Haryana State Board HBSE 10th Class Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.2 Textbook Exercise Questions and Answers.

## Haryana Board 10th Class Maths Solutions 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 71th term of the AP :

Solution :

(i) We have, a = 7, d = 3, n = 8, an = ?

We know that 71th term of an AP is : a_{n} = a + (n – 1)d

⇒ a_{8} = 7 + (8 – 1) × 3

⇒ a_{8} = 7 + 7 × 3

⇒ a_{8}= 7 + 21

⇒ a_{8} = 28

Hence, a_{8} = 28.

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

We know that 7th term of an AP is a_{n} = a + (n – 1) d

⇒ 0 = – 18 + (10 – 1) × d

⇒ 0 = – 18 + 9d

⇒ 9d = 18

⇒ d = \(\frac{18}{9}\) = 2

Hence, d = 2.

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

We know that n term of an AP is

a_{n} = a + (n – 1)d

⇒ – 5 = a + (18 – 1) × (- 3)

⇒ – 5 = a + 17 × (- 3)

⇒ – 5 = a – 51

⇒ a = 51 – 5 = 46

Hence, a = 46.

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

We know that 71th term of an AP is

a_{n} = a + (n – 1) d

⇒ 3.6 = – 18.9 + (n – 1) × 2.5

⇒ 3.6 = – 18.9 + 2.5n – 2-5

⇒ 3.6 = – 21.4 + 2.5n

⇒ 2.5n = 3.6 + 21.4

⇒ 2.5n = 25

⇒ n = \(\frac{25}{2.5}\)

⇒ n = 10

Hence, n = 10.

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

We know that nth term of an AP is

a_{n} = a + (n – 1) d

⇒ a_{105} = 3.5 + (105 – 1) × 0

⇒ a_{105} = 3.5 + 0

⇒ a_{105} = 3.5

Hence, a_{105} = 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 sequence of AP is : 10, 7, 4, …

Here a = 10, d = 7 – 10 = – 3, n = 30

We know that 77th term of an AP is a_{n} = a + (n – 1) d

⇒ a_{30} = 10 + (30 – 1) × (- 3)

⇒ a_{30} = 10 + 29 × (- 3)

⇒ a_{30} = 10 – 87

a_{30} = – 77

Hence, the correct option is (C).

(ii) The given sequence is : – 3, – \(\frac{1}{2}\), 2, ………………..

Here a = – 3,

d = – \(\frac{1}{2}\) – (- 3)

= – \(\frac{1}{2}\) + 3

= \(\frac{-1+6}{2}=\frac{5}{2}\)

and n= 11

We know that nth term of an AP is a_{n} = a + (n – 1) d

⇒ a_{11} = – 3 + (11 – 1) × \(\frac{5}{2}\)

⇒ a_{11} = – 3 + 10 × \(\frac{5}{2}\)

⇒ a_{11} = – 3 + 25

⇒ a_{11} = 22.

Hence, the correct option is (B).

Question 3.

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

(i) 2, ___, 26

(ii) ___, 13, ___, 3

(iii) 5, ___, ___, 91

(iv) – 4, ___, ___, ___, ___, 6

(v) ___, 38, ___, ___, ___, – 22.

Solution :

(i) We have, a = 2, a_{3} = 26

Now, a_{3} = 26

⇒ a + (3 – 1)d = 26

⇒ a + 2d = 26

⇒ 2 + 2d = 26 [∵ a = 2]

⇒ 2d = 26 – 2

⇒ d = \(\frac{24}{2}\) = 12.

Then missing term = a_{2} = a + d = 2 + 12 = 14

Hence, the missing term = 14.

(ii) We have, a_{2} = 13

⇒ a + d = 13 …(1)

and a_{4} = 3.

⇒ a + (4 – 1)d = 3

⇒ a + 3d = 3 ……………..(2)

Subtracting equation (2) from equation (1), we get

Putting the value of d in equation (1), we get

a – 5 = 13

a = 13 + 5

⇒ a = 18

⇒ 3rd missing term (a_{3}) = a + 2d

⇒ a_{3} = 18 + 2 × (- 5)

⇒ a_{3} = 18 – 10

⇒ a_{3} = 8.

Hence, 1st and 3rd missing terms are 18, 8.

