Gujarat Board GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2

Question 1.

Fill in the blanks in the following table. Given that a is the first term d the common difference and a the nth term of the AP.

Solution:

(i) a = 7, d = 3, n = 8, a = _{n} ?

than a_{n} = a + (n – 1) d

a_{n} = 7 + (8 – 1) 3

a_{n} = 7 + 7 x 3

a_{n} = 7 + 21

a_{n} = 28

(ii) a = – 18, d = ?, n = 10, a_{n} = 0

a_{n} = a + (n – 1) d

0 = – 18 + (10 – 1)d

= 18 = 9d

d = 2

(iii) a = ?, d = – 3, n = 18, a_{n} = – 5

a_{n} = a + (n – 1) d

– 5 = a + (18 – 1)(- 3)

– 5 = a + (- 51)

51 – 5 = a

a = 46

(iv) a = – 18.9, d = 2.5, n = ?, a_{n} = 3.6

a_{n} = a + (n – 1) d

3.6 = – 18.9 + (n – 1) x 2.5

3.6 + 18.9 = 2.5 (n – 1)

22.5 = 2.5 (n – 1)

n – 1 = \(\frac { 22.5 }{ 2.5 } \)

n – 1 = \(\frac { 225 }{ 25 } \)

n – 1 = 9

n = 9 + 1

n = 10

(v) a = 3.5, d = 0, n = 105, a_{n} = ?

a_{n} = a + (n – 1) d

a_{n} = 3.5 + (105 – 1) 0

a_{n} = 3.5

Question 2.

Choose the correct choice in the following and justify.

(i) 30th term of the AP: 10, 7, 4,….. is

(A) 97 (B) 77 (C) – 77 (D) – 87

(ii) 11th term of the AP: – 3, –\(\frac { 1 }{ 2 } \), 2, ….. is

(A) 28 (B) 22 (C) – 38 (D) 48 – \(\frac { 1 }{ 2 } \)

Solution:

(i) The given AP is 10, 7, 4, …

a = 10, d = 7 – 10 = – 3

n = 30

a_{n} = a + (n — 1) d

a_{30} = 10 + (30 – 1) x (- 3)

= 10 + 29 x (- 3)

= 10 – 87

a_{30} = – 77

hence the correct choice is (C).

(ii) In the given AP,

a = – 3

and

d = – \(\frac { 1 }{ 2 } \) – ( – 3) = \(\frac { -1 }{ 2 } \) + 3 = \(\frac { 5 }{ 2 } \)

a_{11} = a + (n – 1)d

= – 3 + (11 – 1) x \(\frac { 5 }{ 2 } \)

= – 3 + 10 x \(\frac { 5 }{ 2 } \)

= – 3 + 25

a_{11} = 22

Hence the correct choice is (B)

Question 3.

In the following APs, find the missing terms in the boxes.

(j) 2, 26

(ii) 13, 3

(iii) 5, 9\(\frac { 1 }{ 2 } \)

(iv) – 4, 6

(v) 38, – 22

Solution:

(i) Let the common difference of given AP is d

a = 2, a_{3} = 26

a_{3} = a + 2d = 26

a + 2d = 26

2 + 2d = 26

2d = 26 – 2

2d = 24

d = 12

Hence second term a_{2} = a + d

a_{2} = 2 + 12 = 14

Hence the’missing term in the box is .

(ii)

Let a be the first term and d be the common difference of given AP.

Then, a + d = 13

a_{4} = a + 3d = 3

d = – 5

Putting value of d in equation (1)

a + d = 13

a + ( – 5) = 13

a = 13 + 5

a = 18

a_{3} = a + 2d

= 18 + 2x ( – 5)

= 18 – 10

Hence the missing terms in the boxes are

and

(iii) Let common difference be d and a = 5

a_{4} = a + 3d = 9\(\frac { 1 }{ 2 } \)

a + 3d = \(\frac { 19 }{ 2 } \)

5 + 3d = \(\frac { 19 }{ 2 } \)

= 3d = \(\frac { 19 }{ 2 } \) – 5

3d = \(\frac { 19 }{ 2 } \)

d = \(\frac { 9 }{ 2×3 } \) = \(\frac { 3 }{ 2 } \)

Now, a_{2} = a + d = 5 + \(\frac { 3 }{ 2 } \)

= \(\frac { 10+3 }{ 2 } \) = \(\frac { 13 }{ 2 } \) = 6\(\frac { 1 }{ 2 } \)

a_{3} = a + 2d = 5 + 2 x \(\frac { 3 }{ 2 } \)

a_{3} = 8

Hence the missing terms are and in the boxes.

(iv) a = – 4 and we let d be the common difference.

