Lesson: Arithmetic Progressions

Exercise 5.1

Question: 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 Rs.

15 for the first km and Rs. 8 for each additional

km.

(ii) The amount of air present in a cylinder when a

vacuum pump removes of the air remaining

1

4

in the cylinder at a time.

(iii) The cost of digging a well after every metre of

digging, when it costs Rs. 150 for the first metre

and rises by Rs. 50 for each subsequent metre.

(iv) The amount of money in the account every year,

when Rs. 10,000 is deposited at compound

interest at per annum.

8%

Solution

(i) For the first km, taxi fare

15 Rs

For the first two km, taxi fare

15 8

23

For the first three km, taxi fare

23 8

31

For the first four km, taxi fare

31 8

39

Thus, the sequence is .

15,

23,

31,

39,...

Common difference for each pair of terms is .

8

Clearly, the common difference is the same for

each pair of terms.

Hence, the list of numbers make an arithmetic

progression.

(ii) Let V litres be the volume of air in the cylinder.

According to the question,

In first stroke, the part of air remaining

4

v

v

4

4

v v

3

4

v

In second stroke, the remaining part of air

3 3 1

4 4 4

v v

3 1

1

4 4

v

3 3

4 4

v

2

3

4

v

In third stroke, the remaining part of air

2 2

3 3 1

4 4 4

v v

2

3 1

1

4 4

v

2

3 3

4 4

v

3

3

4

v

Thus, the sequence is

3

,

4

v

2

3

,

4

v

3

3

,...

4

v

Clearly, the common difference is not the same

for each pair of terms.

Hence, the list of numbers does not make an

arithmetic progression.

(iii) The cost involved in digging first metre

150

The cost involved in digging first two metres

150 50

200

The cost involved in digging first three metres

200 50

250

Thus, the sequence is .

150,

200,

250,...

Common difference for each pair of terms is 50.

Clearly, the common difference is same for each

pair of terms.

Hence, the list of numbers makes an arithmetic

progression.

(iv) The formula for compound interest is as follows:

Compound interest

1

100

n

r

p

Where, is the principal amount, is the rate

p

r

of interest, and is the number of years.

n

According to the question,

Interest in the first year

8

10000 1

100

Interest in the second year

2

8

10000 1

100

Interest in the third year

3

8

10000 1

100

Thus, the sequence is

8

1 ,

100

2

8

10000 1 ,

100

3

8

10000 1 ,...

100

Clearly, the common difference is not same for

each pair of terms.

Hence, the list of numbers does not make an

arithmetic progression.

Question: 2

Write first four terms of the AP, when the first term

a

and the common difference are given as follows:

d

(i)

10,a

10d

(ii)

2, a

0d

(iii)

4,a

3 d

(iv)

1, a

1

2

d

(v)

1.25, a

0.25 d

Solution

(i) Given:

10,a

10d

Let the sequence is .

1

,a

2

,a

3

,a

4

,...a

1

10 a a

2 1

a a d

10 10

20

3 2

a a d

20 10

30

4 3

a a d

30 10

40

Thus, the first four terms of the sequence are 10,

20, 30, 40.

(ii) Given:

2, a

0d

Let the sequence is .

1

,a

2

,a

3

,a

4

,...a

1

a a

2

2 1

a a d

2 0

2

3 2

a a d

2 0

2

4 3

a a d

2 0

2

Thus, the first four terms of the sequence are -2,

-2, -2, -2.

(iii) Given:

4,a

3 d

Let the sequence is .

1

,a

2

,a

3

,a

4

,...a

1

a a

4

2 1

a a d

4 3

1

3 2

a a d

1 3

2

4 3

a a d

2 3

5

Thus, the first four terms of the sequence are 4,

1, -2, -5.

(iv) Given:

1, a

1

2

d

Let the sequence is .

1

,a

2

,a

3

,a

4

,...a

1

a a

1

2 1

a a d

1

1

2

1

2

3 2

a a d

1 1

2 2

0

4 3

a a d

1

0

2

1

2

Thus, the first four terms of the sequence are

.

1,

1

,

2

0,

1

2

(v) Given:

1.25, a

0.25 d

Let the sequence is

1

,a

2

,a

3

,a

4

,...a

1

a a

1.25

2 1

a a d

1.25 0.25

1.50

3 2

a a d

1.50 0.25

1.75

4 3

a a d

1.75 0.25

2.00

Thus, the first four terms of the sequence are

.

1.25,

1.50,

1.75,

2.00

Question: 3

For the following APs, write the first term and the

common difference:

(i)

3,

1,

1,

3,...

(ii)

5,

1,

3,

7,...

(iii)

1

,

3

5

,

3

9

,

3

13

,...

3

(iv)

0.6,

1.7,

2.8,

3.9,...

Solution

(i) Given AP:

3,

1,

1,

3,...

Compare the given AP with

,a

,a d

2 ,a d

3 ,....a d

First term .

3a

Now, calculate the common difference.

d a d a

1 3

2

Hence, the first term is and the common

3

difference is .

2

(ii) Given AP:

5,

1,

3,

7,...

Compare the given AP with

,a

,a d

2 ,a d

3 ,....a d

First term .

5 a

Now, calculate the common difference.

d a d a

1 5

4

Hence, the first term is and the common

5

difference is .

4

(iii) Given AP:

1

,

3

5

,

3

9

,

3

13

,...

3

Compare the given AP with

,a

,a d

2 ,a d

3 ,....a d

First term .

1

3

a

Now, calculate the common difference.

d a d a

5 1

3 3

4

3

Hence, the first term is and the common

1

3

difference is .

4

3

(iv) Given AP:

0.6,

1.7,

2.8,

3.9,...

Compare the given AP with

,a

,a d

2 ,a d

3 ,...a d

First term .

0.6a

Now, calculate the common difference.

d a d a

1.7 0.6

1.1

Hence, the first term is and the common

0.6

difference is .

1.1

Question: 4

Which of the following are APs? If they form an AP,

find the common difference and write three more

d

terms.

(i)

2,

4,

8,

16,...

(ii)

2,

5

,

2

3,

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

,

2

1

,

2

1

,

2

1

,

2

,...

(ix)

1,

3,

9,

27,...

(x)

,a

2 ,a

3 ,a

4 ,...a

(xi)

,a

2

,a

3

,a

4

,...a

(xii)

2,

8,

18,

32,...

(xiii)

3,

6,

9,

12,...

(xiv)

2

1 ,

2

3 ,

2

5 ,

2

7 ,...

(xv)

2

1 ,

2

5 ,

2

7 ,

73,...

Solution

(i) Given:

2,

4,

8,

16,...

Compare the sequence with AP

1

,a

2

,a

3

,a

..., .

n

a

1

2,a

2

4,a

3

8,a

4

16a

Now, find the common difference.

2 1

4 2 a a

2

3 2

8 4 a a

4

4 3

16 8 a a

8

Clearly, the common difference is not same for

each pair of terms.

Hence, the given sequence does not form an AP.

(ii) Given:

2,

5

,

2

3,

7

,...

2

Compare the sequence with AP

1

,a

2

,a

3

,a

..., .

n

a

.

1

2,a

2

5

,

2

a

3

3,a

4

7

2

a

Now, find the common difference.

2 1

5

2

2

a a

5 4

2

1

2

3 2

5

3

2

a a

6 5

2

1

2

4 3

7

3

2

a a

7 6

2

1

2

Clearly, the common difference is same for each

pair of terms.

Thus, the given sequence forms an AP with the

common difference .

1

2

d

The next three terms will be:

5 4

a a d

7 1

2 2

8

2

4

6 5

a a d

1

4

2

8 1

2

9

2

7 6

a a d

9 1

2 2

10

2

5

Hence, the next three terms are and .

4,

9

2

5

(iii) Given:

1.2,

3.2,

5.2,

7.2,...

Compare the sequence with AP

1

,a

2

,a

3

,a

..., .

n

a

1

1.2, a

2

3.2, a

3

5.2, a

4

7.2 a

Now, find the common difference.

2 1

3.2 1.2 a a

2

3 2

5.2 3.2 a a

2

Clearly, the common difference is same for each

pair of terms.

Thus, the given sequence forms an AP with the

common difference .

2 d

The next three terms will be:

5 4

a a d

7.2 2

9.2

6 5

a a d

9.2 2

11.2

7 6

a a d

11.2 2

13.2

Hence, the next three terms are

9.2,

11.2,

and

13.2

(iv) Given:

10,

6,

2,

2,...

