Exercise 5.2

1. Fill in the blanks in the following table, given that a is the first term, d the common difference and an the nth term of the AP:

Solution:

Note : The formula for the nth term of an AP is, an = a + (n - 1)d.

a	d	n	an
7	3	8
-18		10	0
	-3	18	-5
-18.9	2.5		3.6
3.5	0	105

2. Choose the correct choice in the following and justify :

(i)

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

Solution:

The AP is 10, 7, 4, …

Note : The formula for the nth term of an AP is, an = a + (n - 1)d.

Here, a = , n =

Common difference d = a2 - a1 = 7 - 10 =

an= a + (n - 1)d

a30 = 10 + (30 - 1)(- 3)

= 10 + ()(- 3)

= 10 -

Thus, 30th term of the AP is - 77.

(ii)

(ii) 11th term of the AP: -3, −12, 2,..., is

Solution:

The AP is - 3, −12 , 2

Note : The formula for the nth term of an AP is, an = a + (n - 1)d.

a = , n =

Common difference d = a2 - a1

= −12 - (- 3)

= −12 +

= (- 1 + 6) 2

d =

an = a + (n - 1)d

a = - 3 + (11 - 1) (52)

= - 3 + 10 × (52)

= - 3 +

a11 =

Thus, the 11th term of the AP is 22.

3. In the following APs, find the missing terms.

(i)

2, _ , 26

Solution:

We know that the nth term of an AP is, an = a + (n - 1)d.

Let the common difference be d

First term a =

Third term = a + (3 - 1)d = a + 2d = + 2d

Third term = (Given)

2 + 2d =

2d = 26 -

2d =

d =

Second term = a +

= 2 + 12 =

The missing term is .

(ii)

_ , 13, _ , 3

Solution:

We know that the nth term of an AP is, an = a + (n - 1)d.

Second term =

a + (2 - 1) d = 13

a + d = ....(1)

Fourth term =

a + (4 - 1) d = 3

a + 3d = ....(2)

Solve equations (1) and (2) to find a and d.

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

a + 3d - (a + d) = 3 - 13

3d - = 3 - 13

2d =

d =

Putting d = -5 in equation (1)

a + (- 5) = 13

a - 5 = 13

a =

Third term = a + (3 - 1)d = a + 2d

= 18 + 2(-5) =

The missing terms are 18 and 8 respectively.

(iii)

5, _ , _ , 912

Solution:

We know that the nth term of an AP is, an = a + (n - 1)d.

First term a =

Fourth term, a4 = 912 =

Fourth term, a4 = a + (4 - 1) d

5 + 3d = 192

6d = 9

d =

Second term = a2 = a + d

= 5 + 32

Third term, a3 = a + 2d

= 5 + 2 × (32)

= 5 + 3 =

The missing terms in the boxes are 132 and 8 respectively.

(iv)

- 4, _ , _ , _ , _ , 6

Solution:

We know that the nth term of an AP is, an = a + (n - 1)d.

First term a =

Sixth term a6 =

a + (6 - 1)d = a + 5d =

- 4 + 5d = 6

5d = 10

d =

Second term a + d = - 4 + 2 =

Third term a + 2d = - 4 + 4 =

Fourth term a + 3d = - 4 + 6 =

Fifth term a + 4d = - 4 + 8 =

The missing terms are - 2, 0, 2, 4.

(v)

_ , 38, _ , _ , _ , - 22

Solution:

We know that the nth term of an AP is, an = a + (n - 1)d.

Let the first term be = a

Common difference = d

Second term = (Given)

a + (2 - 1)d = a + = 38 ....(1)

Sixth term = (Given)

a + (6 - 1)d = a + = - 22 ....(2)

On Subtracting equation (1) from equation (2) for a and d, we get

a + 5d - (a + d) = -22 - 38

4d =

d =

Put the value of d in equation (1)

a - 15 = 38

a =

Third term = a + (3 - 1)d = a + 2d

= 53 + 2(-15) =

Fourth term = a + 3d

= 53 + 3(-15) =

Fifth term = a + 4d

= 53 + 4(-15) =

The missing terms are 53, 23, 8, - 7.

