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Class 10 RD Sharma Solutions – Chapter 9 Arithmetic Progressions – Exercise 9.3
  • Difficulty Level : Medium
  • Last Updated : 17 Dec, 2020

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

(i) -5, -1, 3, 7, …………

Solution: 

Given sequence is -5, -1, 3, 7, …………

∴ First term is -5

And, common difference = a2 – a1 = -1 – (-5) = 4

∴ Common difference is 4



(ii) 1/5, 3/5, 5/5, 7/5, ……

Solution: 

Given sequence is 1/5, 3/5, 5/5, 7/5, ……

∴ First term is 1/5

And, common difference = a2 – a1 = 3/5 – (1/5) = 2/5

∴ Common difference is 2/5

(iii) 0.3, 0.55, 0.80, 1.05, …………

Solution:

Given sequence is 0.3, 0.55, 0.80, 1.05, …………

∴ First term is 0.3



And, common difference = a2 – a1 = 0.55 – (0.3) = 0.25

∴ Common difference is 0.25

(iv) -1.1, -3.1, -5.1, -7.1, …………..

Solution :

Given sequence is -1.1, -3.1, -5.1, -7.1, …………..

∴ First term is -1.1

And, common difference = a2 – a1 = -3.1 – (-1.1) = -2.0

∴ Common difference is -2.0

Problem 2: Write the arithmetic progression when first term a and common difference d are as follows:

(i) a = 4, d = -3

Solution:

Given:

a1 = 4 and d= -3



Therefore, arithmetic progression is: a, a + d, a + 2d, a + 3d, ……

⇒ 4, 4 – 3, 4 + 2(-3), 4 + 3(-3), ……

⇒ 4, 1, – 2, – 5, – 8 ……..

∴ A.P will be 4, 1, – 2, – 5, – 8 ……..

(ii) a = -1, d = 1/2

Solution:

Given:

a1 = -1, d= 1/2

Therefore, arithmetic progression is: a, a + d, a + 2d, a + 3d, ……

⇒ -1, -1 + 1/2, -1, 2½, -1 + 3½, …

⇒ -1, -1/2, 0, 1/2

∴ A.P will be -1, -1/2, 0, 1/2

(iii) a = -1.5, d = -0.5

Solution:

Given:

a1 = -1.5, d = -0.5

Therefore, 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, …….

∴ A.P will be – 1.5, – 2, – 2.5, – 3, …….

Problem 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 150 and rises by 20 for each succeeding metre.

Solution:

Given:

Cost of digging a well for the first metre = ₹150

Therefore,

Cost for the second metre = ₹150 + ₹20 = ₹170

Cost for the third metre = ₹170 + ₹20 = ₹190

Cost for the fourth metre = ₹190 + ₹20 = ₹210

Since, 20 is added fo each succeeding metre

Hence, the sequence will be (In rupees) and is A.P 

150, 170, 190, 210, ………..

∴ The given sequence is in A.P and common difference is 20

(ii) The amount of air present in the cylinder when a vacuum pump removes each time 1/4 of the remaining in the cylinder.

Solution:

Air removed for the first time = (1 x 1/4) = 1/4

Remaining air = 1 – 1/4 = 3/4

Air removed for the second time = (3/4 x 1/4) = 3/16

Remaining air = 3/4 – 3/16 = 9/16

Air removed for the third time = (9/16 x 1/4) = 9/64

Remaining air = 9/16 – 9/64 = 27/64

∴The sequence will be 1, 3/4, 9/16, 27/64

Here, a2 – a1 = 3/4 – (1) = -1/4

a3 – a2 = 9/16 – (3/4) = -3/16

Since, the successive difference of list is not same

∴ The given sequence is not in A.P

(iii) Divya deposited Rs 1000 at compound interest at the rate of 10% per annum. The amount at the end of first year, second year, third year, …, and so on.