Compare the sequence with AP

1

,a

2

,a

3

,a

..., .

n

a

1

10, a

2

6, a

3

2, a

4

2a

Now, find the common difference.

2 1

6 10 a a

4

3 2

2 6 a a

4

4 3

2 2 a a

4

Clearly, the common difference is same for each

pair of terms.

Thus, the given sequence forms an AP with the

common difference .

4d

The next three terms will be:

5 4

a a d

5

2 4 a

5

6a

6 5

a a d

6

6 4 a

6

10a

7 6

a a d

7

10 4 a

7

14a

Hence, the next three terms are and .

6,

10,

14

(v) Given:

3,

3 2,

3 2 2,

3 3 2,...

Compare the sequence with AP

1

,a

2

,a

3

,.a

..., .

n

a

1

3,a

2

3 2, a

3

3 2 2, a

4

3 3 2 a

Now, find the common difference.

2 1

3 2 3 a a

2

3 2

3 2 2 3 2 a a

2

4 3

3 3 2 3 2 2 a a

2

Clearly, the common difference is same for each

pair of terms.

Thus, the given sequence forms an AP with the

common difference .

2d

The next three terms will be:

5 4

a a d

5

3 3 2 2 a

5

3 4 2 a

6 5

a a d

6

3 4 2 2 a

6

3 5 2 a

7 6

a a d

7

3 5 2 2 a

7

3 6 2 a

Hence, the next three terms are

3 4 2,

and .

3 5 2,

3 6 2

(vi) Given:

0.2,

0.22,

0.222,

0.2222,...

Compare the sequence with AP

1

,a

2

,a

3

,a

..., .

n

a

1

0.2,a

2

0.22,a

3

0.222,a

4

0.2222a

Now, find the common difference.

2 1

0.22 0.2 a a

0.02

3 2

0.222 0.22 a a

0.002

4 3

0.2222 0.222 a a

0.0002

Clearly, the common difference is not same for

each pair of terms.

Hence, the given sequence does not form an AP.

(vii) Given:

0,

4,

8,

12,...

Compare the sequence with AP

1

,a

2

,a

3

,a

..., .

n

a

1

0,a

2

4, a

3

8, a

4

12 a

Now, find the common difference.

2 1

4 0 a a

4

3 2

8 4 a a

4

Clearly, the common difference is same for each

pair of terms.

Thus, the given sequence forms an AP with the

common difference .

4 d

The next three terms will be:

5 4

a a d

5

12 4 a

5

16 a

6 5

a a d

6

16 4 a

6

20 a

7 6

a a d

7

20 4 a

7

24 a

Hence, the next three terms are and

16,

20,

.

24

(viii) Given:

1

,

2

1

,

2

1

,

2

1

,...

2

Compare the sequence with AP

1

,a

2

,a

3

,a

..., .

n

a

1

1

,

2

a

2

1

,

2

a

3

1

,

2

a

4

1

2

a

Now, find the common difference.

2 1

1 1

2 2

a a

0

3 2

1 1

2 2

a a

0

4 3

1 1

2 2

a a

0

Clearly, the common difference is same for each

pair of terms.

Thus, the given sequence forms an AP with the

common difference .

0d

The next three terms will be:

5 4

a a d

5

1

0

2

a

5

1

2

a

6 5

a a d

6

1

0

2

a

6

1

2

a

7 6

a a d

7

1

0

2

a

7

1

2

a

Hence, the next three terms are and

1

,

2

1

,

2

.

1

2

(ix) Given:

1,

3,

9,

27,...

Compare the sequence with AP

1

,a

2

,a

3

,a

..... .

n

a

1

1,a

2

3,a

3

9,a

4

27a

Now, find the common difference.

2 1

3 1 a a

2

3 2

9 3 a a

6

4 3

27 9 a a

18

Clearly, the common difference is not same for

each pair of terms.

Hence, the given sequence does not form an AP.

(x) Given:

,a

2 ,a

3 ,a

4 ,...a

Compare the sequence with AP

1

,a

2

,a

3

,a

..., .

n

a

1

,a a

2

2 ,a a

3

3 ,a a

4

4a a

Now, find the common difference.

2 1

2 a a a a

a

3 2

3 2 a a a a

a

4 3

4 3 a a a a

a

Clearly, the common difference is same for each

pair of terms.

Thus, the given sequence forms an AP with the

common difference .

d a

The next three terms will be:

5 4

a a d

5

4 a a a

5

5a a

6 5

a a d

6

5 a a a

6

6a a

7 6

a a d

7

6 a a a

7

7a a

Hence, the next three terms are and .

5 ,a

6 ,a

7a

(xi) Given:

,a

2

,a

3

,a

4

,...a

Compare the sequence with AP

1

,a

2

,a

3

,a

..., .

n

a

1

,a a

2

2

,a a

3

3

,a a

4

4

a a

Now, find the common difference.

2

2 1

a a a a

1 a a

3 2

3 2

a a a a

2

1 a a

4 3

4 3

a a a a

3

1 a a

Clearly, the common difference is not same for

each pair of terms.

Hence, the given sequence does not form an AP.

(xii) Given:

2,

8,

18,

32,...

Compare the sequence with AP .

1

,a

2

,a

3

,.a

...,

n

a

1

2,a

2

8,a

3

18,a

4

32a

Now, find the common difference.

2 1

8 2 a a

2 2 2

2

3 2

18 8 a a

3 2 2 2

2

4 3

32 18 a a

4 2 3 2

2

Clearly, the common difference is the same for

each pair of terms.

Thus, the given sequence forms an AP with the

common difference .

2d

The next three terms will be:

5 4

a a d

5

32 2 a

5

4 2 2 a

5

5 2a

6 5

a a d

6

5 2 2 a

6

6 2a

7 6

a a d

7

6 2 2 a

7

7 2a

Hence, the next three terms are and

5 2,

6 2

.

7 2

(xiii) Given:

3,

6,

9,

12,...

Compare the sequence with AP

1

,a

2

,a

3

,a

..., .

n

a

1

3,a

2

6,a

3

9,a

4

12a

Now, find the common difference.

2 1

6 3 a a

3 2 3

3 2 1

3 2

9 6 a a

3 3 2

3 3 2

4 3

12 9 a a

2 3 3

3 2 3

Clearly, the common difference is not same for

each pair of terms.

Hence, the given sequence does not form an AP.

(xiv) Given:

2

1 ,

2

3 ,

2

5 ,

2

7 ,...

Compare the sequence with AP

1

,a

2

,a

3

,a

..., .

n

a

2

1

1 ,a

2

2

3 ,a

2

3

5 ,a

2

4

7a

Now, find the common difference.

2 2

2 1

3 1 a a

9 1

8

2 2

3 2

5 3 a a

25 9

16

2 2

4 3

7 5 a a

49 25

24

Clearly, the common difference is not same for

each pair of terms.

Hence, the given sequence does not form an AP.

(xv) Given:

2

1 ,

2

5 ,

2

7 ,

73,...

Compare the sequence with AP .

1

,a

2

,a

3

,a

...,

n

a

2

1

1 ,a

2

2

5 ,a

2

3

7 ,a

4

73a

Now, find the common difference.

2 2

2 1

5 1 a a

25 1

24

2 2

3 2

7 5 a a

49 25

24

2

4 3

73 7 a a

73 49

24

Clearly, the common difference is the same for

each pair of terms.

Thus, the given sequence forms an AP with the

common difference .

24d

The next three terms will be:

5 4

a a d

5

73 24 a

5

97a

6 5

a a d

6

97 24 a

6

121a

7 6

a a d

7

121 24 a

7

145a

Hence, the next three terms are and .

97,

121,

145

Exercise 5.2(20)

Question: 1

Fill in the blanks in the following table, given that is

a

the first term, is the common difference and is

d

n

a

the term of the AP:

th

n

a

d

n

n

a

(i)

7

3

8

…

(ii)

18

…

10

0

(iii)

…

3

18

5

(iv)

18.9

2.5

…

3.6

(v)

3.5

0

105

…

Solution

(i) Given:

7,a

3,d

8n

By the formula of the term.

th

n

1

n

a a n d

Put the values of and .

,a d

n

7 8 1 3

n

a

7 7 3

7 21

28

Hence, the term of the AP is .

th

n

28

(ii) Given:

18, a

0,

n

a

10n

By the formula of the term.

th

n

1

n

a a n d

Put the values of and .

