is not an A.P
(iii) Cost of digging for rst metre = 150
Cost of digging for rst 2 metres = 150 + 50 = 200
Cost of digging for rst 3 metres = 200 + 50 = 250
Cost of digging for rst 4 metres = 250 + 50 = 300
Clearly, 150, 200, 250, 300 forms an A.P. because every term is 50 more than the preceding
term.

(iv) We know that if Rs P is deposited at r% compound interest per annum for n years, our
n

money will be P(1 + r

100
) after n years.
Therefore, after every year, our money will be
2 3 4
8 8 8 8
10000 (1 + ) , 10000(1 + ) , 10000(1 + ) , 10000(1 + )
100 100 100 100

Clearly, adjacent terms of this series do not have the same difference
between them. Therefore, this is not an A.P.

Page : 99 , Block Name : Exercise 5.1

Q2 Write rst four terms of the AP, when the rst term a and the common difference d are
given as follows:
(i) a = 10, d = 10
(ii) a = –2, d = 0
(iii) a = 4, d = – 3
(iv) a = – 1, d = 1/2
(v) a = – 1.25, d = – 0.25

Answer. a = 10, d = 10
Let the series a , a , a , a , a …
1 2 3 4 5

a1 = a = 10

a2 = a1 + d = 10 + 10 = 20

a3 = a2 + d = 20 + 10 = 30

a4 = a3 + d = 30 + 10 = 40

a5 = a4 + d = 40 + 10 = 50

Therefore , the series will be 10,20,30,40,50..
First four terms of this A.P. will be 10, 20, 30, and 40.

(ii) a = −2, d = 0
Let the series a a , a , a …
1, 2 3 4

a1 = a = −2

a2 = a1 + d = −2 + 0 = −2

a3 = a2 + d = −2 + 0 = −2

a4 = a3 + d = −2 + 0 = −2

Therefore the series will be −2, −2, −2, −2 …
First four terms of this AP. Will be −2, −2, −2 and − 2

(iii) a = 4, d = −3
Let the series be a , a , a
1 2 3r a4 …

a1 = a = 4

a2 = a1 + d = 4 − 3 = 1

a3 = a2 + d = 1 − 3 = −2

a4 = a3 + d = −2 − 3 = −5

(iv) a = −1, d =
1

2

Let the series be a , a , a , a … 1 2 3 4

a1 = a = −1

1 1
a2 = a1 + d = −1 + = −
2 2

1 1
a3 = a2 + d = − + = 0
2 2

1 1
a4 = a3 + d = 0 + =
2 2

Clearly the series will be
1 1
−1, − , 0,
2 2

Four terms of this A.P. will be −1, − 1

2
, 0 and
1

2

(v) a = −1.25, d = −0.25
Let the series be a , a , a a … 1 2 3r 4

a1 = a = −1.25

a2 = a1 + d = −1.25 − 0.25 = −1.50

a3 = a2 + d = −1.50 − 0.25 = −1.75

a4 = a3 + d = −1.75 − 0.25 = −2.00

Clearly the series will be 1.25, −1.50, −1.75, −2.00 … … .
First four terms of this A.P will be −1.25, −1.50, −1.75 and − 2.00

Page : 99 , Block Name : Exercise 5.1

Q3 For the following APs, write the rst term and the common difference:
(i) 3, 1, −1, −3 …

(ii) − 5, −1, 3, 7 …

5 9 13
(iii) 3, , , …
3 3 3

(iv) 0.6, 1.7, 2.8, 3.9 …

Answer. (i) 3, 1, −1, −3 …
Here, rst term, a = 3
Common difference, d = Second term — First term
= 1-3=-2

(ii) −5, −1, 3, 7 …
Here, rst term, a = —5
Common difference, d = Second term — First term
= (−1) − (−5) = −1 + 5 = 4

(iii) 1

3
,
5

3
,
9

3
,
13

3
…

Here, rst term a = 1

3

Common difference, d = Second term — First term

5 1 4
= − =
3 3 3

(iv) 0.6, 1.7, 2.8, 3.9 …
Here, rst term , a=0.6
Common difference, d = second term - First term
= 1.7 − 0.6

= 1.1

Page : 99 , Block Name : Exercise 5.1

Q4 Which of the following are APs? If they form an AP, nd the common difference d and write
three more terms.

(i) 2, 4, 8, 16, …
(ii) 2, , 3, , …
5

2
7

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, …
(viii) − , − , − , − , …
1 1 1 1

2 2 2 2

(ix) 1, 3, 9, 27, …
(x) a, 2a, 3a, 4a, …
(xi) a, a , a , a , …
2 3 4

(xii) √2, √8, √18, √32, …
(xiii) √3, √6, √9, √12, …
(xiv) 1 , 3 , 5 , 7 , …
2 2 2 2

(xv) 1 , 5 , 7 , 73, …
2 2 2

Answer. Missing

Page : 99 , Block Name : Exercise 5.1

Exercise 5.2

Q1 Fill in the blanks in the following table, given that a is the rst term, d the common
difference and an the nth term of the AP

Answer. (I) a = 7, d = 3, n = 8, a n =?

We know that ,
For an A.P a = a + (n − 1)d
n

= 7 + (8 − 1)3

= 7 + (7)3

= 7 + 21 = 28

Hence a n = 28

(II) Given that
a = −18, n = 10, an = 0, d =?

We know that
a = −18, n = 10, an = 0, d =?

We know that
an = a + (n − 1)d

0 = −18 + (10 − 1)d

18 = 9d

18
d = = 2
9

Hence common difference , d=2

(III) Given that
d = −3, n = 18, an = −5

We know that

an = a + (n − 1)d

−5 = a + (18 − 1)(−3)

−5 = a + (17)(−3)

−5 = a − 51

a = 51 − 5 = 46

Hence a=46

(IV) a = −18.9, d = 2.5, a n = 3.6, n =?

We know that
an = a + (n − 1)d

3.6 = −18.9 + (n − 1)2.5

3.6 + 18.9 = (n − 1)2.5

22.5 = (n − 1)2.5

22.5
(n − 1) =
2.5

n − 1 = 9

n = 10

Hence, n=10

(V) a = 3.5, d = 0, n = 105, a n =?

