# RD Sharma Solutions for Class 10 Maths Chapter 9 Arithmetic Progression Exercise 9.2

**1. For the following arithmetic progressions write the first term a and the common difference d:**

(i) - 5, -1, 3,7, ........

(i) - 5, -1, 3,7, ........

**(ii) 1/5, 3/5, 5/5, 7/5, .......**

**(iii) 0.3, 0.55, 0.80, 1.05, ......**

**(iv) -1.1, -3.1, -5.1, -7.1, .....**

**Solution**

a is the first term and d is the common difference.

Arithmetic progression is a, a +d, a + 2d + a + 3d, .....

(i) -5, -1, 3,7, .....

Given arithmetic series is

-5, -1, 3,7, .....

This is in the form of a, a + d, a + 2d + a + 3d, .... by comparing these two

a = -5, a + d = 1, a + 2d = 3, a + 3d= 7, ....

First term (a) = -5

By subtracting second and first term, we get

(a + d) - (a) = d

⇒ -1-(-5) = d

⇒ 4 = d

Common difference (d) = 4

(iii) 0.3, 0.55, 0.80, 1.05, .......

Given arithmetic series,

0.3, 0.55, 0.80,1.05, .....

General arithmetic series

a, a +d, a + 2d, a + 3d,....

By comparing,

a = 0.3, a + d = 0.5, a + 2d = 0.80, a + 3d = 1.05

First term(a) = 0.3

By subtracting first term from second term. We get

d = (a + d) - (a)

⇒ d = 0.55 - 0.3

⇒ d = 0.25

Common difference (d) = 0.25

(iv) -1.1, -3.1, -5.1, -7.1,....

General series is

a, a + d, a + 2d, a+ 3d, .....

By comparing this two, we get

a = -1.1, a + d= -3.1, a + 2d= -5.1, a + 3d = -71

First term (a) = -1.1

Common difference (d) = (a + d) - (a)

= -3.1 - (-1.1)

Common difference (d) = -2

**2. Write the arithmetic progressions write first term a and common difference d are as follows: (i) a = 4, d = -3(ii) a = -1, d = 1/2(iii) a = -1.5, d = -0.5**

**Solution**

If first term (a) = a and common difference = d, then the arithmetic series is, a, a + d, a + 2d, a + 3d, .....

(i) a = 4, d = -3

Given first term (a) = 4

Common difference(d) = -3

Then arithmetic progression is,

a, a+ d, a + 2d, a+ 3d, .....

⇒ 4, 4 - 3, a + 2(-3), 4 + 3(-3).....

⇒ 4, 1, -2; -5, -8,....

(ii) Given , First term(a) = -1

Common difference(d) = 1/2

Then arithmetic progression is,

⇒ a, a+d, a+ 2d, a+ 3d, .....

(iii) Given, First term(a) = -1.5

Common difference(d) = -0.5

Then arithmetic progression is

⇒ a, a + d, a +2d, a + 3d,....

⇒ -1.5, -1.5 - 0.5, -1.5+2(-0.5), -1.5 + 3(-0.5)

⇒ -1.5, -2, -2.5, -3,....

Then required progression is

-1.5, -2, -2.5, -3,....

**3. In which of the following situations, the sequence of numbers formed will form an A.P.?(i) The cost of digging a well for the first metre is Rs 150 and rises by Rs 20 for each succeeding metre. (ii) The amount of air present in the cylinder when a vacuum pump removes each time 1/4 of their remaining in the cylinder. **

**Solution**

(i) Given,

Cost of digging a well for the first meter(c_{1}) = Rs. 150.

Cost rises by Rs. 20 for each succeeding meter

Then,

Cost of digging for the second meter (c_{2})= Rs. 150 + Rs 20 = Rs 170

Cost of digging for the third meter (c_{3}) = Rs. 170 + Rs 20 = Rs 190

Thus, costs of digging a well for different lengths are 150, 170, 190, 210,......

Clearly, this series in in A.P.

With first term (a) = 150, common difference(d) = 20

(ii) Given,

**4. Show that the sequence defined by a _{n} = 5n - 7 is an A.P., find its common difference. **

**Solution**

_{n}= 5n - 7

n

^{th}term of given sequence (a

_{n}) = 5n - 7

(n+1)

^{th}term of given sequence (a

_{n}+ 1) - a

_{n}= (5n - 2) - (5n - 7)

= 5

∴ d = 5

**OR**

a_{n} = 5n - 7

Let n= 1, 2, 3, 4...., we get

a_{1} = 5 × 1 - 7 = 5 - 7 = -2

a_{2} = 5 × 2 - 7 = 10 - 7 = 3

a_{3} = 5 × 3 - 7 = 15 - 7 = 8

a_{4} = 5 × 4 - 7 = 20 - 7 = 13,.....

