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

## BSEB Bihar Board Class 10th 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_{n} the n^{th} term of the AP:

Solution:

Blanks may be filled as under:

(i) a_{n} – a + (n – 1)d = 7 + (8 – 1)3

= 7 + 7 × 3 = 7 + 21 = 28

(ii) a_{n} = a + (n – 1)d So, 0 = – 18 + (10 – 1)d

or 18 = 9d or d = \(\frac{18}{9}\) = 2

(iii) a_{n} = a + (n – 1)d So, – 5 = a + (18 – 1)(- 3)

or – 5 = a + 17(- 3) or – 5 = a – 51

or a = – 5 + 51 = 46

(iv) a_{n} = a + (n – 1)d So, 3.6 = – 18.9 + (n – 1)25

or 3.6 + 18.9 = (n – 1)2.5 or 22.5 = (n – 1)2.5

or n – 1 = \(\frac{22.5}{2.5}\) or n – 1 = 9

or n = 9 + 1 = 10

(v) a_{n} = a + (n – 1)d = 3.5 + (105 – 1) × 0

= 3.5 + 0 = 3.5

Question 2.

Choose the correct choice in the following and justify:

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

(A) 97

(B) 77

(C) – 77

(D) – 87

(ii) 11th term of the – 3, – \(\frac{1}{2}\) + 3 = \(\frac{-1+6}{2}\) = \(\frac{5}{2}\) and n = 11.

We know that a_{n} = a + (n – 1)d

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

= – 3 + 10 × \(\frac{5}{2}\) = – 3 + 25 = 22

So, (B) is the correct choice.

Question 3.

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

(i) 2, , 26

(ii)

, 13, , 3

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

(iv) – 4, , , , , 6

(v) , 38, , , , – 22.

Solution:

(i) Let 2, and 26 be a, (a + d) and (a + 2d) respectively.

∴ a = 2 and a + 2d = 26

So, 2 + 2d = 26 or 2d = 26 – 2 = 24

or d = \(\frac{24}{2}\) = 12

Thus, the missing term = a + d = 2 + 12 = 14.

(ii) Let , 13, and 3 be a, a + d, a + 2d and a + 3d respectively.

∴ a + d = 13 ……………. (1)

and a + 3d = 3 ……………. (2)

So, (2) – (1) gives 2d = – 10 or d = – 5

Putting d = – 5 in (1), we get

a – 5 = 13 or a = 13 + 5 = 18

∴ The missing terms are a, i.e., 18

and a + 2d, i.e., 18 + 2(- 5) = 18 – 10 = 8.

(iii) Let 5, , and 9\(\frac{1}{2}\) be a, a + d, a + 2d and a + 3d respectively.

∴ a = 5

and a + 2d, i.e; 5 + 2 × \(\frac{3}{2}\) = 5 + 3 = 8.

(iv) Let – 4, ,, , and 6 be a, a + d, a + 2d, a + 3d, a + 4d and a + 5d respectively.

∴ a = – 4 ………………. (1)

and a + 5d = 6 ……………… (2)

So, (2) – (1) gives 5d = 10 or d = 2

The missing terms are

a + d = – 4 + 2 = – 2,

a + 2d = – 4 + 2 × 2 = – 4 + 4 = 0,

a + 3d = – 4 + 3 × 2 = – 4 + 6 = 2,

and a + 4d = – 4 + 4 × 2 = – 4 + 8 = 4.

(v) Let , 38, , , and – 22 be a, a + d, a + 2d, a + 3d, a + 4d and a + 5d respectively.

∴ a + d = 38 ……….. (1)

and a + 5d = – 22 ……………. (2)

So, (2) – (1) gives 4d = – 60 or d = – 15

Putting d = – 15 in (1), we get

a – 15 = 38 or a = 38 + 15 = 53

∴ The missing terms are a, i.e., 53,

a + 2d, i.e., 53 + 2 × (- 15) = 53 – 30 = 23,

a + 3d, i.e., 53 + 3 × (- 15) = 53 – 45 = 8,

and a + 4d, i.e., 53 + 4 × (- 15) = 53 – 60 = – 7

Question 4.

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

Solution:

We have:

a = 3, d = 8 – 3 – 5.

Let 78 be the nth term of the given AP. Then,

a_{n} = 78 or a + (n – 1 )d = 78

∴ 3 + (n – 1)5 = 78 or (n – 1)5 = 78 – 3

or 5(n – 1) = 75 or n – 1 = 15

or n = 15 + 1 i.e., n = 16

Thus, 78 is the 16th term of the given AP.

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) Clearly, it forms an AP with first term a – 3 and common difference d = 13 – 7 = 6.

Let there be n terms in the given AP.

Then, nth term = 205.

So, a + (n – 1)d = 205 or 7 + (n – 1)6 = 205

or 6(n – 1) = 205 – 7 or 6(n – 1) = 198

or n – 1 = \(\frac{198}{6}\) = 33 or n = 33 + 1 = 34

Thus, the given AP contains 34 terms.