(iii) We have, a = 5

a_{4} = 9\(\frac{1}{2}\)

a +(4 – 1) d = \(\frac{19}{2}\)

a + 3d = \(\frac{19}{2}\) …………..(1)

5 + 3d = \(\frac{19}{2}\) [∵ a = 5]

3d = \(\frac{19}{2}\) – 5

3d = \(\frac{19-10}{2}\)

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

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

So, 2nd missing term (a_{2}) = a + d

⇒ a_{2} = 5 + \(\frac{3}{2}\)

⇒ a_{2} = \(\frac{13}{2}\)

⇒ a_{2} = 6\(\frac{1}{2}\)

and 3rd missing term (a_{3}) = a + 2d

⇒ a_{3} = 5 + 2 × \(\frac{3}{2}\)

a_{3} = 5 + 3

a_{3} = 8

Hence, 2nd and 3rd missing terms are 6 \(\frac{1}{2}\), 8.

(iv) We have a = – 4 and a_{6} = 6

a + (6 – 1)d = 6

a + 5d = 6

– 4 + 5d = 6 [:a = – 4]

5d = 6 + 4

5d = 10

d = \(\frac{10}{2}\) = 2

d = 2

2nd missing term (a_{2}) = a + d

a_{2} = – 4 + 2

a_{2} = – 2

3rd missing term (a_{3}) = a + 2d

a_{3} = – 4 + 2 × 2

a_{3} = – 4 + 4

a_{3} = 0

4th missing term (a_{4}) = a + 3d

a_{4} = – 4 + 3 × 2

a_{4} = – 4 + 6

a_{4} = 2

5th missing term (a_{5}) = a + 4d

⇒ a_{5} = -4 + 4 × 2

⇒ a_{5} = – 4 + 8

⇒ a_{5} =4

Henae, 2nd, 3rd, 4th, 5th missing terms are – 2, 0, 2, 4.

(v) We have, a_{2} = 38

⇒ a + d = 38 ……………(1)

and a_{6} = – 22

⇒ a + (6 – 1)d = – 22

⇒ a + 5d = – 22 …………..(2)

Subtracting equation (2) from (1), we get

⇒ d = – 15

Substituting the value of d in equation (1), we

a – 15 = 38

⇒ a = 38 + 15

⇒ a = 53

So, 1st missing term = 53

3rd missing term (a_{3}) = a + 2d

⇒ a_{3} = 53 + 2 × (- 15)

⇒ a_{3} = 53 – 30

⇒ a_{3} = 23

4th missing term (a_{4}) = a + 3d

⇒ a_{4} = 53 + 3 × (- 15)

⇒ a_{4} = 53 – 45

⇒ a_{4} = 8

5th missing term (a_{5}) = a + 4d

⇒ a_{5} = 53 + 4 × (- 15)

⇒ a_{5} = 53 – 60

⇒ a_{5}= – 7

Hence, 1st, 3rd, 4th, 5th missing terms are: 53, 23, 8, – 7.

Question 4.

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

Solution :

The given sequence of AP is : 3, 8, 13, 18, …

Here, a = 3, d = a_{2} – a_{1} = 8 – 3 = 5, a_{n} = 78

We know that nth term of AP is

a_{n} = a + (n – 1 )d.

⇒ 78 = 3 + (n – 1) × 5

⇒ 78 = 3 + 5n – 5

⇒ 78 = – 2 + 5n

⇒ 78 + 2 = 5 n

⇒ 5n = 80

⇒ n = \(\frac{80}{2}\) = 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) The given sequence of AP is : 7, 13, 19, …, 205

Here, a = 7, d = a_{2} – a_{1} = – 13 – 7 = 6, a_{n} = 205

We know that term of AP is

a_{n} = a + (n – 1) d

⇒ 205 = 7 + (n – 1) × 6

⇒ 205 = 7 + 6n – 6

⇒ 205 = 1 + 6n

⇒ 6n = 205 – 1

⇒ 6n = 204

n = \(\frac{204}{6}\) = 34

Hence, the number of terms = 34.