We know that
an = a + (n − 1)d

an = 3.5 + (105 − 1)0

an = 3.5 + 104 × 0

an = 3.5

Hence, an = 3.5

Page : 105 , Block Name : Exercise 5.2

Q2 Choose the correct choice in the following and justify
(I) 30th term of the A.P: 10, 7, 4, , is
A. 97 B. 77 C. − 77 D. − 87
1

(II) 11 th
term of the A.P.
−3,−
2
,2
is
1

A 28B.22C. −38D. −48
2

Answer. Given that
A.P. 10,7,4….
First term a=10
Common difference ,d = a 2 − a1 = 7 − 10

=-3

We know that, an = a + (n − 1)d

a30 = 10 + (30 − 1)(−3)

a30 = 10 + (29)(−3)

a30 = 10 − 87 = −77

Hence the correct answer is C

II. Given that A.P. −3, −
1
, 2, …
2

First term a= -3
Common difference d = a 2 − a1
1
= − − (−3)
2

1 5
= − + 3 =
2 2

We know that
an = a + (n − 1)d

5
a11 = −3 + (11 − 1) ( )
2

5
a11 = −3 + (10) ( )
2

a11 = −3 + 25

a11 = 22

Hence, the answer is B.

Page : 106 , Block Name : Exercise 5.2

Q3 In the following APs nd the missing term in the boxes
Answer. I. 2, □, 26
For this A.P.,
a=2
a3 = 26

We know that a n = a + (n − 1)d

a3 = 2 + (3 − 1)d

26 = 2 + 2d

24 = 2d

d = 12

a2 = 2 + (2 − 1)12

= 14

Therefore 14 is the missing term

II. □, 13, □, 3
For this A.P.
a2 = 13 and

a4 = 3

We know that a n = a + (n − 1)d

a2 = a + (2 − 1)d

13 = a + d(1)

a4 = a + (4 − 1)d

3 = a + 3d(I I )

On subtracting (I) from (II) we obtain
−10 = 2d

d = −5

From equation (I) we obtain
13 = a + (−5)

a = 18

a3 = 18 + (3 − 1)(−5)

= 18 + 2(−5) = 18 − 10 = 8

Therefore the missing terms are 18 and 7 respectively .

III. 5, □, □, 9
1

2

For this A.P.
a = 5

1 19
a4 = 9 =
2 2

We know that
an = a + (n − 1)d

a4 = a + (4 − 1)d

19
= 5 + 3d
2

19
− 5 = 3d
2

9
= 3d
2

3
d =
2
3 13
a2 = a + d = 5 + =
2 2

3
a3 = a + 2d = 5 + 2 ( ) = 8
2

Therefore the missing terms are 13/2 and 8 respectively

IV. −4, □. □. □. □.6
For this A.P.,
a = −4 and

a6 = 6

We know that

an = a + (n − 1)d

a6 = a + (6 − 1)d

6 = −4 + 5d

10 = 5d

d = 2

a2 = a + d = −4 + 2 = −2

a3 = a + 2d = −4 + 2(2) = 0

a4 = a + 3d = −4 + 3(2) = 2

a5 = a + 4d = −4 + 4(2) = 4

Therefore ,the missing terms are -2,0,2 and 4 respectively .

V. □, 38, □, □, □, −22
For this A.P.
a2 = 38

a6 = −22

We know that
an = a + (n − 1)d

a2 = a + (2 − 1)d

38 = a + d(1)

a6 = a + (6 − 1)d

−22 = a + 5d(2)

On subtracting equation (1) from (2) we obtain
−22 − 38 = 4d

−60 = 4d

d = −15

a = a2 − d = 38 − (−15) = 53

a3 = a + 2d = 53 + 2(−15) = 23

a4 = a + 3d = 53 + 3(−15) = 8

a5 = a + 4d = 53 + 4(−15) = −7

Therefore the missing terms are 53,23, 8 and -7 respectively.

Page : 106 , Block Name : Exercise 5.2

Q4 Which term of the AP : 3, 8, 13, 18, . . . ,is 78?

Answer. 3, 8, 13, 18, …
For this A.P.,
a=3
d = a2 − a1 = 8 − 3 = 5

Let nth term of this A.P. be 78

an = a + (n − 1)d

78 = 3 + (n − 1)5

75 = (n − 1)5

(n − 1) = 15

n = 16

Hence, 16tn term of this A.P. is 78.

Page : 106 , Block Name : Exercise 5.2

Q5 Find the number of terms in each of the following APs
I. 7, 13, 19, … , 205
II. 18, 15 , 13, … , −47
1

2

Answer. I. 7, 13, 19, … , 205
For this A.P. ,
a = 7

d = a2 − a1 = 13 − 7 = 6

Let there are n terms in this A.P.
an = 205

We know that
an = a + (n − 1)d

Therefore 205 = 7 + (n − 1)6
198 = (n − 1)6

33 = (n − 1)

n = 34

Therefore this given series has 34 terms in it.

II. 18, 15 1
, 13, … , −47
2

For this A.P.,
a = 18

1
d = a2 − a1 = 15 − 18
2

31 − 36 5
d = = −
2 2

Let there are n terms in this A.P.
Therefore a = −47 and we know that
n

an = a + (n − 1)d

5
−47 = 18 + (n − 1) (− )
2

5
−47 − 18 = (n − 1) (− )
2

5
−65 = (n − 1) (− )
2

−130
(n − 1) =
−5

(n − 1) = 26

n = 27

Therefore, this given A.P. has 27 terms in it.

Page : 106 , Block Name : Exercise 5.2

Q6 Check whether -150 is a term of hw A,.P. 11, 8, 5, 2, …

Answer. For this A.P.,
a = 11

d = a2 − a1 = 8 − 11 = −3

Let -150 be the n th
term of this A.P.
We know that
an = a + (n − 1)d

−150 = 11 + (n − 1)(−3)

−150 = 11 − 3n + 3

−164 = −3n

164
n =
3

Clearly , n is not an integer
Therefore -150 is not a term of this A.P.