(ii) Let there be n terms in the given AP.

18, 15 \(\frac{1}{2}\), 13, …, – 47. Clearly, it forms an AP with first term a = 18 and common difference d = 15\(\frac{1}{2}\) – 18 = \(\frac{31}{2}\) – 18 = \(\frac{31-36}{2}\) = \(\frac{-5}{2}\). Then, nth term = – 47.

So, a + (n – 1)d = – 47 or 18 + (n – 1)(- \(\frac{5}{2}\)) = – 47

or (- \(\frac{5}{2}\)) (n – 1) = – 47 – 18 = – 65

or n – 1 = – 65 × \(\frac{-2}{5}\) = – 13 × (- 2) = 26

Thus, there are 27 terms in the given AP.

Question 6.

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

Solution:

Here, a_{2} – a_{1} = 8 – 11 = – 3

a_{3} – a_{2} = 5 – 8 = – 3

a_{4} – a_{3} = 2 – 5 = – 3

As a_{n+1} – a_{n} is same every time, so the given list of numbers is an AP.

Now, a = 11, d = – 3.

Let – 150 be the nth term of the given AP.

we know that a_{n} = a + (n – 1)d

So, – 150 = 11 + (n – 1)(- 3)

or – 3(n – 1) = – 150 – 11 = – 161

or n – 1 = \(\frac{161}{3}\) or n = \(\frac{161}{3}\) + 1 = \(\frac{164}{3}\)

But n should be a positive integer. So, our assumption was wrong and so – 150 is not a term of the given AP.

Question 7.

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

Solution:

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

Now, a_{n} = a + (n – 1)d

∴ a_{11} = a + 10d = 38 …………….. (1)

and a_{16} = a + 15d = 73 …………….. (2)

Subtracting (1) from (2), we get

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

and then from (1),

a + 10 × 7 = 38

or a = 38 – 70 = – 32

∴ a_{31} = a + 30d = – 32 + 30 × 7

= – 32 + 210 = 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:

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

Now, a_{n} = a + (n – 1)d

∴ a_{3} = a + 2d = 12 …………… (1)

and a_{50} = a + 49d = 106 …………….. (2)

Subtracting (1) from (2), we get

47d = 94 or d = \(\frac{94}{47}\) = 2

and then from (1),

a + 2 × 2 = 12 or a = 12 – 4 = 8

∴ a_{29} = a + 28d = 8 + 28 × 2

= 8 + 56 = 64

Question 9.

If the 3rd and 9th terms of an AP are 4 and – 8 respectively, which term of this AP is 0?

Solution:

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

a_{3} = a + 2d = 4 ………….. (1)

a_{9} = a + 8d = – 8 ………….. (2)

Subtracting (1) from (2), we get

6d = – 12 or d = \(\frac{-12}{6}\) = – 2

and then from (1),

a + 2 × (- 2) = 4 or a + 4 + 4 = 8

Let a_{n} = 0, So, a + (n – 1)d = 0

or 8 + (n – 1)(- 2) = 0 or (n – 1)(- 2) = – 8

or n – 1 = \(\frac{-8}{-2}\) = 4 or n = 4 + 1 = 5

Thus, 5th term of the AP is 0.

Question 10.

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

Solution:

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

It is given that a_{17} – a_{10} = 7 or (a + 16d) – (a + 9d) = 7

or 7d = 7 or d = 1

Thus, the common difference is 1.

Question 11.

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

Solution:

Here, a = 3, d = 15 – 3 = 12. Then,

a_{54} = a + 53d = 3 + 53 × 12

= 3 + 636 = 639

Let an be 132 more than its 54th term

i.e., a_{n} = a_{54} + 132 = 639 + 132 = 771

So, a + (n – 1)d = 771 or 3 + (n – 1)12 = 771

or 12(n – 1) = 771 – 3 or 12(n – 1) = 768

or n – 1 = \(\frac{768}{12}\) = 64 or n = 64 + 1 = 65

Thus, 65th term of the AP is 132 more than its 54th term.

Question 12.

Two APs 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 be a_{1}, a_{2}, a_{3}, …………… a_{n}, …………. and b_{1}, b_{2}, b_{3}, …………, b_{n}, ………….

Also, let d be the common difference of each of the two APs. Then,

a_{n} = a_{1} + (n – 1 )d and b_{n} = b_{1} + (n – 1)d

or a_{n} – b_{n} = [a_{1} + (n – 1)d] – [b_{1} + (n – 1)d]

or a_{n} – b_{n} = a_{1} – b_{1} for all natural numbers n

or a_{100} – b_{100} = a_{1} – b_{1} = 100 [Given]

Now, a_{1000} – b_{1000} = a_{1} – b_{1}

So, a_{1000} – b_{1000} = 100 [∵ a_{1} – b_{1} = 100]

Hence, the difference between 1000th terms is the same as the difference between 100th terms, i.e., 100.

Question 13.

How many three-digit numbers are divisible by 7?

