## RD Sharma Class 10 Solutions Chapter 9 Arithmetic Progressions Ex 9.3

### RD Sharma Class 10 Solutions Arithmetic Progressions Exercise 9.3

Question 1.

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

(i) -5, -1, 3, 7, …………

(ii) \(\frac { 1 }{ 5 }\) , \(\frac { 3 }{ 5 }\) , \(\frac { 5 }{ 5 }\) , \(\frac { 7 }{ 5 }\) , ……

(iii) 0.3, 0.55, 0.80, 1.05, …………

(iv) -1.1, -3.1, -5.1, -7.1, …………..

Solution:

(i) -5, -1, 3, 7, …………

Question 2.

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

(i) a = 4, d = -3

(ii) a = -1, d= \(\frac { 1 }{ 2 }\)

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

Solution:

(i) First term (a) = 4

and common difference (d) = -3

Second term = a + d = 4 – 3 = 1

Third term = a + 2d = 4 + 2 x (-3) = 4 – 6 = -2

Fourth term = a + 3d = 4 + 3 (-3) = 4 – 9 = -5

Fifth term = a + 4d = 4 + 4 (-3) = 4 – 12 = -8

AP will be 4, 1, -2, -5, -8, ……….

Question 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.

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

(iii) Divya deposited ₹ 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. [NCERT Exemplar]

Solution:

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

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

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

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

The sequence will be (In rupees)

150, 170, 190, 210, ………..

Which is an A.P.

Whose = 150 and d = 20

(ii) Let air present in the cylinder = 1

(iii) Amount at the end of the 1st year = ₹ 1100

Amount at the end of the 2nd year = ₹ 1210

Amount at the end of 3rd year = ₹ 1331 and so on.

So, the amount (in ₹) at the end of 1st year, 2nd year, 3rd year, … are

1100, 1210, 1331, …….

Here, a_{2} – a_{1} = 110

a_{3} – a_{2} = 121

As, a_{2} – a_{1} ≠ a_{3} – a_{2}, it does not form an AP

Question 4.

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, \(\frac { 1 }{ 4 }\) , \(\frac { 3 }{ 2 }\) , ……..

(iv) -1, – \(\frac { 5 }{ 6 }\) , – \(\frac { 2 }{ 3 }\) , ………..

Solution:

Question 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, ……….

a_{1} = a + b

a_{2} = a + 2b

a_{3} = a + 3b

a_{4} = a + 4b

a_{5} = a + 5b

We see that it is an A.P. whose common difference is b and a for any real value of a and b

as a_{2} – a_{1} = a + 2b – a – b = b

a_{3} – a_{2} = a + 3b – a – 2b = b

a_{4} – a_{3} = a + 4b – a – 3b = b

and a_{5} – a_{4} = a + 5b – a – 4b = b

Question 6.

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

Solution:

Question 7.

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

(i) 51, 59, 67, 75, …….

(ii) 75, 67, 59, 51, ………

(iii) 1.8, 2.0, 2.2, 2.4, …….

(iv) 0, \(\frac { 1 }{ 4 }\) , \(\frac { 1 }{ 2 }\) , \(\frac { 3 }{ 4 }\) , ………..

(v) 119, 136, 153, 170, ………..

Solution:

### Exercise 9.3

#### RD Sharma Class 10 Solutions

- Chapter 9 Arithmetic Progressions Ex 9.1
- Chapter 9 Arithmetic Progressions Ex 9.2
- Chapter 9 Arithmetic Progressions Ex 9.3
- Chapter 9 Arithmetic Progressions Ex 9.4
- Chapter 9 Arithmetic Progressions Ex 9.5