Page : 106 , Block Name : Exercise 5.2

Q7 Find the 31 term of an AP whose 11th term is 38 and the 16th term is 73.
st

Answer. Given that,
a11 = 38

a16 = 73

We know that,
an = a + (n − 1)d

a11 = a + (11 − 1)d

38 = a + 10d(1)

Similarly,

a16 = a + (16 − 1)d

73 = a + 15d(2)

On subtracting (1) from (2) we obtain
35 = 5d

d = 7

From an equation (1)
38 = a + 10 × (7)

38 − 70 = a

a = −32

a31 = a + (31 − 1)d

= −32 + 30(7)

= −32 + 210

=178
Hence 31 term is 178
st

Page : 106 , Block Name : Exercise 5.2

Q8 An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th
term

Answer. Given that,
a3 = 12

a50 = 106

We know that,
an = a + (n − 1)d

a3 = a + (3 − 1)d

12 = a + 2d(1)

Similarly a 50 = a + (50 − 1)d

106 = a + 49d(11)

On subtracting (I) from (II) we obtain
94 = 47d

d = 2

From equation (I) we obtain
12 = a + 2(2)

a = 12 − 4 = 8

a29 = a + (29 − 1)d

a29 = 8 + (28)2

a29 = 8 + 56 = 64

Therefore 29 th
term is 64.

Page : 106 , Block Name : Exercise 5.2

Q9 If the 3 and the 9th terms of an A.P. are 4 and Which term of this A.P. is zero. — 8
rd

respectively.

Answer. Given that,
a3 = 4

a9 = −8

We know that
an = a + (n − 1)d

a3 = a + (3 − 1)d

4 = a + 2d(1)

a9 = a + (9 − 1)d

−8 = a + 8d(11)

On subtracting equation (I )f rom(I I ), we obtain.
−12 = 6d

d = −2

From equation (I), we obtain

4 = a + 2(−2)

4 = a − 4

a = 8
th
Let n term of this A.P. be zero.

an = a + (n − 1)d

0 = 8 + (n − 1)(−2)

0 = 8 − 2n + 2

2n = 10

n = 5

Hence 5
th
term of this A.P. is 0.

Page : 106 , Block Name : Exercise 5.2

Q10 If the 3rd and the 9th terms of an AP are 4 and – 8 respectively, which term of this AP is
zero?

Answer. We know that,
For an A. P . , an = a + (n − 1)d

a17 = a + (17 − 1)d

a17 = a + 16d

Similarly a 10 = a + 9d

It is given that

a17 − a10 = 7

(a + 16d) − (a + 9d) = 7

7d = 7

d = 1

Therefore the common difference is 1

Page : 106 , Block Name : Exercise 5.2

Q11 Which term of the A.P. 3, 15, 27, 39, . will be 132 more than its 54th term?

Answer. Given A.P. is 3, 15, 27, 39, …
a = 3

d = a2 − a1 = 15 − 3 = 12

a54 = a + (54 − 1)d

= 3 + (53)(12)

= 3 + 636 = 639

132 + 639 = 771

We have to nd out the term of this A.P. which is 771.
th
Let n term be 771.

an = a + (n − 1)d

771 = 3 + (n − 1)12

768 = (n − 1)12

(n − 1) = 64

n = 65

th th
Therefore, 65 term was 132 more than 54 term.

Alternatively,
th th
Let n term be 132 more than 54 term.

132
n = 54 +
12

th
= 54 + 11 = 65 term

Page : 106 , Block Name : Exercise 5.2

Q12 Two APs have the same common difference. The difference between their 100th terms is
100, what is the difference between their 1000th terms?

Answer. Let the rst term of these A.P.s be al and a2 respectively and the common difference
of these A.P.s be d.
For rst A.P.

a100 = a1 + (100 − 1)d

= a1 + 99d

a1000 = a1 + (1000 − 1)d

a1000 = a1 + 999d

For second A.P.

a100 = a2 + (100 − 1)d

= a2 + 99d

a1000 = a2 + (1000 − 1)d

= a2 + 999d

Given that,difference between
th
100 term of these A.P.s = 100

Therefore, (a1 + 99d) − (a2 + 99d) = 100

a1 − a2 = 100(1)
th
Difference between 1000 terms of these A.P.s

(a1 + 999d) − (a2 + 999d) = a1 − a2

From equation (1)

This difference, a1 − a2 = 100

Hence the difference between 1000 th
terms of these A.P. will be 100.

Page : 106 , Block Name : Exercise 5.2

Q13 How many three-digit numbers are divisible by 7?

Answer. First three-digit number that is divisible by 7= 105
Next number = 105 + 7 = 112
Therefore 105, 112, 119, …
All are three digit numbers which are divisible by 7 and thus, all these are terms of an A.P.
having rst term as 105 and common difference as 7. The maximum possible three-digit
number is 999. When we divide it
by 7, the remainder will be 5. Clearly, 999 — 5 = 994 is the maximum possible three-digit
number that is divisible by 7.
The series is as follows.
105, 112, 119, …., 994
Let 994 be the nth term of this A.P.

a = 105

d = 7

an = 994

n =?

an = a + (n − 1)d

994 = 105 + (n − 1)7

889 = (n − 1)7

n = 128

Therefore 128 three digit numbers are divisible by 7.

Page : 106 , Block Name : Exercise 5.2

Q14 How many multiples of 4 lie between 10 and 250?
Answer. First multiple of 4 that is greater than 10 is 12. Next will be 16.
Therefore, 12, 16, 20, 24,
All these are divisible by 4 and thus, all these are terms of an A.P. with rst term as 12 and
common difference as 4.
When we divide 250 by 4, the remainder will be 2. Therefore, 250 — 2 = 248 is divisible by 4.
The series is as follows.
12, 16, 20, 24, …., 248
Let 248 be the nth term of this AP.
a = 12

d = 4

an = 248

an = a + (n − 1)d

248 = 12 + (n − 1)4

236
= n − 1
4

n = 60

Therefore there are 60 multiples of 4 between 10 and 250

Page : 106 , Block Name : Exercise 5.2

Q15 For what value of n, are the nth terms of two APs: 63, 65, 67, . . . and 3, 10, 17, . . . equal?