Solution: 

Given:

Divya deposited Rs 1000 at compound interest of 10% p.a

So, the amount at the end of first year is = 1000 + 0.1(1000) = Rs 1100

And, the amount at the end of second year is = 1100 + 0.1(1100) = Rs 1210

And, the amount at the end of third year is = 1210 + 0.1(1210) = Rs 1331

Here, a2 – a1 = 1210 – 1100 = 110

a3 – a2 = 1331 – 1210 = 121

Since, the successive difference of list is not same

∴ The given sequence is not in A.P

Problem 4: Find the common difference and write the next four terms of each of the following arithmetic progressions:

(i) 1, -2, -5, -8, ……..

Solution:

Given:

First term = 1, Second term = -2

Common difference = -2 – (1) = -3

Now,

Fifth term = -8 + (-3) = -11

Sixth term = -11 + (-3) = -14

Seventh term = -14 + (-3) = -17

Eighth term = -17 + (-3) = -20

(ii) 0, -3, -6, -9, ……

Solution:

Given:

First term = 0, Second term = -3

Common difference = -3 – (0) = -3

Now,

Fifth term = -9 + (-3) = -12

Sixth term = -12 + (-3) = -15

Seventh term = -15 + (-3) = -18

Eighth term = -18 + (-3) = -21

(iii) -1, 1/4, 3/2, ……..

Solution:

Given:

First term = -1, Second term = 1/4

Common difference = 1/4 – (-1) = 5/4

Now,

Fifth term = 3/2 + (5/4) = 11/4

Sixth term = 11/4 + (5/4) = 4

Seventh term = 4 + (5/4) = 21/4

Eighth term = 21/4 + (5/4) = 26/4

(iv) -1, – 5/6, – 2/3, ………..

Solution:

Given:

First term = -1, Second term = -5/6

Common difference = -5/6 – (-1) = 1/6

Now,

Fifth term = -2/3 + (1/6) = -1/2

Sixth term = -1/2 + (1/6) = -1/3

Seventh term = -1/3 + (1/6) = -1/6

Eighth term = -1/6 + (1/6) = 0

Problem 5: 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: 

an = a + nb

Let n= 1, 2, 3, 4, 5, ……….

a1 = a + b

a2 = a + 2b

a3 = a + 3b

a4 = a + 4b

a5 = a + 5b

Now,

a2 – a1 = (a + 2b) – (a + b) = b

a3 – a2 = (a + 3b ) – (a + 2b) = b

a4 – a1 = (a + 4b) – (a + 3b) = b

a5 – a4 = (a + 5b) – (a + 4b) = b

Hence, proved

Problem 6: Find out which of the following sequences are arithmetic progressions. For those which are arithmetic progressions, find out the common difference

(i) 3, 6, 12, 24, ………………………………

Solution:

Now,

a2  – a1 = 6 – (3) = 3

a3 – a1 = 12 – (6) = 6

a4 – a1 = 24 – (12) = 12

Since, the successive difference of list is not same

∴ The given sequence is  not in A.P 

(ii) 0, -4, -8, -12, ……………………………………………………

Solution:

Now,

a2  – a1 = -4 – (0) = -4

a3 – a2 = -8 – (-4) = -4

a4 – a3 = -12 – (-8) = -4

Since, the successive difference of list is same

∴ The given sequence is in A.P and common difference is -4

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

Solution:

Now,

a2  – a1 = 1/4 – (1/2) = -1/4

a3 – a2 = 1/6 – (1/4) = -1/12

a4 – a3 = 1/8 – (1/6) = -1/24

Since, the successive difference of list is not same

∴ The given sequence is not in A.P

 (iv) 12, 2, -8, -18, ……………………………..