,

n

a a

n

0 18 10 1 d

0 18 9 d

18

9

d

2d

Hence, the common difference is .

2

(iii) Given:

3, d

5,

n

a

18n

By the formula of the term.

th

n

1

n

a a n d

Put the values of and .

,d

n

a

n

5 18 1 3 a

5 17 3 a

5 51 a

46a

Hence, the first term of the AP is .

46

(iv) Given:

18.9, a

2.5,d

3.6

n

a

By the formula of the term.

th

n

1

n

a a n d

Put the values of and .

,a d

n

a

3.6 18.9 1 2.5 n

3.6 18.9 1 2.5 n

22.5

1

2.5

n

1 9 n

9 1 n

10n

Hence, .

10n

(v) Given:

3.5,a

0,d

105n

By the formula of the term.

th

n

1

n

a a n d

Put the values of and

,a

d

n

3.5 105 1 0

n

a

3.5 0

n

a

3.5

n

a

Hence, the term of the AP is .

th

n

3.5

Question: 2

Select the correct option in the following and justify

your choice:

(i) term of the AP: is

th

30

10,

7,

4,...,

(A)

97

(B)

77

(C)

77

(D)

87

(ii) term of the AP: is

th

11

3,

1

,

2

2,...,

(A)

28

(B)

22

(C)

38

(D)

1

48

2

Solution

(i) Given AP:

10,

7,

4,...

Compare the given AP with the AP

1

,a

2

,a

3

,a

.

.....

n

a

1

10,a

2

7,a

3

4a

First term

1

a a

10

Common difference

2 1

d a a

7 10

3

To find the 30

th

term, use the formula of the

th

n

term.

1

n

a a n d

30

30 1 a a d

10 30 1 3

10 29 3

10 87

77

Hence, the term of the AP is .

th

30

77

(ii) Given AP:

3,

1

,

2

2,...

Compare the given AP with the AP

1

,a

2

,a

3

,.a

.

....

n

a

1

3 a

2

1

2

a

3

2a

First term

1

a a

3

Common difference

2 1

d a a

1

3

2

1 6

2

5

2

To find the term, use the formula of the

th

11

th

n

term.

1

n

a a n d

11

11 1 a a d

5

3 11 1

2

10 5

3

2

3 5 5

22

Hence, the term of the AP is .

th

11

22

Question: 3

In the following APs, find the missing terms in the

boxes:

(i)

2,

,

26

(ii)

,

13 ,

3

(iii)

5,

,

,

1

9

2

(iv)

4,

,

,

,

,

6

(v)

,

38 ,

,

,

22

Solution

(i) Given:

2,

,

26

and

1

a a

2

3

26a

By the formula of the term,

th

n

1

n

a a n d

3

3 1 a a d

Put the values of and .

a

3

a

26 2 3 1 d

26 2 2 d

2 26 2 d

24

2

d

The second term .

2

2 1 a a d

Put the values of and .

a

d

2

2 2 1 12 a

2

2 1 12 a

2

14a

Thus, the missing terms is .

14

(ii) Given:

,

13 ,

3

From the given AP,

and

2

13,a

4

3a

By the formula of the term,

th

n

1

n

a a n d

2

2 1 a a d

.............................................(1)

13 a d

4

4 1 a a d

...............................................(2)

3 3 a d

Subtract equation (1) from (2).

3 13 3 a d a d

10 2 d

10

2

d

5 d

From equation (1).

13 5 a

13 5 a

18a

Thus, the first term is .

18

3

3 1 a a d

Put the values of and .

a

d

3

3 1 a a d

18 2 5

18 10

8

Thus, the third term is .

8

Hence, the missing terms are and .

18

8

(iii) Given:

5,

,

,

1

9

2

From the given AP,

,

1

a a

5

4

1

9

2

a

By the formula of the term,

th

n

1

n

a a n d

4

4 1 a a d

1

9 5 4 1

2

d

19

5 3

2

d

19

3 5

2

d

19 10

3

2

d

9

3

2

d

9

2 3

d

3

2

d

The second term, .

2

a a d

Put the values of

and .

a

d

2

3

5

2

a

10 3

2

13

2

The third term,

3 2

a a d

Put the values of and .

2

a

d

3

13 3

2 2

a

3

16

2

a

3

8a

Thus, the missing terms are and .

13

2

8

(iv) Given:

4,

,

,

,

,

6

From the given AP,

,

1

a a

4

6

6a

By the formula of the term,

th

n

1

n

a a n d

6

6 1 a a d

Put the values of and .

6

a

a

6 4 6 1 d

6 4 5 d

5 6 4 d

5 10d

10

5

d

2d

The second term, .

2

a a d

Put the values of and .

a

d

2

4 2 a

2

2 a

The third term, .

3 2

a a d

Put the values of and .

2

a

d

3

2 2 a

3

0a

The fourth term, .

4 3

a a d

Put the values of and .

3

a

d

4

0 2 a

4

2a

The fifth term, .

5 4

a a d

Put the values of and .

4

a

d

5

2 2 a

5

4a

Thus, the missing terms are .

2,

0,

2,

4

(v) Given:

,

38 ,

,

,

22

From the given AP,

and

2

38a

6

22 a

By the formula of the term,

th

n

1

n

a a n d

2

2 1 a a d

… (1)

38 a d

6

6 1 a a d

… (2)

22 5 a d

Subtract equation (1) from (2).

22 38 5 a d a d

60 4 d

60

4

d

15 d

From equation (1).

38 15 a

38 15 a

53a

Thus, the first term is .

53

The third term,

3 2

a a d

Put the values of and .

2

a

d

3

38 15 a

3

23a

The fourth term,

4 3

a a d

Put the values of and .

3

a

d

4

23 15 a

4

8a

The fifth term,

5 4

a a d

Put the values of and .

4

a

d

5

8 15 a

5

7 a

Thus, the missing terms are .

53,

23,

8,

7

Question: 4

Which term of the AP: is ?

3,

8,

13,

18,...,

78

Solution

The given AP is

3,

8,

13,

18,...,

Compare the given AP with the AP .

1

,a

2

,a

3

,.a

....

n

a

.

1

3,a

2

8,a

3

13a

The first term,

1

a a

3

Common difference

2 1

d a a

8 3

5

Formula for the term

th

n

1

n

a a n d

The term is

th

n

78

So,

78

n

a

1 78 a n d

3 1 5 78 n

1 5 78 3 n

5 5 75 n

5 75 5 n

80

5

n

16n

Hence, 78 is the 16

th

term of the AP

3,

8,

13,

18,...

Question: 5

Find the number of terms in each of the following

APs:

(i)

7,

13,

19,

...,205

(ii)

18,

1

15 ,

2

13,

..., 47

Solution

(i) Given AP:

7,

13,

19,.

..,205

Compare the given AP with the AP

1

,a

2

,a

3

,a

.

.....

n

a

1

7,a

2

13,a

3

19,a

205

n

a

The first term

1

a a

7

Common difference

2 1

d a a

13 7

6

By the formula of the term.

th

n

1

n

a a n d

205 7 1 6 n

205 7 1 6 n

198 1 6 n

198

1

6

n

1 33 n

33 1 n

34n

Thus, the given AP has terms.

34

(ii) Given AP:

18,

1

15 ,

2

13,...,

47

Compare the given AP with the AP

1

,a

2

,a

3

,a

.

.....

n

a

1

18,a

2

1

15 ,

2

a

3

13,a

47

n

a

The first term

1

a a

18

Common difference

2 1

d a a

1

15 18

2

31

18

2

31 36

2

5

2

By the formula of the term.

th

n

1

n

a a n d

5

47 18 1

2

n

5

47 18 1

2

n

5 1 2 65 n

2 65

1

5

n

1 26 n

26 1 n

27n

Thus, the given AP has terms.

27

Question: 6

Check whether or not is a term of the AP

150

11,

8,

5,

2,...

Solution

The given AP is .

11,

8,

5,

2,...

Compare the given AP with the AP .

1

,a

2

,a

3

,a

...,

n

a

1

11,a

2

8,a

3

5,a

4

2a

The first term,

1

a a

11

Common difference

2 1

d a a

8 11

3

The term of the given AP is

th

n

150

By the formula of the term.

th

n

1

n

a a n d

150 11 1 3 n

150 11 3 1 n

3 3 161 n

3 161 3 n

3 164n

164

3

n

is not an integer.