63, 65, 67, …

a = 63

Answer. d = a 2 − a1 = 65 − 63 = 2

th
n term of this A.P. = an = a + (n − 1)d

an = 63 + (n − 1)2 = 63 + 2n − 2

an = 61 + 2n(1)

3, 10, 17, …

a = 3

d = a2 − a1 = 10 − 3 = 7

th
n term of this A.P. = 3 + (n − 1)7

an = 3 + 7n − 7

an = 7n − 4(2)

It is given that, nth term of these A.P.s are equal to each other. Equating both these equations,
we obtain
61 + 2n = 7n − 4

61 + 4 = 5n

5n = 65

n = 13

Therefore 13 th
terms of both these A.P.s are equal to each other.

Page : 106 , Block Name : Exercise 5.2

Q16 Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12

= a3 = 16

a + (3 − 1)d = 16

a + 2d = 16(1)

Answer. a 7 − a5 = 12

[a + (7 − 1)d] − [a + (5 − 1)d] = 12

(a + 6d) − (a + 4d) = 12

2d = 12

d = 6

From equation (1), we obta in

a + 2(6) = 16

a + 12 = 16

a = 4

Therefore, A.P. will be

4, 10, 16, 22, …

Page : 106 , Block Name : Exercise 5.2

Q17 Find the 20th term from the last term of the AP : 3, 8, 13, . . ., 253.

Answer. Given A.P is

a = 253

d = 248 − 253 = −5

n = 20

a20 = a + (20 − 1)d

a20 = 253 + (19)(−5)

a20 = 253 − 95

a = 158

Therefore 20th term from last term is 158

Page : 107 , Block Name : Exercise 5.2

Q18 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 rst three terms of the AP.

We know that,

an = a + (n − 1)d

a4 = a + (4 − 1)d

a4 = a + 3d
Answer.
Similarly,

a8 = a + 7d

a6 = a + 5d

a10 = a + 9d

Given that, a4 + a8 = 24

a + 3d + a + 7d = 24

2a + 10d = 24

a + 5d = 12(1)

a6 + a10 = 44

a + 5d + a + 9d = 44

2a + 14d = 44

a + 7d = 22(2)

On subtracting equation (1) from (2), we obtain

2d = 22 − 12

2d = 10

d = 5

From equation (1), we obtain

a + 5d = 12

a + 5(5) = 12

a + 25 = 12

a = −13

a2 = a + d = −13 + 5 = −3

a3 = a2 + d = −8 + 5 = −3

Therefore, the first three terms of this A.P. are − 13, −8, and − 3 .

Page : 107 , Block Name : Exercise 5.2

Q19 Subba Rao started work in 1995 at an annual salary of
5000 and receivedan ∈ crementof 200 each year. In which year did his income reach `

7000?

Answer. It can be observed that the incomes that Subba Rao obtained in various years are in
A.P. as every year, his salary is increased by Rs 200.
Therefore, the salaries of each year after 1995 are 5000, 5200, 5400,
Here, a = 5000
d = 200
Let after nth year, his salary be Rs 7000.
Therefore, an a = a + (n − 1)d
n

7000 = 5000 + (n - 1) 200
200(n - 1) = 2000
(n - 1) = 10
n = 11
Therefore, in 11th year, his salary will be Rs 7000.

Page : 107 , Block Name : Exercise 5.2

Q20 Ramkali saved ₹ 5 in the rst week of a year and then increased her weekly savings by
₹1.75. If in the nth week, her weekly savings become ₹20.75, nd n.

Answer. Given that,
a = 5

d = 1.75

an = 20.75

n =?

an = a + (n − 1)d

20.75 = 5 + (n − 1)1.75

15.75 = (n − 1)1.75

15.75 1575
(n − 1) = =
1.75 175

63
= = 9
7

n − 1 = 9

n = 10

Hence, n is 10.

Page : 107 , Block Name : Exercise 5.2

Exercise 5.3

Q1 Find the sum of the following A.P.
(i) 2, 7, 12 ,. ..., to 10 terms.
(ii) - 37, 33 — 29 ,..., to 12 terms
(iii) 0.6, 1.7 28 ,.... to 100 terms
(iv) , , , … … , to 11 terms
1

15
1

12
1

10

Answer. (i) 2, 7, 12, … , to 10 terms
For this A.P
a = 2

d = a2 − a1 = 7 − 2 = 5

n = 10

We know that,
n
Sn = [2a + (n − 1)d]
2
10
S19 = [2(2) + (10 − 1)5]
2

= 5[4 + (9) × (5)]

= 5 × 49 = 245

(ii) −37, −33, −29, … , to 12 terms
For this A.P.
a = −37

d = a2 − a1 = (−33) − (−37)

= −33 + 37 = 4

n = 12

We know that,
n
Sn = [2a + (n − 1)d]
2

12
S12 = [2(−37) + (12 − 1)4]
2

= 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

d = a2 − a1 = 1.7 − 0.6 = 1.1

n = 100

We know that,
n
Sn = [2a + (n − 1)d]
2
100
S100 = [2(0.6) + (100 − 1)1.1]
2

= 50[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.
1
a =
15

n = 11

1 1
d = a2 − a1 = −
12 15

5 − 4 1
= =
60 60

We know that,
n
Sn = [2a + (n − 1)d]
2

11 1 1
S11 = [2 ( ) + (11 − 1) ]
2 15 60

11 2 10 11 4 + 5
= [ + ] = [ ]
2 15 60 2 30

11 2 1 11 4 + 5
= [ + ] = [ ]
2 15 6 2 30

11 2 1 11 4 + 5
= [ + ] = [ ]
2 15 6 2 30

11 9 33
= ( )( ) =
2 30 20

Page : 112 , Block Name : Exercise 5.3

Q2 Find the sums given below :
(i) 7 + 10 + 14 + … + 84
1

2

(ii) 54 + 32 + 30 + … + 10
(iii) −5 + (−8) + (−11) + … + (−230)

1

Answer. (i) 7 + 10
2 +14 + … … … + 84

For this A.P
a = 7

l = 84

1 21 7
d = a2 − a1 = 10 − 7 = − 7 =
2 2 2
th
Let 84 be the n term of this A.P.