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

Solution:

The nth term of an AP is, an = a + (n - 1)d.

Common difference d = a2 - a1

First term: a =

Second term: a + d = 8

Common difference: d = 8 - 3 =

an = a + (n - 1)d

an = 78, n = ?

3 + (n - 1) 5 =

5(n - 1) = 78 - 3

n - 1 =

n =

78 is the 16th term of the given AP.

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

(i)

7, 13, 19,..., 205

Solution:

The nth term of an AP is, an = a + (n - 1)d.

Here, an is the nth term, a is the first term, d is the common difference and n is the number of terms.

a =

d = 13 - 7 =

an =

n = ?

an = a + (n - 1)d

a + (n - 1)d =

+ (n - 1) = 205

(n - 1) 6 = 205 -

(n - 1) 6 =

n - 1 =

Number of terms in the given Arithmetic Progression n is 34.

(ii)

18, 1512, 13 ,..., -47

Solution:

The nth term of an AP is, an = a + (n - 1)d.

a =

an =

n = ?

d = 1512 - 18 = 312 - 18 =

an = a + (n - 1)d

- 47 = a + (n - 1) d

- 47 = 18 + (n - 1) (−52)

- 65 = (−52) n + 52

52 n = 52 + 65

52 n = 1352

n = (1352) × (25) =

The number of terms in the given AP is 27.

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

Solution:

The nth term of an AP is, an = a + (n - 1)d.

Here,an is the nth term, a is the first term, d is the common difference and n is the number of terms.

11, 8, 5, 2 ... are in AP

First term a =

Common difference, d = 8 - 11 =

Let -150 be a term of the given AP.

an = a + (n - 1)d

a + (n - 1)d =

+ (n - 1)(- 3) = -150

11 - 3n + = -150

-3n =

n =

n is not a positive integer.

– 150 is not a term of the AP : 11, 8, 5, 2 . . . because n should be a positive integer.

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

Solution:

The nth term of an AP is, an = a + (n - 1)d.

Here, a is the first term, d is the common difference and n is the number of terms.

an = a + (n - 1)d

a11 =

a + (11 - 1) d =

a + = 38 ...(1)

a16 =

a + = 73 ...(2)

By subtracting equation (1) from equation (2) we get,

a + 15d - ( a + 10d ) = 73 -

5d =

d =

Putting the value of d in equation (1),

a + 10(7) = 38

a = 38 -

31st terms is, a31 = a + (31 - 1)d

= - 32 + 30 × 7

= - 32 +

The 31st term of the AP is 178.

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

Solution:

The nth term of an AP is, an = a + (n - 1)d.

Here, a is the first term, d is the common difference and n is the number of terms.

Third term of AP is a + (3 - 1)d = a +

a + 2d = .... (1)

Last term =

Thus, 50th term =106 [Since, n = 50]

a + (50 - 1)d =

a + = 106........ (2)

By solving equations (1) and (2) for the values of a and d,

a + 49d - (a + 2d) = 106 -

47d =

d =

Putting d = 2 in equation (1)

a + 2 × 2 = 12

a + = 12

a = 12 -

a =

29th term of the AP is a29 = a + (29 - 1)d

a29 = 8 + 28 × 2

a29 = 8 +

a29 =

Thus, 29th term of the AP is 64.

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

Solution:

The nth term of an AP is, an = a + (n - 1)d.

Here, a is the first term, d is the common difference and n is the number of terms.

3rd term of the AP =

a3 = a + (3 - 1)d = 4

a + = 4 .... (1)

9th term of the AP =

a9 = a + (9 - 1)d =

a + = - 8 ....(2)

Solving (1) and (2) for a and d

a + 2d - (a + 8d) = 4 - (- 8)

- = 12

d =

Putting d = - 2 in equation (1)

a + 2 × (- 2) = 4

a - = 4

a =

Now, by using the values of a and d, we will find the term for which the value is 0.

a + (n - 1)d = 0

8 + (n - 1)(-) = 0

n - 1 =

n =

Thus, 5th term of the AP will be 0.