Then a_{6} = a + (n – 1) d

a_{6} = – 4 + (6 – 1) x d

6 = – 4 + 5d

10 = 5d

d = 2

a_{2} = a + d = – 4 + 2 = – 2

a_{3} = a + 2d = – 4 + 2 x 2

= 4 + 4 = 0

a_{4} = a + 3d = – 4 + 3 x 2

= – 4 + 6 = 2

a_{5} = a + 4d = – 4 + 4 x 2

= – 4 + 8 = 4

Hence the missing terms in the boxes are

(v) Let common difference be d.

a_{2} = a + d = 38

= a + d = 38 ……………(1)

a_{6} = a + 5d = – 22 ……………. (2)

d = – 15

Putting value of d in equation (1)

a + d = 38

= a – 15 = 38

a = 38 + 15

a = 53

a_{3} = a + 2d

a_{3} = 53 + 2 ( – 15)

a_{3} = 53 – 30

a_{3} = 23

a_{4} = a + 3d

a_{4} = 53 + 3 x ( – 15)

a_{4} = 53 – 45 = 8

a_{5} = a + 4d

a_{5} = 53 + 4 x ( – 15)

a_{5} = 53 – 60

a_{5} = – 7

Hence the missing terms in the boxes are

and

Question 4.

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

Solution:

The given AP is 3, 8, 13, 18, ……………

a = 3

d = 8 – 3 = 5

Let the term of given AP is 78.

Hence a_{n} = a + (n – 1) d

78 = 3 + (n – 1) x 5

75 = (n – 1) x 5

n – 1 = \(\frac { 75 }{ 5 } \)

n – 1 = 15

n = 16

Hence 16th term of the given AP is 78.

Question 5.

Find the number of terms in each of the following APs.

(i) 7 13, 19, …, 205

(ii) 18, 15\(\frac { 1 }{ 2 } \),13,…,- 47

Solution:

(i) Given AP is 7, 13, 19, ………….., 205

a = 7, d = 13 – 7 = 6

Let n be the number of terms

then a_{n} = 205

a_{n} = a + (n – 1) d

205 = 7 + (n – 1) 6

= 205 – 7 = (n – 1) x 6

198 = (n – 1) x 6

\(\frac { 198 }{ 6 } \) = n – 1

= n – 1 = 33

n = 34

Hence the number of terms in the given AP is 34.

(ii) Given AP is 18, 15\(\frac { 1 }{ 2 } \), 13, …,- 47

a = 18, d = \(\frac { 31 }{ 2 } \) – 18 = \(\frac { -5 }{ 2 } \)

Let a_{n} = – 47

then a_{n} = a + (n – 1) d

– 47 = 18 + (n – 1) \(\frac { -5 }{ 2 } \)

– 47 = 18 – \(\frac { -5 }{ 2 } \) (n – 1)

7 – 18 = \(\frac { -5 }{ 2 } \)(n – 1)

– 65 = \(\frac { -5 }{ 2 } \)(n – 1)

\(\frac { 65×2 }{ 5 } \) = n – 1

26 = n – 1

= n – 1 = 26

= n = 26 + 1

n = 27

Hence the number of terms in the given AP is 27.

Question 6.

Check whether – 150 is a term of the AP: 11, 8, 5, 2, ………….

Solution:

The given list of numbers is 11, 8, 5, 2, ……..

a_{2} – a_{1} = 8 – 11 = – 3

Here a = 11, d = – 3

Let – 150 be the a_{n} term of the given AP.

then

a_{n} = a + (n – 1)d

– 150 = 11 + (n – 1)( – 3)

– 150 = 11 – 3n + 3

3n = 150 + 14

n = \(\frac { 164 }{ 3 } \)

But n must be positive integer therefore – 150 is not a term of given AP.

Question 7.

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

Solution:

Let a be the first term and cl is the common difference of the AP.

11th term, a_{11} = 38 [a_{11} = a + (11 – 1)d]

a + 10 d = 38 ……………….. (1)

And 16th term = 73

d = \(\frac { 35 }{ 5 } \) = 7

d = 7

Putting value of d an equation (1)

a + 10 d = 38

a + 10 x 7 = 38

a + 70 = 38

a = 38 – 70

a = – 32

there fore a_{n} = a + (n – 1)d

a_{31} = – 32 + (31 – 1) x 7

a_{31} = – 32 + 210

a_{31} = 178

Hence 31 term of the AP is 178,

Question 8.

An AP cvneists of 50 terms ofwhich third tarms is 12 and last term is 106. Find the 29th terms.

Solution:

Let a be the first term and d be the common difference of AP.