I = a + (n − 1)d

7
84 = 7 + (n − 1)
2

7
77 = (n − 1)
2

22 = n − 1

n = 23

We know that
n
Sn = (a + l)
2
23
Sn = [7 + 84]
2
23 × 91 2093
= =
2 2
1
= 1046
2

(ii) 34 + 32 + 30 + … … … + 10
For this A.P.
a=34
d = a2 − a1 = 32 − 34 = −2

I = 10

th
Let 10 be the n term of this A.P.

I = a + (n − 1)d

10 = 34 + (n − 1)(−2)

−24 = (n − 1)(−2)

12 = n − 1

n = 13
n
Sn = (a + l)
2

13
= (34 + 10)
2

13×44
= = 13 × 22
2

13×44
= = 13 × 22
2

= 286

(iii) (−5) + (−8) + (−11) + … … … + (−230)
For this A.P.
a = −5

l = −230

d = a2 − a1 = (−8) − (−5)

= −8 + 5 = −3
th
Let − 230 be the n term of this A.P.

I = a + (n − 1)d

−230 = −5 + (n − 1)(−3)

−225 = (n − 1)(−3)

(n − 1) = 75

n = 76
n
And, Sn = (a + l)
2
76
= [(−5) + (−230)]
2

= 38(−235)

= −8930

Page : 112 , Block Name : Exercise 5.3

Q3 In an AP :
(i) Given a = 5, d = 3, a n = 50, find n and Sn

(ii) Given a = 7, a = 35, find d and S
13 13

(iii) Given a = 37, d = 3, find a and S
12 12

(iv) Given a = 15, S = 125, find d and a
3 10 10

(v) Given d = 5, S = 75, find a and a
9 9

(vi) Given a = 2, d = 8, S = 90, find n and a
n n

(vii) Given a = 8, a = 62, s = 210, find n and d
n n

(viii) Given a = 4, d = 2, S = −14, find n and a
n n

(ix) Given a = 3, n = 8, S = 192, find d
(x) Given I = 28, S = 144 and there are total 9 nd a

Answer. (i) Given that a = 5, d = 3, a n = 50

As an = a + (n − 1)d

∴ 50 = 5 + (n − 1)3

45 = (n − 1)3

15 = n − 1

n = 16
n
Sn = [a + an ]
2
16
S16 = [5 + 50]
2

= 8 × 55

= 440

(ii) Given that a = 7, a 13 = 35

As an = a + (n − 1)dr

∴ a13 = a + (13 − 1)d

35 = 7 + 12d

35 − 7 = 12d

28 = 12d

7
d =
3
n
Sn = [a + an ]
2
n
S13 = [a + a13 ]
2

13
= [7 + 35]
2

13 × 42
= = 13 × 21
2

= 273

(iii) Given that a 12 = 37, d = 3

As an = a + (n − 1)d

a12 = a + (12 − 1)3

37 = a + 33

a = 4
n
Sn = [a + an ]
2

12
Sn = [4 + 37]
2

Sn = 6(41)

Sn = 246

(iv) Given that a 3 = 15, S10 = 125

As an = a + (n − 1)d

a3 = a + (3 − 1)d

15 = a + 2d(i)
n
Sn = [2a + (n − 1)d]
2

10
S19 = [2a + (10 − 1)d]
2

125 = 5(2a + 9d)

25 = 2a + 9d

On multiplying equation (I) by 2, we obtain
30 = 2a + 4d ...(iii)
on subtracting equation (iii) from (ii), we obtain
−5 = 5d

d = −1

From equation (

15 = a + 2(−1)

15 = a − 2

a = 17

a10 = a + (10 − 1)d

a10 = 17 + (9)(−1)

a10 = 17 − 9 = 8

(v) Given that d = 5, S 9 = 75
n
Sn = [2a+(n−1)d]
As 2

9
S0 = [2a + (9 − 1)5]
2

9
75 = (2a + 40)
2

25 = 3(a + 20)

25 = 3a + 60

3a = 25 − 60

−35
a =
3

an = a + (n − 1)d

a9 = a + (9 − 1)(5)

−35
= + 8(5)
3

−35
= + 40
3

−35+120 85
= =
3 3

(vi) Given that a = 2, d = 8, S n = 90
n
As Sn = [2a + (n − 1)d]
2

n
90 = [4 + (n − 1)8]
2

90 = n[2 + (n − 1)4]

90 = n[2 + 4n − 4]

2
90 = n(4n − 2) = 4n − 2n
2
4n − 2n − 90 = 0

2
4n − 20n + 18n − 90 = 0

4n(n − 5) + 18(n − 5) = 0

(n − 5)(4n + 18) = 0

Either n − 5 = 0 or 4n + 18 = 0
18
n=− −9
n = 5 or 4 =
2

However, n can neither be negative nor fractional.
Therefore, n = 5

an = a + (n − 1)d

a5 = 2 + (5 − 1)8

= 2 + (4)(8)

= 2 + 32 = 34

(vii) Given that a = 8, a n = 62, Sn = 210
n
Sn = [a + an ]
2

n
210 = [8 + 62]
2

n
210 = (70)
2

n = 6

an = a + (n − 1)d

62 = 8 + (6 − 1)d

62 − 8 = 5d

54 = 5d

54
a =
5

(viii) Given that a n = 4, d = 2, Sn = −14

an = a + (n − 1)d

4 = a + (n − 1)2

4 = a + 2n − 2

a + 2n = 6

a = 6 − 2n(i)
n
Sn = [a + an ]
2

n
−14 = [a + 4]
2

−28 = n(a + 4)