Solution:

105 is the first three-digit number divisible by 7 and 994 is the last 3-digit number divisible by 7. Thus, we have to determine the number of terms in the list 105, 112, 119, …, 994.

Clearly, it forms an AP with first term a = 105

and common difference d = 112 – 105 = 7.

Let there be n terms in the AP. Then, nth term = 994.

As a_{n} = a + (n – 1)d, therefore

105 + (n – 1)7 = 994 or 7(n – 1) = 994 – 105

or 7(n – 1) = 889 or n – 1 = \(\frac{889}{7}\) = 127

or n = 127 + 1 = 128

Hence, there are 128 numbers of three digit which are divisible by 7.

Question 14.

How many multiples of 4 lie between 10 and 250?

Solution:

We observe that 12 is the first natural number between 10 and 250, which is a multiple of 4 (i.e., divisible by 4).

Also, when we divide 250 by 4, the remainder is 2. Therefore, 250 – 2 = 248 is the largest natural number divisible by 4 (i.e., multiple of 4) and lying between 10 and 250.

Thus, we haye to find the number of terms in an AP with first term = 12, last term = 248, and common difference = 4 (as the numbers are divisible by 4).

Let there be n terms in the AP. Then

a_{n} = 248 So, 12 + (n – 1)4 = 248

or 4(n – 1) = 248 – 12 or 4(n – 1) = 236

or n – 1 = \(\frac{236}{4}\) = 59 or n = 59 + 1 = 60

Hence, there are 60 multiples of 4 between 10 and 250.

Question 15.

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

Solution:

If nth terms of the APs 63, 65, 67, … and 3, 10, 17, … are equal, then

63 + (n – 1)2 = 3 + (n – 1)7

[∵ In 1st AP, a = 63, d = 65 – 63 = 2 and in 2nd AP, a = 3, d = 10 – 3 = 7]

or 7(n – 1) – 2(n – 1) = 63 – 3 or (n – 1)(7 – 2) = 60

or 5(n – 1) = 60 or n – 1 = \(\frac{60}{5}\) = 12

or n = 12 + 1 = 13

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

Question 16.

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

Solution:

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

Here, a_{3} = 16 and a_{7} – a_{5} = 12

So, a + 2d = 16 ……….. (1)

and (a + 6d) – (a + 4d) = 12 or 2d = 12

or d = 6 ……………. (2)

Using (2) in (1), we get

a + 2 × 6 = 16. or a = 16 – 12 = 4

Thus, the required AP is 4, 4 + 6, 4 + 2 × 6, 4 + 3 × 6, … i.e., 4, 10, 16, 22, ……

Question 17.

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

Solution:

We have:

l = Last term = 253

and d = Common difference = 8 – 3 = 5

∴ 20th term from the end = l – (20 – 1)d = l – 19d

= 253 – 19 × 5

= 253 – 95 = 158

Question 18.

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 first three terms of the AP.

Solution:

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

Here, a_{4} + a_{8} = 24 or (a + 3d) + (a + 7d) = 24

or 2a + 10d = 24 or a + 5d = 12 ……….. (1)

and a_{6} + a_{10} = 44 or (a + 5d) + (a + 9d) = 44

or 2a + 14d = 44 or a + 7d = 22 ……… (2)

Subtracting (1) from (2), we get

2d = 10 or d = 5

and then from (1),

a + 25 = 12 or a = – 13

The first three terms are a, (a + d) and (a + 2d)

Putting values of a and d, we get – 13, (- 13 + 5) and (- 13 + 2 × 5)

i.e., – 13, – 8 and – 3.

Question 19.

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

Solution:

The annual salaries drawn by Subba Rao in the years 1995, 1996, 1997, etc. are Rs 5000, Rs 5200, Rs 5400, …, Rs 7000.

The list of these numbers is 5000, 5200, 5400, …, 7000.

It forms an AP. [∵ a_{2} – a_{1} = a_{3} – a_{2} = 200]

Let a_{n} = 7000

So, 7000 = a + (n – 1)d

or 7000 = 5000 + (n – 1)(200)

or 200(n – 1) = 7000 – 5000

or n – 1 = \(\frac{2000}{200}\) = 10 or n = 10 + 1 = 11

Thus, in the 11th year (i.e., in 2005) of his service, Subba Rao drew an annual salary of Rs 7000.

Question 20.

Ramkali saves Rs 5 in the first week of a year and then increased her weekly savings by Rs 1.75. If in the nth week, her weekly savings become Rs 20.75, find n.

Solution:

Ramkali’s savings in the subsequent weeks are respectively Rs 5, Rs 5 + Rs 1.75, Rs 5 + 2 × Rs 1.75, Rs 5 + 3 × Rs 1.75, …

And in the nth week her savings will be Rs 5 + (n – 1) × Rs 1.75

or 5 + (n – 1) × 1.75 = 20.75

or (n – 1) × 1.75 = 20.75 – 5 = 15.75

∴ n – 1 = \(\frac{15.75}{1.75}\) = 9

or n = 9 + 1 = 10