(ii) The given sequence of AP is 18, 15\(\frac{1}{2}\), 13, ……….. – 47

Here, a = 18, d = a_{2} – a_{1}

= 15\(\frac{1}{2}\) – 18

= \(\frac{31}{2}\) – 18

= \(\frac{31-36}{2}=-\frac{5}{2}\)

d = – \(\frac{5}{2}\)

and a_{n} = – 47

We know that nth term of AP is a_{n} = a + (n – 1)d

\(-\frac{135}{2}=-\frac{5}{2} n\)

n = \(-\frac{135}{2} \times \frac{2}{-5}\) = 27

Hence, the number of terms = 27.

Question 6.

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

Solution :

The given sequence of AP is : 11, 8, 5, 2,…

Here, a = 11

d = a_{2} – a_{1}

= 8 – 11 = – 3 and

a_{n} = – 150

We know that nth term of AP is

a_{n} = a + (n – 1) d

– 150 = 11 + (n – 1) x (- 3)

– 15o = 11 – 3n + 3

– 150 = 14 – 3n

150 – 14 = – 3n

– 3n = – 164

n = \(\frac{-164}{-3}=54 \frac{2}{3}\)

∵ n is a natural number, so it is not possible.

Hence, – 150 is not a term of given AP.

Question 7.

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

Solution :

We have,

a_{11} = 38

⇒ a + (11 – 1)d = 38

⇒ a + 10d = 38 ……………(1)

and a_{16} = 73

⇒ a + (16 – 1)d = 73

⇒ a + 15d = 73 ……………(2)

Subtracting equation (2) from equation (1), we get

Putting the value of d in equation (1), we get

a + 10 × 7 = 38

⇒ a + 70 = 38

⇒ a = 38 – 70 = – 32

We know that nth term of AP is :

a_{n} = a + (n – 1 )d

⇒ a_{331} = – 32 + (31 – 1) × 7

⇒ a_{31} = – 32 + 30 × 7

⇒ a_{31} = – 32 + 210

⇒ a_{31} = 178

Hence, 31st term of AP = 178.

Question 8.

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

Solution :

We have,

n = 50

a_{3} = 12

⇒ a+ (3 – 1)d = 12

⇒ a + 2d = 12 ………….(1)

and last term (a_{50}) = 106

⇒ a + (50 – 1 )d = 106

⇒ a + 49d = 106 ……………(2)

Subtracting equation (2) from equation (1), we get

Putting the value of d in equation (1), we get

a + 2 × 2 = 12

⇒ a + 4 = 12

⇒ a = 12 – 4

⇒ a = 8

We know that nth term of AP is a_{n} = a + (n – 1 )d

a_{29} = 8 + (29 – 1) × 2

a_{29} = 8 + 28 × 2

a_{29} = 8 + 56

a_{29} = 64.

Hence, 29th term of AP = 64.

Question 9.

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

Solution :

We have, a_{3} = 4

⇒ a + (3 – 1)d = 4

⇒ a + 2d = 4 ………….(1)

and a_{9} = – 8

⇒ a + (9 – 1)d = – 8

⇒ a + 8d = – 8 ………….(2)

Subtracting equation (2) from equation (1), we get

putting the value of d in equation (1), we get

a + 2 × (- 2) = 4

⇒ a – 4 = 4

so a = 4 + 4

⇒ a = 8

We know that nth term of AP is :

a_{n} = a + (n – 1)d

0 = 8 + (n – 1) × (- 2)

[∵ given that a_{n} = 0]

⇒ 0 = 8 – 2n + 2

0 = 10 – 2n

⇒ 2n = 10

n = \(\frac{10}{2}\) = 5

Hence, 5th term of AP is zero.

Question 10.

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

Solution :

We have,

a_{17} – a_{10} = 7

⇒ a + (17 – 1)d – [a + (10 – 1)d] = 7

⇒ a + 16d – a – 9d = 7

⇒ 7d = 7

⇒ d = \(\frac{7}{7}\)

⇒ d = 1

Hence, common difference = 1.

Question 11.

Which term of the AP : 3, 15, 27, 39, … will be 132 more than its 54th, term?