−28 = n(6 − 2n + 4){ From equation (i)}

−28 = n(−2n + 10)

2
−28 = −2n + 10n

2
2n − 10n − 28 = 0

2
n − 5n − 14 = 0

2
n − 7n + 14 = 0

2
n − 7n + 2n − 14 = 0

n(n − 7)(n + 2) = 0

Either n − 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 obtain

a = 6 − 2n

a = 6 − 2(7)

= 6 − 14

= −8

(ix) Given that a = 3, n = 8, S = 192
n
Sn = [2a + (n − 1)d]
2

8
192 = [2 × 3 + (8 − 1)d]
2

192 = 4[6 + 7d]

48 = 6 + 7d

42 = 7d

d = 6

(x) Given that I = 28, S = 144 and there are total of 9 terms

n
Sn = (a + l)
2

9
144 = (a + 28)
2

(16) × (2) = a + 28

32 = a + 28

a = 4

Page : 112 , Block Name : Exercise 5.3

Q4 How many terms of the AP : 9, 17, 25, . . . must be taken to give a sum of 636?

Answer. Let there be n terms of this A.P.
For this A.P., a = 9

d = a2 − a1 = 17 − 9 = 8
n
Sn = [2a + (n − 1)d]
2

n
636 = [2 × a + (n − 1)8]
2

n
636 = [18 + (n − 1)8]
2

636 = n[9 + 4n − 4]

636 = n(4n + 5)

2
4n + 5n − 636 = 0

2
4n + 53n − 48n − 636 = 0

n(4n + 53) − 12(4n + 53) = 0

(4n + 53)(n − 12) = 0

Either 4n + 53 = 0 or n − 12 = 0
−53
n = or n = 12
4

n cannot be 4 . As the number of terms can neither be negative nor fractional, therefore, n = 12
only.

Page : 113 , Block Name : Exercise 5.3

Q5 The rst 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.

Answer. Given that ,
a = 5

I = 45

Sn = 400
n
Sn = (a + l)
2

n
400 = (5 + 45)
2

n
400 = (50)
2

n = 16

I = a + (n − 1)d

45 = 5 + (16 − 1)d

40 = 15d

40 8
d = =
15 3

Page : 113 , Block Name : Exercise 5.3

Q6 The rst 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?

Answer. Given that
a = 17

l = 350

d = 9

Let there be n terms in the A.P.

I = a + (n − 1)d

350 = 17 + (n − 1)9

333 = (n − 1)9

(n − 1) = 37

n = 38

n
Sn = (a + l)
2

38
⇒ Sn = (17 + 350) = 19(367) = 6973
2

Thus this A.P. contains 38 terms and the terms of this A.P. is 6973

Page : 113 , Block Name : Exercise 5.3

Q7 Find the sum of rst 22 terms of an AP in which d = 7 and 22nd term is 149

d = 7

a22 = 149

S22 =?

an = a + (n − 1)d
Answer.
a22 = a + (22 − 1)d

149 = a + 21 × 7

149 = a + 147

a = 2

n
Sn = (a + an )
2
22
= (2 + 149)
2

= 11(151) = 1661

Page : 113 , Block Name : Exercise 5.3

Q8 Find the sum of rst 51 terms of an AP whose second and third terms are 14 and 18
respectively

Answer. Given that
a2 = 14

a3 = 18

d = a3 − a2 = 18 − 14 = 4

a2 = a + d

14 = a + 4

a = 10
n
Sn = [2a + (n − 1)d]
2
51
S51 = [2 × 10 + (51 − 1)4]
2
51
= [20 + (50)(4)]
2

51(220)
= = 51(110)
2

=5610

Page : 113 , Block Name : Exercise 5.3

Q9 If the sum of rst 7 terms of an AP is 49 and that of 17 terms is 289, nd the sum of rst n
terms.

Answer. Given that,
S7 = 49

S17 = 289
n
Sn = [2a + (n − 1)d]
2

7
S7 = [2a + (7 − 1)d]
2

79 = (a + 3d)

a + 3d = 7(i)

17
S17 = [2a + (17 − 1)d]
2

17
289 = [2a + 16d]
2

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
n
Sn = [2a + (n − 1)d]
2
n
= [2(1) + (n − 1)(2)]
2
n
= (2 + 2n − 2)
2
n
= (2n)
2
2
=n

Page : 113 , Block Name : Exercise 5.3

Q10 Show that a , a … , a , … form an AP where an is de ned as below
1 2 n

(i) a = 3 + 4n
n

(li) a = 9 − 5n
n

Also nd the sum of the rst 15 terms in each case.

(i)an = 3 + 4n

a1 = 3 + 4(1) = 7

Answer. a 2 = 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

, ak+1 − ak is same every time. Therefore, this is an AP with common difference as 4 and rst
term as 7.

n
Sn = [2a + (n − 1)d]
2
15
S15 = [2(7) + (15 − 1)4]
2
15
= [(14) + 56]
2
15
= (70)
2

=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

(ii) a 4 = 9 − 5 × 4 = 9 − 20 = −11

It can be observed that

a2 − a1 = −1 − 4 = −5

a3 − a2 = −1 − 4 = −5

a3 − a2 = −11 − (−1) = −5

ie , a
⋅ k+1 − a is same every time. Therefore, this is an A.P. with common difference as —5 and
k

rst term as 4.
n
Sn = [2a + (n − 1)d]
2
15
S15 = [2(4) + (15 − 1)(−5)]
2
15
= [8 + 14(−5)]
2
15
= (8 − 70)
2
15
= (−62) = 15(−31)
2

= − 465

Page : 113 , Block Name : Exercise 5.3

Q11 If the sum of the rst n terms of an AP is 4n − n , what is the rst term (that is S )?
2
1

What is the sum of rst two terms? What is the second term? Similarly nd the 3rd, the 10 th

and the nth terms.