Third terni a_{3} = 12

a + 2d = 12

last term = 106 (l = 50th)

l = a + (n – 1)d

106 = a + (50 – 1) d

106 = a + 49d

a + 49d = 106 ………. (2)

Solving equations (1) and (2), we get

47d = 94

d = \(\frac { 94 }{ 47 } \)

Putting value of din equation (1)

a + 2d = 12

a + 2 x 2 = 12

a = 12 – 4

a = 8

Therefore 29th term

a_{29} = a + 28d

= 8 + 28 x 2

= 8 + 56 = 64

Question 9.

If the 3rd and the Wit terms of an AP are 4 and -8 respectively, which term of this AP is zero?

solution:

Let a and d be the first term and common difference of an AP.

Third term a_{3} = 4

a + 2d = 4

(a_{n} = a + (n – 1)d)

Ninth term a_{9} = – 8

a + 8d = – 8 …………… (2)

Solving equations (1) and (2),

d = – 2

Putting value old in equation (2),

a + 8d = – 8

a + 8 x (- 2) = – 8

a – 16 = – 8

a = 8

Let the the nth term of the AP be zero

then a + (n – 1)d = 0

(a_{n} = a + (n – 1)d)

8 + (n – 1)( – 2) = 0

8 – 2n + 2 = 0

2n = 10

n = 5

Hence 5’ term of the AP is zero.

Question 10.

The 17th term of an AP exceeds 10th term by 7. Find the common difference.

Solution:

Let a be the first term and d be the common difference of the AP.

17th term = a + 16d

(a + (17 – 1) d)

and 10th term = a + 9d

According to problem

a_{17} = a_{10} + 7

a + 16d = a + 9d +7

7d = 7

d = 1

Hence common difference is 1.

Question 11.

Which term of the AP: 3, 15, 27, 39, ….. will be 132 more than its 84th term?

Solution:

Given AP is 3, 15, 27, 39, ……….

Here a = 3 and d = 15 – 3 = 12

4th term = a + 53d (a + (54 – 1)d)

According to problem

nth term = 54th term + 132

a + (n – 1)d = a + 53d + 132

3 + (n – 1) x 12 = 3 + 53 x 12 + 132

(n – 1) x 12 = 636 + 132

n – 1 = \(\frac { 768 }{ 12 } \)

= n – 1 = 64

n = 65

Hence 65th term will be 132 more than 54th term.

Question 12.

Two APs have the same common difference The difference between their 100th term is 100. What is the difference between 1000th terms?

Solution:

Let a and a be the Gral term of two APs respectively and ci be the saine cximrnon difference of the two APs.

Then 100th term of first AP = a_{1} + 99d

(a_{100} = a_{1} + (100 – 1) d)

100th term of second AP = a_{2} + 99d

According to problem

a_{1} + 99d (a_{2} + 99d) = 100

a_{1} – a_{2} = 100

Now we have

1000th term of lint AP = a_{1} + 999d

and 1000th term of second AP

= a + 999d

Hence the difference of 1000th terms of both AP’s

a_{1} + 999d – (a_{2} + 999d)

= a_{1} – a_{2}

= 100

Hence the difference of 1000th term of both the APs is also 100.

Question 13.

How many three.digit numbers are divisible by 7 ?

Solution:

Three digit numbers which are divisible by 7

are: 105, 112, 119, 126, …………, 994

a_{2} – a_{1} = 112 – 105 = 7

a_{3} – a_{2} = 119 – 112 = 7

a_{4} – a_{3} = 126 – 119 = 7

Since difference, between two consecutive terms is 7. So the three-digit numbers which an divisible by 7 are in AP.

a = 105

last term = 994

nth term = l

l = a + (n – 1) d

994 = 105 + (n – 1) x 7

994 – 105 = (n – 1) x 7

(n – 1) x 7 = 889

n – 1 = \(\frac { 889 }{ 7 } \)

n – 1 = 127

n = 128

Hence, 128 three digit numbers are divisible by 7.

Question 14.

How many multiples of 4 Lie between 10 and 250?

Solution:

Multiples at 4 which lies between 10 and 250

are 12, 16, 20, 24, ……….., 248

a_{2} – a_{1} = 16 – 12 = 4

a_{3} – a_{2} = 20 – 16 = 4

Since difference between two consecutive terms is 4 i.e. same so these numbers form an AP.

Here a = 12, and l = 248

l = a + (n – 1) d

12 + (n – 1) x 4 = 248

(n – 1) x 4 = 248 – 12

n – 1 = 59

n = 60

Hence 60 multiples of 4 between 10 and 250.

Question 15.

For what value of is, are the e terms of two

APs 63, 65, 67, ….. and 3, 10, 17, ……….. equal?