164

3

n

Thus, is not a term of the given AP.

150

Question: 7

Find the term of an AP whose term is 38 and

st

31

th

11

the term is 73.

th

16

Solution

By the formula for the term of an AP,

th

n

1

n

a a n d

… (1)

11

11 1 a a d

… (2)

16

16 1 a a d

According to the question,

and

11

38a

16

73a

Put these values in equation (1) and (2).

11 1 38 a d

… (3)

10 38 a d

16 1 73 a d

… (4)

15 73 a d

Subtract equation (3) from (4).

15 10 73 38 a d a d

5 35d

7d

Put the value of in equation (3).

d

10 7 38 a

38 70 a

32 a

The 31

st

term of the AP is

31

31 1 a a d

Put the value of and

a

d

31

32 31 1 7 a

31

32 31 1 7 a

32 210

178

Thus, the 31

st

term of the given AP is .

178

Question: 8

An AP consists of 50 terms of which the term is 12

rd

3

and the last term is 106. Find the term.

th

29

Solution

By the formula for the term of an AP,

th

n

1

n

a a n d

… (1)

3

3 1 a a d

… (2)

50

50 1 a a d

According to the question,

3 1 12 a d

… (3)

2 12 a d

50 1 106 a d

… (4)

49 106 a d

Subtract equation (3) from (4).

49 2 106 12 a d a d

47 94d

94

47

d

2d

Put the value of in equation (3).

d

2 12 a d

2 2 12 a

12 4 a

8a

The term of the AP is .

th

29

29

29 1 a a d

Put the value of and .

a

d

29

8 29 1 2 a

8 28 2

8 56

64

Thus, the term of the AP is .

th

29

64

Question: 9

If the 3

rd

and the 9

th

terms of an AP are 4 and

8

respectively, which term of this AP is zero?

Solution

By the formula for the term of an AP,

th

n

1

n

a a n d

… (1)

3

3 1 a a d

… (2)

9

9 1 a a d

According to the question,

3 1 4 a d

… (3)

2 4 a d

9 1 8 a d

… (4)

8 8 a d

Subtract equation (3) from (4).

8 2 8 4 a d a d

6 12 d

12

6

d

2 d

Put the value of in equation (3).

d

2 2 4 a

4 4 a

4 4 a

8a

Let the AP have its term as zero.

th

n

1 0 a n d

Put the values of and .

a

d

8 1 2 0 n

8 2 2 0 n

2 8 2 n

2 10n

10

2

n

5n

Thus, the term of the given AP is zero.

th

5

Question: 10

The 17

th

term of an AP exceeds its 10

th

term by 7. Find

the common difference.

Solution

By the formula for the term of an AP,

th

n

1

n

a a n d

................(1)

17

17 1 a a d

.................(2)

10

10 1 a a d

According to the question,

17 10

7 a a

From equations (1) and (2)

17 1 10 1 7 a d a d

16 9 7 d d

7 7d

7

7

d

1d

Thus, the common difference of the AP is .

1

Question: 11

Which term of the AP: will be 132 more

3,

15,

27,

39,...

than its 54

th

term?

Solution

The given AP is

3,

15,

27,

39,...

Compare the given AP with the AP

1

,a

2

,a

3

,a

...,

n

a

1

3,a

2

15,a

3

27,a

4

39a

The first term,

1

a a

3

Common difference

2 1

d a a

15 3

12

To find the term, use the formula of the

th

54

th

n

term.

1

n

a a n d

54

54 1 a a d

Put the values of and .

a

d

54

3 54 1 12 a

3 53 12

3 636

639

Let term be the required term which is 132 more

th

n

than the term

th

54

54

132

n

a a

639 132

n

a

771

n

a

By the formula of the term.

th

n

1

n

a a n d

Put the value of and

,

n

a a

d

771 3 1 12 n

771 3 1 12 n

768

1

12

n

1 64 n

64 1 n

65n

Thus, the term of the given AP will be 132 more

th

65

than its term.

th

54

Question: 12

Two APs have the same common difference. The

difference between their 100

th

terms is 100, what is

the difference between their 1000

th

terms?

Solution

Let and be the first terms of the two APs, and

1

a

2

a

d

is their common difference.

By the formula for the term of an AP,

th

n

1

n

a a n d

In the case of the first AP

100 1

100 1 a a d

1000 1

1000 1 a a d

For the case of the second AP

100 2

100 1 a a d

1000 2

1000 1 a a d

According to the question,

The term of first AP the term of second AP

th

100

th

100

100

1 2

100 1 100 1 100 a d a d

… (1)

1 2

100 a a

Now, calculate the difference between the 1000

th

term of the two APs.

1 2 1 2

1000 1 1000 1 a d a d a a

From equation (1),

1 2

100 a a

Thus, the difference between the 1000

th

term of the

two APs is .

100

Question: 13

How many three-digit numbers are divisible by 7?

Solution

The first three-digit number which can be divided by

7 is 105.

The second number

105 7

112

All other numbers can be found by adding 7 to the

subsequent numbers.

Thus, the sequence of three-digit numbers that can

be divided by 7 can be written as:

105,

112,

119,...

This sequence forms an AP as the common difference

is same for all pair of terms.

7d

Compare the given AP with the AP

1

,a

2

,a

3

,a

...,

n

a

1

105,a

2

112,a

3

119a

the first term,

1

105 a a

When the biggest three-digit number, 999 is divided

by 7, it gives remainder 5.

Thus, is the biggest three-digit number

999 5 994

that can be divided by 7.

Thus is the last term of the AP.

994

The resultant AP is .

105,

112,

119,

...,994

Let the term of the AP is .

th

n

994

By the formula for the term of an AP,

th

n

1

n

a a n d

994 105 1 7 n

994 105 1 7 n

889 1 7 n

889

1

7

n

1 127 n

127 1 n

128n

Hence, 128 three-digit numbers can be divided by 7.

Question: 14

How many multiples of 4 lie between 10 and 250?

Solution

The first multiple of 4 which is greater than 10 is .

12

The second multiple

12 4

16

All other multiples of 4 can be found by adding 4 to

the subsequent numbers.

Thus, the sequence of the multiples of 4 greater than

10 can be written as:

12,

16,

20,

24,...

This sequence forms an AP as the common difference

, which is same for all pair of terms.

4d

Compare the given AP with the AP

1

,a

2

,a

3

,a

...,

n

a

1

12,a

2

16,a

3

20a

First term

1

a a

12

When, 250 is divided by 4, it gives remainder 2.

Thus, is the biggest multiple of 4 less

250 2 248

than 250.

Thus is the last term of the AP.

248

The resultant AP is

12,

16,

20,

24,...,

248

Let the term of the AP be

th

n

248

By the formula for the term of an AP,

th

n

1

n

a a n d

248 12 1 4 n

248 12 1 4 n

1 4 236 n

236

1

4

n

1 59 n

59 1 n

60n

Thus, 60 multiples of 4 exist between and .

10

250

Question: 15

For what value of are the terms of two Aps:

n

th

n

63,

and equal?

65,

67,...

3,

10,

17,...

Solution

For the first AP

63,65,67,...

1

63,a

2

65,a

3

67a

The first term,

1

a a

63

Common difference

2 1

d a a

65 63 2

By the formula of the term.

th

n

1

n

a a n d

.........(1)

63 1 2

n

a n

For the second AP

3,

10,

17,...

1

3,a

2

10,a

3

17a

First term

1

a a

3

Common difference

2 1

d a a

10 3

7

By the formula of the term.

th

n

1

n

a a n d

.........(2)

3 1 7

n

a n

According to the question,

The term of the first AP The term of the

th

n

th

n

second AP.

From equation (1) and (2).

63 1 2 3 1 7 n n

63 3 1 7 1 2 n n

60 7 7 2 2 n n

60 2 7 5 n

5 65n

65

5

n

13n

Thus, for , the terms of the two given APs

13n

th

n

are equal.

Question: 16

Determine the AP whose third term is 16 and the 7

th

term exceeds the 5

th

term by 12.

Solution

By the formula for the term of an AP,

th

n

1

n

a a n d

3

3 1 a a d

7

7 1 a a d

5

5 1 a a d

According to the question,

3 1 16 a d

...........(1)

2 16 a d

And

7 5

12 a a

Thus,

7 1 5 1 12 a d a d

6 4 12 d d

2 12d

12

2

d

6d

Put the value of in equation (1).

d

2 6 16 a

12 16 a

16 12 a

4a

Now,

4 6 a d

10

2 4 2 6 a d

4 12

16

The AP will be

,a

,a d

2 ,a d

3 ,...a d

Thus, the required AP is

4,

10,

16,...