-2, 3, 8, 13,.... is an A.P. whose common difference is 5 as

a_{2} - a_{1} = 3-(-2) = 3 + 2 = 5

a_{3} - a_{2} = 8-3 = 5

a_{4} - a_{3} = 13 - 8 = 5

**5. Show that the sequence defined by a**

_{n}= 3n^{2}- 5 is not an A.P.**Solution**

Given sequence is:

_{n}= 3n

^{2}- 5

Let n = 1, 2, 3, 4,....we get

a

_{1}= 3(1)

^{2}- 5 = 3×1 - 5 = 3-5 = -2

a

_{2}= 3(2)

^{2}- 5 = 3×4 - 5 = 12-5 = 7

a

_{3}= 3(3)

^{2}- 5 = 3×9 - 5 = 27-5 = 22

a

_{4}= 3(4)

^{2}- 5 = 3×16 - 5 = 48-5 = 43

The sequence will be -2, 7, 22, 43,....

We see that it has no common difference.

It is not an A.P.

**6. The general term of a sequence is given by a**

_{n}= -4n + 15. Is the sequence an A.P. ? If so, find its 15th term and the common difference.**Solution**

Let n = 1, 2, 3, 4, 5,.... then

a

_{1}= -4×1 + 15 = -4+15 = 11

a

_{2}= -4×2 + 15 = -8+15 = 7

a

_{3}= -4×3 + 15 = -12+15 = 3

a

_{4}= -4×4 + 15 = -16+15 = -1

a

_{5}= -4×5 + 15 = -20+15 = -5

We see that first term is 11 and common difference is -4

a

_{2}- a

_{1}= 7-11 = -4

a

_{3}- a

_{2}= 3-7 = -4

a

_{4}- a

_{3}= -1-3 = -4

a

_{5}- a

_{4}= -5-(-1) = -5+1 = -4

Yes, it is an A.P.

Now, 15th term = n

_{15}= -4×15 + 15 = -60+15 = -45

**7. Write the sequence with nth term :**

_{n}= 3 + 4n

_{n}= 5 + 2n

_{n}= 6 – n

_{n}= 9 – 5n

**Show that all of the above sequences form A.P.**

**Solution**

(i) nth term

a_{n} = 3 + 4n

Let n = 1, 2, 3, 4,.... we get

a_{1} = 3 + 4×1 = 3+4 = 7

a_{2} = 3 + 4×2 = 3+8 = 11

a_{3} = 3 + 4×3 = 3+12 = 15

a_{4} = 3 + 4×4 = 3+16 = 19

We see that the sequence is 7, 4, 15, 19,....

Here, common difference a_{2} - a_{1} = 11 - 7 = 4 is same for the sequences.

Therefore, this sequence is an A.P

(ii) nth term, an = 5 + 2n

Let n = 1, 2, 3, 4,.... we get

a_{1} = 5 + 2×1 = 5+2 = 7

a_{2} = 5 + 2×2 = 5+4 = 9

a_{3} = 5 + 2×3 = 5+6 = 11

a_{4} = 5 + 2×4 = 5+8 = 13

We see that the sequence is 7, 9, 11, 13,....

Here, common difference a_{2} - a_{1} = 9 - 7 = 2 is same for the sequences.

Therefore, this sequence is an A.P

(iii) nth term, an = 6 - n

Let n = 1, 2, 3, 4, 5,....we get,

a_{1} = 6 - 1 = 5

a_{2} = 6 - 2 = 4

a_{3} = 6 - 3 = 3

a_{4} = 6 - 4 = 2

a_{5} = 6 - 5 = 1

We see that the sequence is 5, 4, 3, 2, 1,....

Here, common difference a_{2} - a_{1} = 4 - 5 = -1 is same for the sequences.

Therefore, this sequence is an A.P.

(iv) nth term, an = 9 - 5n

Let n = 1, 2, 3, 4, 5,.... then

a_{1} = 9 - 5×1 = 9-5 = 4

a_{2} = 9 - 5×2 = 9-10 = -1

a_{3} = 9 - 5×3 = 9-15 = -6

a_{4} = 9 - 5×4 = 9-20 = -11

a_{5} = 9 - 5×5 = 9-25 = -16

We see that the sequence is 4, -1, -6, -11, -16,....

Here, common difference a_{2} - a_{1} = -1 - 4 = -5 is same for the sequences.

Therefore, this sequence is an A.P.

**8. Find the common difference and write the next four terms of each of the following arithmetic progressions: **

**(i) 1, -2, -5, -8, .......(ii) 0, -3, -6, -9, .......(iii) -1, 1/4, 3/2,.......(iv) -1, -5/6, -2/3,.......**

**Solution**

**8. Prove that no matter what the real numbers a and b are, the sequence with nth term a + nb is always an A.P. What is the common difference ? **

**Solution**

Given sequence (a_{n}) = an + 6n

n^{th} term (a_{n}) = a + nb

(n +1)^{th} term (a_{n+1}) = a + (n+1)b.

