RD Sharma Class 10 Solutions Chapter 9 Arithmetic Progressions Exercise 9.3 (Updated for 2024)

RD Sharma Class 10 Solutions Chapter 9 Exercise 9.3: This exercise puts the basic concept of A.P. to the test. The nth term of the A.P. and problems of identifying the common difference are thoroughly practiced here. Students will find the RD Sharma Class 10 Solutions prepared by subject experts at Kopykitab to be a very useful resource for reference and exam preparation. For full solutions to questions, check the RD Sharma Solutions for Class 10 Maths Chapter 9 Exercise 9.3 PDF provided below.

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RD Sharma Class 10 Solutions Chapter 9 Exercise 9.3

Access answers to RD Sharma Solutions Class 10 Maths Chapter 9 Exercise 9.3- Important Question with Answers

Question 1.
For the following arithmetic progressions write the first term a and the common difference d :
(i) -5, -1, 3, 7, …………
(ii) 15 , 35 , 55 , 75 , ……
(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= 12
(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 meter is ₹ 150 and rises by ₹ 20 for each succeeding meter.
(ii) The amount of air present in the cylinder when a vacuum pump removes each time 14 of the remaining in the cylinder each.

(iii) Divya deposited ₹ 1000 at compound interest at the rate of 10% per annum. The amount at the end of the first year, second year, third year, …, and so on. [NCERT Exemplar]
Solution:
(i) Cost of digging a well for the first meter = ₹ 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, a₂ – a₁ = 110
a₃ – a₂ = 121
As, a₂ – a₁ ≠ a₃ – a₂, 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, 14 , 32 , ……..
(iv) -1, – 56 , – 23 , ………..
Solution:

Question 5.
Prove that no matter what the real numbers a and b are, the sequence with the 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
We see that it is an A.P. whose common difference is b and a for any real value of a and b
as a₂ – a₁ = a + 2b – a – b = b
a₃ – a₂ = a + 3b – a – 2b = b
a₄ – a₃ = a + 4b – a – 3b = b
and a₅ – a₄ = a + 5b – a – 4b = b

Question 6.
Find out which of the following sequences are arithmetic progressions. For those who 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, 14 , 12 , 34 , ………..
(v) 119, 136, 153, 170, ………..
Solution:

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