Solution:

The given sequence of AP is: 3, 15, 27, 39, ……………

Here, a = 3

d = a_{2} – a_{1} = 15 – 3 = 12

Let nth term of given AP be 132 more than its 54th term

i. e. a_{n} = a_{54} + 132

⇒ a + (n – 1 )d = 3 + (54 – 1) × 12 + 132

3 + (n – 1) × 12 = 3 + 53 × 12 + 132

3 + (n – 1) × 12 = 3 + 636 + 132

⇒ 3 + 12n – 12 = 771

⇒ 12n – 9 = 771

⇒ 12n = 771 + 9

⇒ 12n = 780

⇒ n = \(\frac{780}{12}\)

⇒ n = 65

Hence 65th term is 132 more than 54th term.

Question 12.

Two AP’s have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?

Solution :

Let the two APs, with same common difference d be

a, a + d, a + 2d, a + 3d, …

and b, b + d, b + 2d, b + 3d, …

According to question,

a_{100} – b_{100} = 100

⇒ a + (100 – 1)d – [b + (100 – 1)d] = 100

⇒ a + 99d – b – 99d = 100

⇒ a – b = 100 ……………..(1)

Now, a_{1000} – b_{1000} = a + (1000 – 1)d – [b + (1000 – 1)d]

⇒ a_{1000} – b_{1000} = a + 999d – b – 999d

⇒ a_{1000} – b_{1000} = a – b

⇒ a_{1000} – b_{1000} = 100

[From equation (1) a – b = 100]

Hence, the difference between 1000th terms = 100.

Question 13.

How many three-digit numbers are divisible by 7?

Solution :

The list of three digit numbers which are divisible by 7 is : 105, 112, 119, 126,…, 994.

Let n numbers be divisible by 7.

It is an AP with a = 105, d = 7, a_{n} = 994.

We know that nth term of AP is

a_{n} = a + (n – 1 )d

⇒ 994 = 105 + (n – 1) × 7

⇒ 994 = 105 + 7n – 7

⇒ 994 = 98 + 7n

⇒ 7n = 994 – 98

⇒ 7n = 896

n = \(\frac{896}{7}\) = 128

Hence, 128 numbers are divisble by 7.

Question 14.

How many multiples of 4 lie between 10 and 250?

Solution :

Let the multiples of 4 lying between 10 and 250 are : 12, 16, 20, …………… 248.

Let the multiples of 4 between 10 and 250 be n.

It is an AP with a = 12, d = 4, a_{n} = 248.

We know that nth term of AP is

a_{n} = a + (n – 1)d

248 = 12 + (n – 1) × 4

248 = 12 + 4n – 4

248 = 8 + 4n

248 = 4n

4n = 240

n = \(\frac{240}{4}\) = 60

Hence, 60 multiples of 4 lie between 10 and 250.

Question 15.

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

Solution:

The first given AP is : 63, 65, 67, …………..

Here, a = 63

d = a_{2} – a_{1}

= 65 – 63 = 2

Let nth terms of given two APs are equal

a_{n} = a + (n – 1)d

⇒ a_{n} = 63 + (n – 1) × 2

⇒ a_{n} = 63 + 2n – 2

a_{n} = 61 + 2n …………….(1)

The second given AP is : 3, 10, 17, …………..

Here a = 3

d = a_{2} – a_{1} = 10 – 3 = 7

a_{n} = a + (n- 1 )d

⇒ a_{n} = 3 + (n – 1) × 7

⇒ a_{n} = 3 + 7n – 7

⇒ a_{n} = – 4 + 7n …………(2)

Since, /ith terms of given two APs are equal, from equations (1) and (2), we get

61 + 2n = – 4 + 7n

⇒ 7n – 2n = 61 + 4

⇒ 5n = 65

⇒ n = \(\frac{65}{5}\)

⇒ n = 13

Hence, 13th terms of given two APs are equal.

Question 16.

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

Solution :

We have,

a_{3} = 16

a + (3 – 1 )d = 16

⇒ a + 2d = 16 …………..(1)

and given, 7th term – 5th term = 12

⇒ a_{7} – a_{5} = 12

⇒ a + (7 – 1 )d – [a + (5 – 1) d] = 12

⇒ a + 6d – a – 4d = 12

⇒ 2d = 12

⇒ d = \(\frac{12}{2}\) = 6

Putting the value of d in equation (1), we get

a + 2 × 6 = 16

⇒ a + 12 = 16

⇒ a = 16 – 12

⇒ a = 4

So, the AP is : a, a + d, a + 2d, a + 3d, ………..

i.e. 4, 4 + 6, 4 + 2 × 6, 4 + 3 × 6, ……………….

i.e. 4, 10, 16, 22, …………..