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

Solution:

The nth term of an AP is, an = a + (n - 1)d.

Here, a is the first term, d is the common difference and n is the number of terms.

a17 = a + (17 - 1)d

a17 = a +

a10 = a + (10 - 1)d

a10 = a +

According to the question, a17 - a10 = (given)

a + 16d - (a + 9d) =

16d - = 7

= 7

d =

The common difference is 1.

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

Solution:

The nth term of an AP is, an = a + (n - 1)d.

Here, a is the first term, d is the common difference and n is the number of terms.

Given AP is 3, 15, 27, 39.

First term a =

Second term a + d =

d = 15 - 3 =

54th term of the AP is

a54 = a + (54 - 1)d

= 3 + × 12

= 3 +

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

We get, 132 + 639 =

an =

an = a + (n - 1)d

771 = + (n - 1)

= (n - 1)12

(n - 1) =

n =

Therefore, the 65th term will be 132 more than the 54th term.

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:

The nth term of an AP is, an = a + (n - 1)d.

Here, a is the first term, d is the common difference and n is the number of terms.

For first A.P., a100 = a1 + (100 - 1) d

a100 = a1 + ------ (1)

a1000 = a1 + (1000 - 1)d

a1000 = a1 + ------ (2)

For second A.P.,

b100 = b1+ (100 - 1)d

b100 = b1 + ------ (3)

b1000 =b1 + (1000 - 1)d

b1000 =b1 + ------ (4)

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

Thus, from equations (1) and (3) we have

(a1 + 99d ) - (b1 + ) = 100

a1 - b1 = ...(5)

Difference between 1000th terms of these A.P.s

Thus, from equations (2) and (4) we have

(a1 + ) - (b1 + 999d ) = a1 - b1

But a1 - b1 = [From equation(5)]

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

13. How many three-digit numbers are divisible by 7?

Solution:

The nth term of an AP is, an = a + (n - 1)d.

Here, a is the first term, d is the common difference and n is the number of terms.

First three-digit number that is divisible by 7 =

Next number = 105 + 7 =

Therefore, the series becomes 105, 112, , ...

Thus, 105, 112, 119, ... is forming an A.P. having the first term as 105 and a common difference of .

When we divide 999 by 7, the remainder will be .

Clearly, 999 − 5 = is the maximum possible three-digit number that is divisible by 7.

Hence the final sequence is as follows:

105, 112, 119, ...,

Let 994 be the nth term of this A.P.

a =

d =

an =

n = ?

We know that the nth term of an A.P. is,an = a + (n - 1)d.

= 105 + (n - 1)7

= (n - 1)7

n - 1 = 8897

n - 1 =

n = 127 +

n =

There are 128 three-digit numbers that are divisible by 7.

14. How many multiples of 4 lie between 10 and 250?

Solution:

Note that:

The multiple of 4 which comes just after 10 is

The multiple of 4 which comes just before 250 is

Therefore, all the multiples of 4 between 10 and 250 are: , , , ,.........248

First Term, a: ; Common Difference, d = ; Last Term, an =

We know that the nth term of an A.P. is,an = a + (n - 1)d

248 = + (n - 1)(4)

248 - = 4 (n - 1)

= 4 (n - 1)

= n - 1

n =

Therefore, there are 60 terms, which means there are 60 multiples of 4 between 10 and 250.

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

Solution:

The nth term of an AP is, an = a + (n - 1)d.

Here, a is the first term, d is the common difference and n is the number of terms.

Let the nth term of the two APs be an and an1

Given that the nth term of the two APs are equal.

In first AP 63, 65, 67, . . ., a = , d = 65 - 63 =

and in second AP 3, 10, 17, . . ., a = , d = 10 - 3 =

Then,

an = an1

+ (n - 1)2 = 3 + (n - 1)......... equation (1)

By Simplifying equation (1)

7(n - 1) - 2(n - 1) = 63 -

7n - - 2n + 2 =

- 5 = 60

n = 655

n =

The 13th term of the two given APs are equal.