Solution:

Now,

a2  – a1 = 2 – (12) = -10

a3 – a2 = -8 – (2) = -10

a4 – a3 = -18 – (-8) = -10

Since, the successive difference of list is same

∴ The given sequence is in A.P

(v) 3, 3, 3, 3, ………………………………………

Solution:

Now,

a2  – a1 = 3 – (3) = 0

a3 – a2 = 3 – (3) = 0

a4 – a3 = 3 – (3) = 0

Since, the successive difference of list is same

∴ The given sequence is in A.P and common difference is 0

(vi) p, p + 90, p + 180, ………………..

Solution:

Now,

a2  – a1 = p+90 – (p) = 90

a3 – a2 = p+180 – (p+90) = 90

Since, the successive difference of list is same

∴ The given sequence is in A.P and common difference is 90

(vii) 1.0, 1.7, 2.4, ………………………….

Solution:

Now,

a2  – a1 = 1.7 – (1.0) = 0.7

a3 – a2 = 2.4 – (1.7) = 0.7

Since, the successive difference of list is same

∴ The given sequence is in A.P and common difference is 0.7

(viii) -225, -425, -625, …………………………….

Solution:

Now,

a2  – a1 = -425 – (-225) = -200

a3 – a2 = -625 – (-425) = -200

Since, the successive difference of list is same

∴ The given sequence is in A.P and common difference is -200

(ix) 10, 10+25, 10+26, 10+27 …………………………………………..

Solution:

Now,

a2  – a1 = 10+25 – (10) = 32

a3 – a2 = 10+26 – (10+25) = 32

a4 – a3 = 10+27 – (10+26) = 64

Since, the successive difference of list is not same

∴ The given sequence is in A.P

(x) a+b, (a+1)+b, (a+1)+(b+1), (a+2)+(b+1), ……………………………….

Solution:

Now,

a2  – a1 = ((a+1) + b) – (a + b) = 1

a3 – a2 = ((a+1)+(b+1)) – ((a+1)+b) = 1

a4 – a3 = ((a+2)+(b+1)) – ((a+1)+(b+1)) = 1

Since, the successive difference of list is same

∴ The given sequence is in A.P and common difference is 1

(xi) 12, 32, 52, 72, ………………………………………………………

Solution: 

Now,

a2  – a1 = (32 – 12) = 8

a3 – a2 = (52 – 32) = 16

a4 – a3 = (72 – 52) = 24

Since, the successive difference of list is not same

∴ The given sequence is not in A.P

(xii) 12, 52, 72, 73, …………………………………………………………..

Solution:

Now,

a2  – a1 = 52 – 12 = 24

a3 – a2 = 72 – 52 = 24

a4 – a3 = 73 – 72 = 24

Since, the successive difference of list is same

∴ The given sequence is in A.P and common difference is 24

Problem 7: Find the common difference of the A.P. and write the next two terms :

(i) 51, 59, 67, 75, …….

Solution:

Given: 

First term = 51, Second term = 59

Common difference = 59 – 51 = 8

Now,

Fifth term = 75 + 8 = 83

Sixth term = 83 + 8 = 91

(ii) 75, 67, 59, 51, ………

Solution:

Given:

 First term = 75, Second term = 67

Common difference = 67 – 75 = -8

Now,

Fifth term = 51 – 8 = 43

Sixth term = 43 – 8 = 35

(iii) 1.8, 2.0, 2.2, 2.4, …….

Solution:

Given:

 First term = 1.8, Second term = 2.0

Common difference = 2.0 – 1.8 = 0.2

Now,

Fifth term = 2.4 + (0.2) = 2.6

Sixth term = 2.6 + (0.2) = 2.8

(iv) 0, 1/4, 1/2, 3/4, ………..

Solution:

Given: 

First term = 0, Second term = 1/4

Common difference = 1/4 – 0 = 1/4

Now,

Fifth term = 3/4 + (1/4) = 1

Sixth term = 1 + (1/4) = 5/4

(v) 119, 136, 153, 170, ………..

Solution:

Given: 

First term = 119, Second term = 136

Common difference = 136 – 119 = 17

Now,

Fifth term = 170 + (17) = 187

Sixth term = 187 + (17) = 204

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