Question: 17

Find the 20

th

term from the last term of the AP

3,

8,

13,

.

...,253

Solution

The given AP is

3,

8,

13,

...,253

Since we need to find the 20

th

term from the last

term, we would first write the series in reverse order.

The AP, in reverse order, would be:

253,.......13,8,3

For the reverse-ordered AP:

First term = 253

Common difference,

3 8 d

5

As per the formula for the n

th

term of an AP,

1

n

a a n d

Therefore,

20

253 20 1 5 a

253 19 5

253 95

158

Thus, the 20

th

term from the last term of the given AP

is 158.

Question: 18

The sum of the fourth and eighth terms of an AP is 24

and the sum of the sixth and tenth terms is 44. Find

the first three terms of the AP.

Solution

By the formula for the term of an AP,

th

n

1

n

a a n d

4

4 1 ......(1)a a d

8

8 1 ......(2)a a d

6

6 1 ......(3)a a d

10

10 1 ......(4)a a d

According to the question,

and

4 8

24a a

6 10

44a a

From equations (1) and (2).

4 1 8 1 24a d a d

2 3 7 24a d d

2 10 24a d

5 12 ......(5)a d

From equations (3) and (4).

6 1 10 1 44a d a d

2 5 9 44a d d

2 5 9 44a d d

7 22 ......(6)a d

Subtract equation (5) from (6).

7 5 22 12a d a d

2 10d

10

2

d

5d

From equation (5).

5 5 12a

25 12a

12 25a

13a

The second term,

2

a a d

13 5

8

The third term,

3 2

a a d

8 5

3

Thus, and are the first three terms of the

13,

8

3

AP.

Question: 19

Subba Rao started work in 1995 at an annual salary of

Rs 5,000 and received an increment of Rs 200 each

year. In which year did his income reach Rs 7,000?

Solution

Since Subba Rao received an increment of Rs 200

each year, his salary in each year after 1995 can be

written as:

5000,

5000 200 ,

5000 200 200 ,...

5000,

5200,

5400,...,

This sequence forms an AP since the common

difference,

, is same for each pair of terms.

200d

In this AP,

1

5000,a

2

5200,a

3

5400a

The first term,

1

5000a a

By the formula for the term of an AP,

th

n

1 ......(1)

n

a a n d

Let the salary of Subba Rao reach in the

7000

th

n

year.

Therefore, .

7000

n

a

Put the values of , and in equation (1).

n

a

a

d

7000 5000 1 200n

7000 5000 1 200n

1 200 2000n

2000

1

200

n

1 10n

10 1n

11n

Hence, the salary of Subba Rao will reach in the

7000

year.

Question: 20

Ramkali saved Rs 5 in the first week of a year and

then increased her weekly savings by Rs 1.75. If in the

week, her weekly savings becomes Rs 20.75, find her

weekly savings.

Solution

Since Ramkali started with saving Rs 5 with an

increment of Rs 1.75 each week, her savings in each

week can be written as:

5,

5 1.75 ,

5 1.75 1.75 ,...,

5,

6.75,

8.5,...,

This sequence forms an AP since the common

difference is same for each pair of terms.

1.75d

In this AP,

1

5,a

2

6.75,a

3

8.5a

The first term

1

5a a

By the formula for the term of an AP,

th

n

1 ......(1)

n

a a n d

According to the question,

In the week, weekly savings of Ramkali become Rs

th

n

20.75.

Thus, .

20.75

n

a

Put the values of and in equation (1).

,

n

a

a

d

20.75 5 1 1.75n

20.75 5 1 1.75n

1 1.75 15.75n

15.75

1

1.75

n

1 9n

9 1n

10n

Hence, .

10n

Exercise 5.3 (20)

Question: 1

Find the sum of the following APs:

(i) to 10 terms.

2,

7,

12,...

(ii) to 12 terms.

37,

33,

29,...

(iii) to 100 terms.

0.6,

1.7,

2.8,...

(iv) to 11 terms.

1

,

15

1

,

12

1

,...

10

Solution

(i) Given: to 10 terms.

For the given AP,

1

2,a a

2

7,a

3

12a

Common difference

2 1

7 2 5d a a

The sum of the first terms is

n

.

2 1

2

n

n

S a n d

Here, .

10n

10

10

2 2 10 1 5

2

S

5 4 9 5

5 4 45

5 49

245

Thus, the sum of the first 10 terms is 245.

(ii) Given AP: to 12 terms.

37,

33,

29,...

For the given AP,

1

37,a a

2

33,a

3

29a

Common difference

2 1

33 37 4d a a

The sum of the first terms is

n

.

2 1

2

n

n

S a n d

Here, .

12n

12

12

2 37 12 1 4

2

S

6 74 11 4

6 74 44

6 30

180

Thus, the sum of the first 12 terms is

180

(iii) Given AP: to 100 terms.

0.6,

1.7,

2.8,...,

For the given AP,

1

0.6,a a

2

1.7,a

3

2.8a

Common difference

2 1

1.7 0.6 1.1d a a

The sum of the first terms is

n

.

2 1

2

n

n

S a n d

Here .

100n

100

100

2 0.6 100 1 1.1

2

S

50 1.2 99 1.1

50 1.2 108.9

50 110.1

5505

Thus, the sum of the first 100 terms is 5505.

(iv) Given AP: to 11 terms.

1

,

15

1

,

12

1

,...,

10

For the given AP,

1

1

,

15

a a

2

1

,

12

a

3

1

10

a

Common difference,

2 1

d a a

1 1

12 15

5 4

60

1

60

The sum of the first terms is

n

.

2 1

2

n

n

S a n d

Here, .

11n

11

11 1 1

2 11 1

2 15 60

S

11 2 10

2 15 60

11 4 2 1 10

2 60

11 8 10

2 60

11 18

2 60

33

20

Thus, the sum of the first 11 terms is

33

20

Question: 2

Find the sums given below:

(i)

1

7 10 14 ... 84

2

(ii)

34 32 30 ... 10

(iii)

5 8 11 ... 230

Solution

(i) Given:

1

7 10 14 ... 84

2

For the given AP,

1

7,a a

2

1

10 ,

2

a

3

14,a

84l

Common difference

2 1

1 21 21 14 7

10 7 7

2 2 2 2

d a a

By the formula for the term of an AP,

th

n

1l a n d

7

84 7 1

2

n

7

84 7 1

2

n

7

77 1

2

n

2

1 77

7

n

1 11 2n

1 22n

22 1n

23n

The sum of the first terms is .

n

2

n

n

S a l

23

23

7 84

2

S

23

91

2

2093

2

Hence, the required sum is .

1

1046

2

(ii) Given:

34 32 30 ... 10

For the given AP,

1

a a

34

2

32a

3

30a

10l

Common difference

2 1

32 34 2d a a

By the formula for the term of an AP,

th

n

1l a n d

10 34 1 2n

10 34 1 2n

24 1 2n

24

1

2

n

1 12n

12 1n

13n

The sum of the first terms is .

n

2

n

n

S a l

13

13

34 10

2

S

13

44

2

13 22

286

Hence, the required sum is .

286

(iii) Given:

5 8 11 ... 230

For the given AP,

1

5,a a

2

8,a

3

11,a

230l

Common difference

2 1

8 5 3d a a

By the formula for the term of an AP,

th

n

1l a n d

230 5 1 3n

230 5 1 3n

225 1 3n

225

1

3

n

1 75n

75 1n

76n

The sum of the first terms is .

n

2

n

n

S a l

76

76

5 230

2

S

38 235

8930

Hence, the required sum is

8930

Question: 3

In an AP:

(i) given , find and .

5,a

3,d

50

n

a

n

n

S

(ii) given , find and .

7,a

13

35a

d

13

S

(iii) given , find and .

12

37,a

3d

a

12

S

(iv) given , find and .

3

15,a

10

125S

d

10

a

(v) given , find and .

5,d

9

75S

a

9

a

(vi) given , find and .

2,a

8,d

90

n

S

n

n

a

(vii) given , find and .

8,a

62,

n

a

210

n

S

n

d

(viii) given , find and .

4,

n

a

2,d

14

n

S

n

a

(ix) given , find .

3,a

8,n

192S

d

(x) given ,and there are total 9 terms.