Common difference (d) = a_{n+1} - a_{n}d = (a +(n +1)b)-(a + nb)

= a + nb + b - a - nb

= b

∴ common difference (d) does not depend on nth value so, given sequence I sin AP with (d) = b

**10. Find out which of the following sequences are arithmetic progressions. For those which are arithmetic progressions, find out the common differences.**

**(i) 3, 6, 12, 24; ......**

**(ii) 0, -4, -8, -12, ......**

**(iii) 1/2, 1/4, 1/6, 1/8,......**

**(iv) 12, 2, -8, -18, .......**

**(v) 3, 3, 3, 3, ......**

**(vi) P, P + 90, P + 80, P + 270, .......where P = (999)**

**(vii) 1.0, 1.7, 2.4, 3.1, ......**

**(viii) -225, -425, -625, -825,.....**

**(ix) 10, 10 + 2**

^{5}, 10 + 2^{6}, 10 + 2^{7}, .....**Solution**

**(i) 3, 6, 12, 24; ......**

General arithmetic progression is a, a + d, a +2d, a + 3d, .......

Common difference (d) = Second term - first term

= (d + d) - a = d (or )

= Third term - second term

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

To check given sequence is in A.P. or not we use this condition.

Second term - First term = Third term - Second term

a_{1} = 3, a_{2} = 6, a_{3} = 12, a_{4} = 24

Second term - First term = 6 - 3= 3

Third term - Second term = 12 - 6 = 6

This two are not equal so given sequence is not in A.P.

**(ii) 0, -4, -8, -12, ......**

In the given sequence

a_{1} = 0, a_{2} = - 4, a_{3} = -8, a_{4} = -12 Check the condition

Second term - first term = third term - second term

a_{2} - a_{1} = a_{3} - a_{2}

⇒ -4 - 0 = -8 - (-4)

⇒ -4 = + 8 + 4

⇒ -4 = -4

Condition is satisfied ∴ given sequence is in A.P. with common difference

d = a_{2} - a_{1} = - 4

**(iii) 1/2, 1/4, 1/6, 1/8,......**

In the given sequence

Condition is not satisfied

∴ Given sequence not in A.P.

**(iv) 12, 2, -8, -18, .........**

In the given sequence

a_{1} = 12, a_{2} = 2, a_{3} = -8, a_{4} = -18

Check the condition

a_{2} - a_{1} = a_{3} - a_{2}

⇒ 2 - 12 = -8 - 2

⇒ -10 = -10

∴ Given sequence is in A.P. with common difference d = 10 **(v) 3, 3, 3, 3, ........**

In the given sequence

a_{1} = 3, a_{2} = 3, a_{3} = 3, a_{4} = 3

Check the condition

a_{2} - a_{1} = a_{3} - a_{2}

⇒ 3 - 3 = 3 - 3

⇒ 0 = 0

∴ Given sequence is in A.P. with common difference d = 0

**(vi) P, P + 90, P + 80, P + 270, .......where P = (999)**

In the given sequence

_{1}= P, a

_{2}= p + 90, a

_{3}= p+180, a

_{4}= P+ 270

Check the condition

a

_{2}- a

_{1}= a

_{3}- a

_{2}

⇒ P + 90 - P = P + 180 - P - 90

⇒ 90 = 180 - 90

⇒ 90 = 90

∴ Given sequence is in A.P. with common difference d = 90

**(vii) 1.0, 1.7, 2.4, 3.1, ......**

a

_{1}= 1.0, a

_{2}= 1.7, a

_{3}= 2.4, a

_{4}= 3.1

_{2}- a

_{1}= a

_{8}- a

_{2}

⇒ 1.7 - 1.0 = 2.4 - 1.7

⇒ 0.7 = 0.7

∴ The given sequence is in A.P with d = 0.7

**(viii) -225, -425, -625, -825,.....**

In the given sequence

a

_{1}= -225, a

_{2}= -425, a

_{3}= -625, a

_{4}= -825

Check the condition

a

_{2}- a

_{1}= a

_{3}- a

_{2}

⇒ -425 + 225 = -625 +425

⇒ -200 = -200

∴ The given sequence is in A.P. with d = -200

**(ix) 10, 10 + 2**

^{5}, 10 + 2^{6}, 10 + 2^{7}, .......In the given sequence

a

_{1}= 10, a

_{2}= 10 + 2

^{5}, a

_{3}= 10 + 2

^{6}, a

_{4}= 10 + 2

^{7}

a

_{2}- a

_{1}= a

_{3}- a

_{2}

⇒ 10 + 2

^{5}- 10 = 10 + 2

^{6}- 10 + 2

^{5}

⇒ 2

^{5}≠ 2

^{6}- 2

^{5}

**11. Justify whether it is true to say that the sequence, having following nth term is an A.P.**(i) a

_{n}= 2n – 1

(ii) a

_{n}= 3n² + 5

(iii) a

_{n}= 1 + n + n²

**Solution**