Answer. Given that

2
Sn = 4n − n

2
First term, a = S1 = 4(1) − (1) = 4 − 1 = 3

Sum of first two terms = S2

2
= 4(2) − (2) = 8 − 4 = 4

Second term, a2 = S2 − S1 = 4 − 3 = 1

d = a2 − a = 1 − 3 = −2

an = a + (n − 1)d

= 3 + (n − 1)(−2)

= 3 − 2n + 2

= 5 − 2n

Therefore, a3 = 5 − 2(3) = 5 − 6 = −1

a10 = 5 − 2(10) = 5 − 20 = −15

th th
Hence, the sum of first two terms is 4. The second term is 1.3 , 10

th
and n terms are − 1, −15, and 5 − 2n respectively.

Page : 113 , Block Name : Exercise 5.3

Q12 Find the sum of the rst 40 positive integers divisible by 6.

Answer. The positive integers that are divisible by 6 are
6, 12, 18, 24
It can be observed that these are making an AP. whose rst term is 6 and common difference is
6.
a = 6

d = 6

S40 =?
n
Sn = [2a + (n − 1)d]
2

40
S40 = [2(6) + (40 − 1)6]
2

= 20[12 + (39)(6)]

= 20(12 + 234)

= 20 × 246

= 4920

Page : 113 , Block Name : Exercise 5.3

Q13 Find the sum of the rst 15 multiples of 8.

Answer. The multiples of 8 are
8, 16, 24, 32..
These are in an A.P., having rst term as 8 and common difference as 8.

Therefore, a = 8
d=8
S15 =?
n
Sn = [2a + (n − 1)d]
2

15
= [2(8) + (15 − 1)8]
2
15
= [16 + 14(8)]
2
15
= (16 + 14(8)]
2

15(128)
= = 15 × 64
2

=960

Page : 113 , Block Name : Exercise 5.3

Q14 Find the sum of the odd numbers between 0 and 50.

Answer. 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

I = 49

I = a + (n − 1)d

49 = 1 + (n − 1)2

48 = 2(n − 1)

n − 1 = 24

n = 25

n
Sn = (a + l)
2
25
S25 = (1 + 49)
2

25(50)
= = (25)(25)
2

=625

Page : 113 , Block Name : Exercise 5.3

Q15 A contract on construction job speci es a penalty for delay of completion beyond a certain
date as follows: rs 200 for the rst 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?

Answer. It can be observed that these penalties are in an A.P. having rst 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(200) + (30 − 1)50]
2

= 15[400 + 1450]

= 15(180)

= 27750

Therefore , the contractor has to pay Rs 27750 as penalty

Page : 113 , Block Name : Exercise 5.3

Q16 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, nd the value
of each of the prizes

Answer. Let the cost of 1 prize be P.
st

Cost of 2nd prize = P — 20
And cost of 3rd prize = P — 40
It can be observed that the cost of these prizes are in an A.P. having common difference as —
20 and rst term as P.
a=P
d = —20
Given that S = 700
7

7
[2a + (7 − 1)d] = 700
2

[2a+(6)(−20)]
= 100
2

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.

Page : 113 , Block Name : Exercise 5.3

Q17 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?

Answer. 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
n
Sn = [2a + (n − 1)d]
2

12
S12 = [2(1) + (12 − 1)(1)]
2

= 6(2 + 11)

= 6(13)

= 78

Therefore, number of trees planted by I section of the classes = 78
Number of trees planted by 3 sections of the classes = 3 x 78 = 234
Therefore, 234 trees will be planted by the students.

Page : 113 , Block Name : Exercise 5.3

Q18 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 Fig. 5.4. What is the
total length of such a spiral made up of thirteen consecutive semicircles( Take π = 22

7
)

Semi-perimeter of circle = nr
π
I1 = n(0.5) = cm
Answer.
2

I2 = n(1) = ncm

3π
I3 = n(1.5) = cm
2

Therefore, I , I , I , , i.e. the lengths of the semi-circles are in an A.P.,
1 2 3

π 3π
, π, , 2π, … … …
2 2

π
a =
2

π π
d = π − =
2 2

S13 =?

We know that the sum of n terms of an a A.P. is given by
n
Sn = [2a + (n − 1)d]
2
13 π π
= [2 ( ) + (13 − 1) ( )]
2 2 2

13 12π
= [π + ]
2 2

13
= ( ) (7π)
2

91π
=
2
91 × 22
= = 13 × 11
2 × 7

=143
Therefore, the length of such spiral of thirteen consecutive semi-circles will be 143 cm.

Page : 113 , Block Name : Exercise 5.3

Q19 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 Fig. 5.5). In how many rows are the 200 logs placed
and how many logs are in the top row?

Answer. 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 placed in n rows.

Sn = 200
n
Sn = [2a + (n − 1)d]
2

n
200 = [2(20) + (n − 1)(−1)]
2

400 = n(40 − n + 1)

400 = n(41 − n)

2
400 = 41n − n

2
n − 41n + 400 = 0

2
n − 16n − 25n + 400 = 0

n(n − 16) − 25(n − 16) = 0

(n − 16)(n − 25) = 0

Either (n − 16) = 0 or n − 25 = 0

n = 16 or n = 25

an = a + (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 25th row is
negative, which is not possible.
Therefore, 200 logs can be placed in 16 rows and the number of logs in the 16th row is 5.

Page : 114 , Block Name : Exercise 5.3

Q20 In a potato race, a bucket is placed at the starting point, which is 5 m from the rst
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 rst potato and the second potato, the total distance (in metres) run by a
competitor is 2 × 5 + 2 × (5 + 3)]

Answer.

The distances of potatoes are as follows.
5, 8, 11, 14...
It can be observed that these distances are in A.P.
d = 8-5=3
n
Sn = [2a + (n − 1)d]
2

10
S10 = [2(5) + (10 − 1)3]
2

= 5[10 + 9 × 3]

= 5(10 + 27) = 5(37)

= 185

As every time she has to run back to the bucket, therefore, the total distance that the
competitor has to run will be two times of it. Therefore, total distance that the competitor will
run = 2 x 185 = 370 m

Page : 114 , Block Name : Exercise 5.3

Exercise 5.4

Q1 Which term of the AP : 121, 117, 113, . . ., is its rst negative term?
[Hint : Find n for a < 0]
n

Answer.
Given A.P. 121, 117, 113 …

a = 121

d = 117 − 121 = −4

an = a + (n − 1)d

= 121 + (n − 1)(−4)

= 121 − 4n + 4

= 125 − 4n

We have to nd the rst negative term of this A.P.