Solution:

First AP is 63. 65,67, ………

Here a = 63, and d = 65 – 63 = 2

nth term a + (n – 1)d

(a_{n} = a + (n – 1) d)

= 63 (n – 1) x 2

Second AP is 3,10, 17, ……….

a = 3 and d = 10 – 3 = 7

nth term = a + (n – 1) d

= 3 + (n – 1) = 7

According to problem nth terms are same.

63 + (n – 1) x 2 = 3 + (n – 1) x 7

63 + 2n – 2 = 3 + 7n – 7

61 + 2o = 7n – 4

– 5n = – 65

n = 13

Hence 13th term of both APa are equal.

Question 16.

Determine the AP whose third term 16 and the 7th term exceeds the 5th, term by 12.

Solution:

Let a and d be the first and the common difference of the AP respectively.

Third term a_{3} = 16

n_{3} = a + 2d

a + 2d = 16

and 7th term = a + 6d

5th term = a + 21d

According the problem

a_{7} = a_{5} + 12

a + 6d = a + 4d + 12

2d = 12

d = 6

Putting value old in eqn.(1)

a + 2d = 16

a + 2 x 6 = 16

a = 4

Hence required AP is.

a, a + d, a + 2d, a + 3d, ……….

4, 4 + 6, 4 + 2 x 6, 4 + 3 x 6, ……….

4, 10, 16, 22, …………

Question 17.

Find the 20 term from the la.t term of the AP: 3, 8, 13, …………. 253.

Solution:

Given AP is

3, 8, 13, ……….., 253

Here a = 3 and d = 8 – 3 = 5

20th term from the last = l – (n – 1)d

= 253 – (120 – 1) x 5

= 253 – 19 x 5

= 253 – 95 = 158

Hence the 20th term from the last term of the given AP is 158.

Alternate method

Given AP is 3, 8, 13, …………. 253

Here a =3 and d = 8 – 3 = 5

l = a + (n – 1)d

253 = 3 + (n – 1) x 5

250 = (n – 1) x 5

n – 1 = \(\frac { 250 }{ 5 } \)

n – 1 = 50

n = 51

term from the last = (51 – 20 + 1) term from the beginning.

= 32 term from the beginning

a_{n} term = a + (n – 1)d

a_{32} = 3 + (32 – 1) x 5

= 3 + 31 x 5

= 3 + 155 = 158

Hence 20th term from the last term of the given AP is 158.

Question 18.

The sum of the 4th and 8th term of an AP is 24 and the sum of the 6th and 10th term is 44. Find the first three term of the AP.

Solution:

Let a be the first term and d be the common difference

4th term = a + 3d [a_{4} = a + (4 – 1)d]

8th term = a + 7d [a_{8} = a + (8 – 1)d]

According to the problem.

a_{4} + a_{8} = 24

a + 3d + a + 7d = 24

2a + 10 d = 24

a + 5d = 12

6th term = a + 5d [a_{6} = a + (6 – 1)d]

10th term = a + 9d [a_{10} = a + (10 – 1)d)

According to problem,

a_{6} + a_{10} = 44

a + 5d + a + 9d = 44

2a + 14d = 44

d = 5

Putting value, old in equ. (1)

a + 5d = 12

a + 5 x 5 = 12

a = 12 – 25

a = – 13

Therefore, first three terms of AP art

a, a + d and a + 2d

i.e.,-13, – 13 + 5 and -13 + 2 x 5

or, – 13, – 8 and -3

Question 19.

Subba Rao started work in 1995 at an annual salary of ₹ 5000 and received an increment of ₹ 200 each year. In which year did his income reach ₹ 7000?

Solution:

Here a = ₹ 5000

d = ₹ 200

l = ₹ 7000

Suppose in n years his income reached ₹ 7000,

Then

l – a + (n – 1)d

7000 = 5000 + (n – 1) 200

(n – 1) x 200 = 7000 – 5000

(n – 1) x 200 = 2000

n – 1 = \(\frac { 2000 }{ 200 } \)

n – 1 = 10

n = 11

In 11th year his salary becomes ₹ 7000.

Question 20.

Ramkali saved ₹ 5 In the Brat week of a year and then increased her weekly savings by ₹ 1.75. If in the nth week her weekly savings becomes ₹ 20,75, find n.

Solution:

Herr

a = ₹ 5

d = ₹ 1.75

a_{n} = 20.75

term of an AP.

a_{n} = a + (n – 1)d

20.75 = 5 + (n – 1)1.75

(n – 1)(1.75) = 15.75

n – 1 = \(\frac { 15.75 }{ 1.75 } \)

n – 1 = \(\frac { 1575 }{ 175 } \)

n – 1 = 9

n = 10

Hence required value of n is 10.