28,l

144S

Find .

a

Solution

(i) Given:

5,a

3,d

50

n

a

By the formula for the term of an AP,

th

n

1

n

a a n d

50 5 1 3n

1 3 50 5n

1 3 45n

45

1

3

n

1 15n

15 1n

16n

The sum of the first

n

terms is

2

n n

n

S a a

.

16

16

5 50

2

S

8 55

440

Hence,

16n

and

440

n

S

.

(ii) Given:

7,a

13

35a

.

By the formula for the

th

n

term of an AP,

1

n

a a n d

13

13 1a a d

35 7 12d

35 7 12d

28

12

d

7

3

d

The sum of the first

n

terms is

2

n n

n

S a a

.

13 13

13

2

S a a

13 7 35

2

13 42

2

13 21

273

Hence,

7

3

d

and

13

273S

.

(iii) Given:

12

37,a

3d

By the formula for the

th

n

term of an AP,

1

n

a a n d

12

12 1

37 11 3

37 33

4

a a d

a

a

a

The sum of the first

n

terms is

2

n n

n

S a a

.

12 12

12

2

S a a

6 4 37

6 41

246

Hence,

4a

and

12

246S

.

(iv) Given:

3

15,a

10

125S

By the formula for the

th

n

term of an AP,

1

n

a a n d

3

3 1a a d

15 2 ......(1)a d

The sum of the first

n

terms is

2 1

2

n

n

S a n d

.

10

10

2 10 1

2

S a d

125 5 2 9a d

125

2 9

5

a d

2 9 25 ......(2)a d

Multiply equation (1) by 2.

30 2 4 ......(3)a d

Subtract equation (3) from (2).

2 9 2 4 25 30a d a d

5 5d

1d

Put the value of

d

in equation (1).

2 1 15a

15 2a

17a

Now,

10

10 1a a d

17 9 1

17 9

8

Hence,

1d

and

10

8a

.

(v) Given:

5,d

9

75S

The sum of the first

n

terms is

2 1

2

n

n

S a n d

.

9

9

2 9 1

2

S a d

9

75 2 8 5

2

a

9

75 2 40

2

a

9 2

75 20

2

a

75 9 20a

25 3 20a

25 3 60a

3 25 60a

35

3

a

By the formula for the

th

n

term of an AP,

1

n

a a n d

9

9 1a a d

35

8 5

3

35

40

3

35 120

3

85

3

Hence,

35

3

a

and

9

85

3

a

.

(vi) Given:

2,a

8,d

90

n

S

The sum of the first

n

terms is

2 1

2

n

n

S a n d

.

90 2 2 1 8

2

n

n

2

90 2 1 4

2

n

n

90 2 4 4n n

2

90 2 4 4n n n

2

4 2 90 0n n

Split the middle term to find the factors of the

above equation.

2

4 20 18 90 0n n n

4 5 18 5 0n n n

5 4 18 0n n

That is,

5 0n

or

4 18 0n

5n

or

4 18n

5n

or

18

4

n

The value of

n

cannot be negative.

Thus,

5n

.

By the formula for the

th

n

term of an AP,

1

n

a a n d

5

2 5 1 8a

2 4 8

2 32

34

Hence,

5n

and

5

34a

.

(vii) Given:

8,a

62,

n

a

210

n

S

The sum of the first

n

terms is

2

n n

n

S a a

.

210 8 62

2

n

8 62 2 210n

70 420n

420

70

n

6n

By the formula for the

th

n

term of an AP,

1

n

a a n d

62 8 6 1 d

62 8 5d

5 54d

54

5

d

Hence,

6n

and

54

5

d

.

(viii) Given:

4,

n

a

2,d

14

n

S

By the formula for the

th

n

term of an AP,

1

n

a a n d

4 1 2a n

4 2 2a n

2 4 2a n

2 6a n

6 2 ......(1)a n

The sum of the first

n

terms is

2

n n

n

S a a

.

14 4

2

n

a

14 2 4n a

4 28n a

Put the value of

a

from equation (1).

6 2 4 28n n

2

6 2 4 28n n n

2

2 10 28 0n n

2

5 14 0n n

Split the middle term to find the factors of the

above equation.

2

7 2 14 0n n n

7 2 7 0n n n

7 2 0n n

That is,

7 0n

or

2 0n

7n

or

2n

Since

n

is always positive,

Thus,

7n

.

Put the value of

n

in equation (1).

6 2 7a

6 14

8

Hence,

7n

and

8a

.

(ix) Given

3,a

8,n

192S

The sum of the first

n

terms is

2 1

2

n

n

S a n d

.

192 4 2 3 7d

192 4 2 3 7d

192

6 7

4

d

7 48 6d

7 42d

42

7

d

6d

Hence,

6d

.

(x) Given

28,l

144S

,and there are total

9

terms.

The sum of the first

n

terms is

2

n

n

S a l

.

Put

9,n

144

n

S

and

28l

.

9

144 28

2

a

144 2

28

9

a

288

28

9

a

28 32a

32 28a

4a

Hence

4a

.

Question: 4

How many terms of the AP:

9,

17,

27,....

must be taken

to give a sum of

636

?

Solution

The given AP is

9,

17,

27,....

.

For the given AP,

1

9,a a

2

17,a

3

25a

Common difference

2 1

17 9 8d a a

Let

n

terms of the AP give a sum of

636

.

The sum of the first

n

terms is

2 1

2

n

n

S a n d

.

636 2 9 1 8

2

n

n

2

636 9 1 4

2

n

n

636 9 4 4n n

2

636 9 4 4n n n

2

4 5 636 0n n

Split the middle term to find the factors of the given

equation.

2

4 53 48 636 0n n n

4 53 12 4 53 0n n n

4 53 12 0n n

That is,

4 53 0 12 0n n or

4 53 12n n or

53

12

4

n n or

Since

n

is always positive,

Hence,

12n

.

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

According to the question,

and5, 45 400

n

a l S

The sum of the first

n

terms is

2

n

n

S a l

.

400 5 45

2

n

400 2 50n

50 800n

800

50

n

16n

The last term of the AP is given by

1l a n d

.

45 5 16 1 d

45 5 15d

15 40d

40

15

d

8

3

d

Hence, the number of terms are

16

and common

difference is

8

3

.

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:

and17, 350 9a l d

Let the AP have a total of

n

terms.

The last term of the AP is given by

1l a n d

.

350 17 1 9n

350 17 9 9n

9 9 333n

9 333 9n

9 342n

342

9

n

38n

The sum of the first

n

terms is

2

n

n

S a l

.

38

17 350

2

n

S

19 367

n

S

6973

n

S

Hence, the AP has total

38

terms and their sum is

6973

.

Question: 7

Find the sum of the first

22

terms of an AP in which

7d

and the

22nd

term is

149

.

Solution

Given:

22

149, 7a d

.

By the formula for the

th

n

term of an AP,

1

n

a a n d

22

22 1a a d

149 21 7a

147 149a

149 147a

2a

The sum of the first

n

terms is

2

n n

n

S a a

.

22 22

22

2

S a a

11 2 149

11 151

1661

Hence, the sum of first

22

terms of the given AP is

1661

.

Question: 8

Find the sum of the first

51

terms of an AP whose

second and third terms are

14

and

18

, respectively.

Solution

Given:

and

2 3

14 18a a

Common difference

3 2

18 14 4d a a

By the formula for the

th

n

term of an AP,

1

n

a a n d

2

2 1a a d

14 4a

14 4a

10a

The sum of the first

n

terms is

2 1

2

n

n

S a n d

.

51

51

2 51 1

2

S a d

51

2 10 50 4

2

51

20 200

2

51 220

2

51 110

5610

Hence, the sum of first

51

terms of the given AP is

5610

.

Question: 9

If the sum of the first

7

terms of an AP is

49

and that

of

17

terms is

289

, find the sum of first

n

terms.

Solution

Given:

and

7 17

49 289S S

The sum of the first

n

terms is

2 1

2

n

n

S a n d

.

For the sum of first

7

terms,

7

7

2 7 1

2

S a d

7

49 2 6

2

a d

2 7

49 3

2

a d

49 7 3a d

49

3

7

a d

3 7 ...... 1a d

For the sum of the first

17

terms,

17

17

2 17 1

2

S a d

17

289 2 16

2

a d

17 2

289 8

2

a d

8 17 289a d

289

8

17

a d

8 17 ...... 2a d

Subtract equation (1) from (2).