Hence, AP is 4, 10, 16, 22, ……………..

Question 17.

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

Solution:

For finding the 20th term from the last term of AP.

We write the given AP in reverse order as follows : 253, …………….. 13, 8, 3.

Here, a = 253

d= 3 – 8 = – 5

We know that nth term of AP is :

a_{n} = a + (n – 1) d

⇒ a_{20} = 253 + (20 – 1) × (- 5)

⇒ a_{20} = 253 + 19 × (- 5)

⇒ a_{20} = 253 – 95

⇒ a_{20} = 158

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

Question 18.

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

Solution:

We have,

a_{4} + a_{8} = 24

⇒ a + (4 – 1 )d + a + (8 – 1 )d = 24

⇒ o + 3d + o + 7d = 24

⇒ 2a + 10d= 24 …………..(1)

⇒ a + 5d = 12

and a_{6} + a_{10} = 44

⇒ a + (6 – 1)d + a + (10 – 1)d = 44

⇒ a + 5d + a + 9d = 44

⇒ 2a + 14d = 44

⇒ a + 7d = 22 …………….(2)

Subtracting equation (2) from equation (1), we get

Putting the value of d in equation (1), we get

a + 5 × 5 = 12

⇒ a + 25 = 12

⇒ a = 12 – 25

⇒ a = – 13

So, the first three terms of AP are : a, a + d, a + 2d.

i.e. – 13, – 13 + 5, – 13 + 2 × 5

i.e. – 13, – 8, – 3.

Hence, the first three terms of AP are – 13, – 8, – 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 :

We have,

Increment of each year = ₹ 200

Income in 1995 = ₹ 5000

Income in 1996 = 5000 + 200 = ₹ 5200

Income in 1997 = 5200 + 200

= ₹ 5400 and so on.

Sequence of salary of Subba Rao is : 5000, 5200, 5400, ………….., 7000

a_{2} – a_{1}= 5200 – 5000 = 200

a_{3} – a_{2} = 5400 – 5200 = 200

a_{2} – a_{1} = a_{3} – a_{2}

∴ The given sequence of salary forms an AP.

Here a = 5000

d= 5200 – 5000 = 200 and a_{n} = 7000

We know that nth term of AP is

a_{n} = a + (n – 1) d

7000= 5000 + (n – 1) × 200

7000 = 5000 + 200n – 200

7000 = 4800 + 200n

200n = 7000 – 4800

200n = 2200

n = \(\frac{2200}{200}\)

n = 11

Hence, in 11th year Subba Rao’s income reached ₹ 7000.

Question 20.

Reunkali saved ₹ 5 in the first week of a year and then increased her weekly savings by ₹ 1.75. If in the nth week, her weekly savings become ₹ 20.75, Find n.

Solution:

We have weekly increment = ₹ 1.75

Saving in first week = ₹ 5.00

Saving in second week = ₹ 5.00 + ₹ 1.75 = ₹ 6.75

Saving in third week= 6.75 + 1.75

= Rs. 8.50 … and so on.

Sequence of savings of Ramkali is : 5.00, 6.75, 8.50, ……………. 20.75.

a_{2} – a_{1} = 6.75 – 5.00 = 1.75

a_{3} – a_{2} = 8.50 – 6.75 = 1.75

a_{2} – a_{1} = a_{3} – a_{2}

∴ The sequence of savings of Ramkali forms an AP

Here, a = 5.00

d= 6.75 – 5.00 = 1.75

a_{n} = 20.75

We know that nth term of AP is

a_{n} = a + (n – 1 )d

⇒ 20.75 = 5 + (n – 1) × 1.75

⇒ 20.75 = 5 + 1.75n – 1.75

⇒ 20.75 = 3.25 + 1.75n

⇒ 1.75n = 20.75 – 3.25

⇒ 1.75n = 17.50

⇒ n = \(\frac{17.50}{1.75}\)

⇒ n = \(\frac{1750}{175}\)

Hence, in 10th week Ramkali’s savings become ₹ 20.75.