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

Solution:

an = a + (n - 1)d is the nth term of an AP, where an is the nth term, a is the first term, d is a common difference and n is the number of terms.

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

a3 = and a7 - a5 =

a + (3 - 1)d =

a + = 16 ... equation(1)

Using a7 - a5 = 12

[a + (7 - 1) d] - [a + (5 - 1) d] =

[a + ] - [a + ] = 12

= 12

d =

By substituting this in equation (1), we obtain

a + 2 × = 16

a + = 16

a =

Therefore, A.P. will be 4, 4 + 6, 4 + 2 × 6, 4 + 3 × 6, ...

Hence, the sequence will be 4, , , , ...

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

Solution:

The nth term of an AP is, an = a + (n - 1)d.

Here, a is the first term, d is the common difference and n is the number of terms.

Given A.P. is 3, 8, 13, ..., 253

As the 20th term is considered from last, so a =

Common difference, d = 3 - 8 = (Considered in reverse order)

We know that the nth term of an A.P. is,an = a + (n - 1)d

So, 20th Term is a20 = a + ( - 1)d

a20 = + (20 - 1)(- 5)

= 253 - ×

= 253 -

20th term from the last term is 158.

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:

The nth term of an AP is, an = a + (n - 1)d.

Here, a is the first term, d is the common difference and n is the number of terms.

Given, a4 + a8 =

(a + ) + (a + ) = 24

⇒ a + d = 24

⇒ a + 5d = ..... Equation(1)

Also, a6 + a10 =

(a + ) + (a + ) = 44

⇒ 2a + d = 44

⇒ a + d = .... Equation(2)

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

(a + 7d ) - (a + 5d) = 22 -

a + 7d - a - 5d =

d = 10

d =

By substituting the value of d = 5 in equation (1), we obtain

a + 5d = 12

a + 5 × = 12

a + = 12

a =

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

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

The first three terms of this A.P. are - 13, , and .

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:

The nth term of an AP is, an = a + (n - 1)d.

Here, a is the first term, d is the common difference and n is the number of terms.

Since, there's an increment of ₹ 200 every year, the incomes received by Subba Rao in the years 1995, 1996, and 1997 are ₹ 5000, ₹ 5200, and ₹ 5400 respectively.

This forms an AP with a = and d =

Let after nth year, his salary be ₹ .

Hence, an =

We know that the nth term of an A.P. is an = a + (n - 1)d

By substituting the values in the above equation, we get

7000 = + (n - 1) 200

(n - 1) = 7000 - 5000

n - 1 = 2000200

n = + 1

n =

Therefore, in the 11th year, his income reached ₹ 7000. This means after 10 years of 1995, that is, 1995 + 10 ⇒

In 2005, his income reached ₹ 7000.

20. Ramkali 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:

The nth term of an AP is, an = a + (n - 1)d.

Here, a is the first term, d is the common difference and n is the number of terms.

From the given data, Ramkali’s savings in the consecutive weeks are ₹ 5, ₹ (5 + 1.75), ₹ (5 + × 1.75), ₹ (5 + × 1.75) ... and so on

Hence, in nth weeks the savings will be, ₹ [5 + (n - 1) × 1.75] = ₹

we have a = , d = , an =

We know that the nth term of an A.P. is an = a + (n - 1)d

20.75 = + (n - 1)1.75

= (n - 1)1.75

(n - 1) = 15.751.75

n - 1 = 1575175

n - 1 =

n =

So, in the 10th week Ramkali's saving will be ₹ 20.75.

Arithmetic Progressions

Glossary

Exercise 5.2

(i)

(ii)

(i)

(ii)

(iii)

(iv)

(v)

(i)

(ii)

Sign in to Innings2

Arithmetic Progressions

Reset Progress

Glossary

Exercise 5.2

(i)

(ii)

(i)

(ii)

(iii)

(iv)

(v)

(i)

(ii)