8 3 17 7a d a d

5 10d

10

5

d

2d

Put the value of

d

in equation (1).

3 2 7a

6 7a

7 6a

1a

The sum of the first

n

terms is

2 1

2

n

n

S a n d

.

2 1 1 2

2

n

n

S n

2

1 1

2

n

n

n n

2

n

Hence, the sum of the first

n

terms is

2

n

.

Question: 10

Show that

1 2 3

, , ,..., ,...

n

a a a a

form an AP where

n

a

is

defined as below:

(i)

3 4

n

a n

(ii)

9 5

n

a n

Also find the sum of the first

15

terms in each case.

Solution

(i) Given:

3 4

n

a n

Put

1,2,3,4n

in the above expression.

1

3 4 1 3 4 7a

2

3 4 2 3 8 11a

3

3 4 3 3 12 15a

4

3 4 4 3 16 19a

The first term

1

7a a

.

To find the common difference.

2 1

11 7 4a a

3 2

15 11 4a a

4 3

19 15 4a a

Clearly, the common difference is the same for

each pair of terms.

Hence, the given sequence forms an AP with the

common difference

4d

.

The sum of the first

n

terms is

2 1

2

n

n

S a n d

.

15

15

2 15 1

2

S a d

15

2 7 14 4

2

15

14 56

2

15

70

2

15 35

525

Hence, the sum of the first

15

terms is

525

.

(ii) Given:

9 5

n

a n

Put

1,2,3,4n

in the above expression.

1

9 5 1 9 5 4a

2

9 5 2 9 10 1a

3

9 5 3 9 15 6a

4

9 5 4 9 20 11a

The first term

1

4a a

.

To find the common difference.

2 1

1 4 5a a

3 2

6 1 5a a

4 3

11 6 5a a

Clearly, the common difference is the same for

each pair of terms.

Hence, the given sequence forms an AP with the

common difference

d

is

5

.

The sum of the first

n

terms is

2 1

2

n

n

S a n d

.

15

15

2 15 1

2

S a d

15

2 4 14 5

2

15

8 70

2

15

62

2

15 31

465

Hence, the sum of the first

15

terms is

465

.

Question: 11

If the sum of the first

n

terms of an AP is

2

4n n

,

what is the first term (that is

1

S

)? What is the sum of

first two terms? What is the second term? Similarly,

find the

3

rd

, the

10

th

and the

th

n

terms.

Solution

The sum of the first

n

terms

2

4

n

S n n

.

2

1

4 1 1S

4 1

3

The first term of AP,

1

a S

.

Thus,

3a

.

Sum of first two terms

2

2

4 2 2S

8 4

4

The second term of AP

2 2 1

a S S

4 3

1

Common difference

2

1 3 2d a a

By the formula for the

th

n

term of an AP,

1

n

a a n d

3

3 1a a d

3 2 2

3 4

1

Similarly,

10

10 1a a d

3 9 2

3 18

15

Now,

1

n

a a n d

3 1 2n

3 2 2n

5 2n

Hence, the first term is

3

and sum of first two terms is

4

.

Also, the second, third, tenth and

th

n

terms are

1

,

1

,

15

, and

5 2n

respectively.

Question: 12

Find the sum of the first

40

positive integers divisible

by

6

.

Solution

The sequence of positive integers divisible by 6 can

be written as:

6,12,18,24,...

For the given sequence,

1 2 3 4

6, 12, 18, 24a a a a a

Find the common difference.

2 1

12 6 6a a

3 2

18 12 6a a

4 3

24 18 6a a

Clearly, the common difference is the same for each

pair of terms.

Hence, the sequence forms an AP with the common

difference

6d

.

The sum of the first

n

terms is

2 1

2

n

n

S a n d

.

40

40

2 40 1

2

S a d

20 2 6 40 1 6

20 12 39 6

20 12 234

20 246

4920

Thus, the sum of the first

40

positive integers

divisible by

6

is

4920

.

Question: 13

Find the sum of the first

15

multiples of

8

.

Solution

The sequence of the multiples of 8 can be written as:

8,16,24,32,...

For the given sequence,

1 2 3 4

8, 16, 24, 32a a a a a

Find the common difference.

2 1

16 8 8a a

3 2

24 16 8a a

4 3

32 24 8a a

Clearly, the common difference is the same for each

pair of terms.

Hence, the sequence forms an AP with the common

difference

8d

.

The sum of the first

n

terms is

2 1

2

n

n

S a n d

.

15

15

2 15 1

2

S a d

15

2 8 15 1 8

2

15

16 14 8

2

15

16 112

2

15

128

2

15 64

960

Thus, the sum of the first

15

multiples of

8

is

960

.

Question: 14

Find the sum of the odd numbers between

0

and

50

.

Solution

The sequence of the odd numbers between

0

and

50

can be written as:

1,3,5,7,9,...,49

For the given sequence,

1

1a a

2

3a

3

5a

4

7a

Find the common difference.

2 1

3 1 2a a

3 2

5 3 2a a

4 3

7 5 2a a

Clearly, the common difference is the same for each

pair of terms.

Hence, the sequence forms an AP with the common

difference

2d

.

The last term of AP,

49l

.

Last term of an AP is given by

1l a n d

.

49 1 1 2n

49 1 1 2n

1 2 48n

48

1

2

n

1 24n

24 1n

25n

Thus, the total number of terms in the AP are

25

.

The sum of the first

n

terms is

2

n

n

S a l

.

25

25

2

S a l

25

1 49

2

25 50

2

25 25

625

Hence, the sum of the 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:

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?

Solution

The sequence of the penalties can be written as:

200,250,300,...

Each term in the above sequence is

50

more than the

previous term.

Thus, the sequence forms an AP with the common

difference

50d

.

From the sequence,

1

200a a

2

250a

3

300a

The sum of the first

n

terms is

2 1

2

n

n

S a n d

.

30

30

2 30 1

2

S a d

15 2 200 30 1 50

15 400 29 50

15 400 1450

15 1850

27750

Hence, the contractor will pay the penalty of Rs

27750.

Question: 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.

Solution

Let the value of the first prize is

x

.

According to the question,

The value of the second prize

20x

The value of the third prize

40x

The sequence formed by these values can be written

as:

, 20, 40,...x x x

Each term in the above sequence is

20

less than the

previous term.

Thus, the sequence forms an AP with the common

difference

20d

.

From the sequence,

1 2 3

, 20, 40a a x a x a x

It is given that the sum of the values of

7

prizes is

700

.

Thus,

7

700S

.

The sum of the first

n

terms is

2 1

2

n

n

S a n d

.

7

7

2 7 1

2

S a d

7

700 2 6

2

x d

7

700 2 6 20

2

x

7

700 2 120

2

x

700 2

2 120

7

x

2 120 100 2x

2 120 200x

2 200 120x

2 320x

320

2

x

160x

Thus, the value of first prize is Rs 160.

The value of second prize

160 20 140

and so on.

Thus, the values of the seven prizes are Rs

160

,

140

,

120

,

100

,

80

,

60

and

40

respectively.

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

The sequence formed by the number of trees

planted by the students of 12 classes can be written

as:

1,2,3,4,...,12

Each term in the above sequence is

1

more than the

previous term.

Thus, the sequence forms an AP with the common

difference

1d

.

From the sequence,

1

1a a

2 3

2, 3a a

The sum of the first

n

terms is

2 1

2

n

n

S a n d

.

12

12

2 12 1

2

S a d

6 2 1 11 1

6 2 11

6 13

78

Each class has three sections, thus, total number of

trees planted by all the sections of 12 classes

78 3 234

.

Question: 18

A spiral is made up of successive semicircles, with

centres alternately at A and B, starting with centre at

A, of radii

cm cm cm cm0.5 ,1.0 ,1.5 ,2.0 ,...

. 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

)

Fig. Exc_5.3_18

Solution

Given that radii of the semicircles are

1

0.5r cm

,

2

1.0r cm

,

3

1.5r cm

4

2.0r cm

The length of the perimeter of a circle is

π2 r

.

Thus, the length of a semicircle

πr

Length of first semicircle

1 1

l r π

0.5=π

2

π

Length of second semicircle

2 2

l r π

1 π

π

Length of first semicircle,

3 3

l r π

1.5 π

2

3π

The sequence formed by the lengths of the

semicircles can be written as:

π π

π

3

, , ,....

2 2

Each term in the above sequence is

π

2

more than the

previous term.