Therefore, an < 0

125 − 4n < 0

125 < 4n

125
n >
4

n > 31.25

Therefore, 32 nd
term will be the rst negative term of this A.P.

Page : 115 , Block Name : Exercise 5.4 (Optional)

Q2 The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the
sum of rst sixteen terms of the AP.

Answer. We know that
an = a + (n − 1)d

a3 = a + (3 − 1)d

a3 = a + 2d

Similarly, a7 = a + 6d

Given that, a3 + a7 = 6

(a + 2d) + (a + 6d) = 6

2a + 8d = 6

a + 4d = 3

a = 3 − 4d(i)

Also, it is given that (a3 ) × (a7 ) = 8

(a + 2d) × (a + 6d) = 8

From equation (i)

(3 − 4d + 2d) × (3 − 4d + 6d) = 8

(3 − 2d) × (3 + 2d) = 8

2
9 − 4d = 8

2
4d = 9 − 8 = 1
2 1
d =
4

1
d = ±
2

1 1
d = or −
2 2

From equation ( i

1
( When d is )
2

a = 3 − 4d

1
a = 3 − 4( )
2

= 3 − 2 = 1

1
( When d is − )
2

1
a = 3 − 4 (− )
2

1
a = 3 − 4 (− )
2

a = 3 + 2 = 5
n
Sn = [2a(n − 1)d]
2

1
( When a is 1 and d is )
2

16 1
S16 = [2(1) + (16 − 1) ( )]
2 2

15
= 8 [2 + ]
2

= 4(19) = 76

1
( When a is 5 and d is − )
2

16 1
S16 = [2(5) + (16 − 1) (− )]
2 2

1
= 8 [10 + (15) (− )]
2

5
= 8( )
2

=20

Page : 115 , Block Name : Exercise 5.4 (Optional)

Q3 A ladder has rungs 25 cm apart. (See gure). The rungs decrease uniformly in length from
45 cm at the bottom to 25 cm at the top. If the top and bottom rungs are 2 m apart, what is
1

2

the length of the wood required for the rungs?
[Hint: number of rungs = ] 250

25

Answer. It is given that the rungs are 25 cm apart and the top and bottom rungs are 2 1

2
m

apart.
Total number of rungs
1
2 ×100
2 250
= + 1 = + 1 = 11
25 25

Now, as the lengths of the rungs decrease uniformly, they will be in an A.P
The length of the wood required for the rungs equals the sum of all the terms of this A.P.
terms of this A.P.

First term, a = 45

Last term, I = 25

n = 11
n
Sn = (a + l)
2

11 11
∴ S10 = (45 + 25) = (70) = 385cm
2 2

Therefore the length of the wood required for the rungs is 385.

Page : 115 , Block Name : Exercise 5.4 (Optional)

Q4 The houses of a row are number consecutively from 1 to 49. Show that there is a value of x
such that the sum of numbers of the houses preceding the house numbered x is equal to the
sum of the number of houses following it.
Find this value of x.
[HintSx−1 = S49 − Sx ]

Answer. The number of houses was 1,2,3 …. 49
It can be observed that the number of houses are in A.P having a as 1 and d also as 1
Let us assume that the number of x house was like this.
th

We know that,

n
Sum of n terms in an A.P. = [2a + (n − 1)d]
2

th
Sum of number of houses preceding x house = Sx−1

(x−1)
= [2a + (x − 1 − 1)d]
2

x−1
= [2(1) + (x − 2)(1)]
2

x−1
= [2 + x − 2]
2

(x)(x−1)
=
2

th
Sum of number of houses following x house = S49 − Sx

49 x
= [2(1) + (49 − 1)(1)] − [2(1) + (x − 1)(1)]
2 2

49 x
= (2 + 49 − 1) − (2 + x − 1)
2 2

49 x
= ( ) (50) − (x + 1)
2 2

x(x+1)
= 25(49) −
2

It is given that these sums are equal to each other.

x(x−1) x+1
= 25(49) − x ( )
2 2

2 2
x x x x
− = 1225 − −
2 2 2 2

2
x = 1225

x = ±35

However, the house numbers are positive integers.

The value of x will be 35 only Therefore, house number 35 is such that the sum of the numbers
of houses preceding the house numbered 35 is equal to the sum of the numbers of the houses
following it.

Page : 115 , Block Name : Exercise 5.4 (Optional)

Q5 A small terrace at a football ground comprises of 15 steps each of which is 50 m long and
built of solid concrete.
Each step has a rise of ¼ m and a tread of ½ m (See gure) calculate the total volume of
concrete required to build the terrace.

Answer.
From the gure it can be observed that
st
1 step is 2m wide,

nd
2 step is 1m wide,

rd 3
3 step is m wide.
2

Therefore, the width of each step is increasing by 1/2m each time whereas their height 4m and
length 50 m remains the same.
Therefore, the Widths of these steps are , 1, , 2, … 1

2
3

2

Volume of concrete in 1
st 1 1 25
step = × × 50 =
4 2 4

Volume of concrete in 2 nd
step =
1

4
× 1 × 50 =
25

2

Volume of concrete in 3rd step 3 rd 1 3 75
step = × × 50 =
4 2 4

It can be observed that the volumes of concrete in these steps are in an A.P.

25 25 75
, , ,…
4 2 4

25
a =
4

25 25 25
d = − =
2 4 4

n
and Sn = [2a + (n − 1)d]
2

15 25 25
S15 = [2 ( ) + (15 − 1) ]
2 4 4

15 25 (14)25
= [ + ]
2 2 4

15 25 175
= [ + ]
2 2 2

15
= (100) = 750
2

Volume of concrete required to build the terrace is 750m 3

Page : 115 , Block Name : Exercise 5.4 (Optional)