Thus, the sequence forms an AP with the common

difference

π

2

d

.

From the sequence,

1

2

a a

π

2

a π

3

3

2

a

π

The sum of the first

n

terms is

2 1

2

n

n

S a n d

.

13

13

2 13 1

2

S a d

13

2 12

2 2 2

π π

13 12

2 2

π

π

13

6

2

π π

13

7

2

π

13 22 22

7

2 7 7

π

13 11

143

Hence, the total length of the spiral made up of

thirteen consecutive semicircles is

cm143

.

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 (see Fig. 5.5). In how may rows

are the 200 logs placed and how many logs are in the

top row?

Fig. Exc_5.3_19

Solution

The sequence formed by the number of logs placed

in rows can be written as:

20,19,18,...

Each term in the above sequence is

1

less than the

previous term.

Thus, the sequence forms an AP.

From the sequence,

1

20,a a

2

19a

3

18a

Common difference

2 1

d a a

19 20

1

Let n number of rows are formed by 200 logs.

The sum of the first

n

terms is

2 1

2

n

n

S a n d

.

200 2 20 1 1

2

n

n

200 2 40 1n n

2

400 40n n n

2

41 400 0n n

Split the middle term to find the factors of the above

equation.

2

16 25 400 0n n n

16 25 16 0n n n

16 25 0n n

That is,

16 0n

or

25 0n

16n

or

25n

By the formula for the

th

n

term of an AP,

1

n

a a n d

To find the number of logs in the 16

th

row.

16

16 1a a d

20 16 1 1

20 15

5

To find the number of logs in the 25

th

row.

25

25 1a a d

20 25 1 1

20 24

4

Number of logs cannot be negative.

Thus,

16n

.

Hence, 16 rows are made to place 200 logs and the

16

th

row will occupy 5 logs.

Question: 20

In a potato race, a bucket is placed at the starting

point, which is 5 m away from the first potato. The

other potatoes are placed 3 m apart in a straight line.

There are ten potatoes in the line (see Fig. 5.6).

Fig. Exc_5.3_20

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?

Solution

Distance of the first potato from the bucket

m5

Distance of the second potato from the bucket

m5 3 8

and, so on.

Thus, the sequence formed by the distance of

potatoes from the bucket can be written as:

5,8,11,14,...

Each term in the above sequence is

3

more than the

previous term.

Thus, the sequence forms an AP with common

difference

d

3

.

From the sequence,

1

5a a

2 3

8, 11a a

The sum of the first

n

terms is

2 1

2

n

n

S a n d

.

Total number of potatoes

10n

.

10

10

2 10 1

2

S a d

5 2 5 10 1 3

5 10 9 3

5 10 27

5 37

185

The competitor runs back to the bucket each time.

Thus, the total distance covered by the competitor

m2 185 370

Exercise 5.4(5) (Optional)

Question: 1

Which term of the AP: 121, 117, 113, … is its first

negative term?

Solution

From the given AP 121, 117, 113, …

1

121a a

,

2

117a

,

3

113a

Common difference

2 1

d a a

117 121

4

By the formula for the

th

n

term of an AP,

1

n

a a n d

121 1 4n

121 4 4n

125 4n

To find the first negative term.

0

n

a

125 4 0n

125 4n

125

4

n

Hence, the first negative term of the AP is its 32

nd

term.

Question: 2

The sum of the third and the seventh terms of an AP

is 6 and their product is 8. Find the sum of the first

sixteen terms of the AP.

Solution

By the formula for the

th

n

term of an AP,

1

n

a a n d

3

3 1a a d

3

2 ......(1)a a d

Similarly,

7

6 ......(2)a a d

According to the first condition of the question,

3 7

6a a

From equation (1) and (2).

2 6 6a d a d

2 8 6a d

4 3a d

3 4 ...... (1)a d

According to the second condition of the question,

3 7

8a a

From equation (1) and (2).

2 6 8a d a d

Put the value of

a

from equation (3).

3 4 2 3 4 6 8d d d d

3 2 3 2 8d d

2 2

3 2 8d

2

9 4 8d

2

4 9 8d

2

4 1d

2

1

4

d

1

2

d

Put in equation (3).

1

2

d

1

3 4

2

a

3 2

1

Put in equation (3).

1

2

d

1

3 4

2

a

3 2

5

The sum of the first

n

terms is:

2 1

2

n

n

S a n d

To find for and :

16

S

1a

1

2

d

16

16

2 16 1

2

S a d

1

8 2 1 16 1

2

1

8 2 15

2

4 15

8

2

8 19

2

4 19

76

Now, find for and :

16

S

5a

1

2

d

16

16

2 16 1

2

S a d

1

8 2 5 16 1

2

15

8 10

2

20 15

8

2

5

8

2

4 5

20

Hence, the sum of the first sixteen terms of the AP, is

76 when

and

1

1

2

a d

, and 20 when

and

1

5

2

a d

.

Question: 3

A ladder has rungs 25 cm apart (see Fig. 5.7). 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 m apart, what is the length of

1

2

2

the wood required for the rungs?

Fig. Exc_5.4_3

Solution

The distance between the rungs

cm25

Distance between top and bottom rungs

m

1

2

2

Now convert m into cm.

1 1

2 2 100 1 100

2 2

m cm m cm

5 100

2

cm

5 50 cm

250 cm

The number of rungs in the ladder,

250

1

25

10 1

=11

We know that the rungs decrease uniformly in length

from 45 cm at the bottom to 25 cm at the top.

Thus, the sequence formed by the lengths of the

rungs will form an AP with

45a

and

25l

.

Number of rungs in the ladder

11

Sum of all the terms of the AP will give the length of

the wood needed for the rungs.

The sum of the first

n

terms is:

2

n

n

S a l

11

45 25

2

n

S

11

70

2

11 35

385

Hence, the length of the wood needed for the rungs

is

cm385

.

Question: 4

A row of houses 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

.

Solution

The sequence formed by the number of houses can

be written as:

1,2,3,...,49

Each term in the above sequence is

1

more than the

previous term.

Thus, the sequence forms an AP with common

difference,

1d

.

From the sequence,

1

1a a

2

2a

3

3a

Let the number of

th

x

house satisfy the given

condition.

The sum of the first

n

terms is

2 1

2

n

n

S a n d

Sum of number of houses preceding the house

numbered

x

1x

S

1

1

2 1 1

2

x

x

S a x d

1

2 1 2 1

2

x

x

1

2 2

2

x

x

1

2

x x

Sum of the number of houses following the house

numbered

x

49 x

S S

49

49

2 1 49 1 1 2 1 1 1

2 2

x

x

S S x

49

2 49 1 2 1

2 2

x

x

49

50 1

2 2

x

x

1

49 25

2

x x

According to the question,

1 49x x

S S S

Thus,

1 1

49 25

2 2

x x x x

2 2

1225

2 2 2 2

x x x x

2 2

1225

2 2

x x

2

1225x

35x

The number of the houses cannot be negative.

Thus

35x

.

Hence, the house numbered 35 is such that,

the sum of the numbers of houses preceding the 35

th

house

The sum of the numbers of the houses

following the 35

th

house.

Question: 5

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

1

4

m and a tread of

1

2

m. (see Fig. 5.8). Calculate the total volume of

concrete required to build the terrace.

Fig. Exc_5.4_5

Solution

From the figure:

The height of each step is

m

1

4

and length is

m50

.

Width of the first step

m

1

2

Width of the second step

m

1 1

1

2 2

and so on.

Thus, the sequence formed by the width of the steps

can be written as:

1 3

,1, ,...

2 2

The volume of concrete used in first step

1 1

50

4 2

25

4

The volume of concrete used in second step

1

1 50

4

25

2

The volume of concrete used in third step

1 3

50

4 2

3 25

4

75

4

Thus, the sequence formed by the volumes of the

steps can be written as:

25 25 75

, , ,...

4 2 4

Each term in the above sequence is

25

4

more than

the previous term.

Thus, the sequence forms an AP with common

difference

25

4

d

.

From the sequence,

1

25

4

a a

2

25

2

a

3

75

2

a

The sum of the first

n

terms is

2 1

2

n

n

S a n d

.

15

15

2 15 1

2

15 25 25

2 15 1

2 4 4

15 25 14 25

2 2 4

15 25 175

2 2 2

15 200

2 2

15

100

2

750

S a d

Hence, the volume of required concrete is

m

3

750

.