RD Sharma Class 10 Solutions Chapter 9 Arithmetic Progressions Exercise 9.3 (Updated for 2021-22)

RD Sharma Class 10 Solutions Chapter 9 Exercise 9.3

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 practised 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, …………
RD Sharma Class 10 Chapter 9 Arithmetic Progressions
Arithmetic Progressions Class 10 RD Sharma

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, ……….
RD Sharma Class 10 Chapter 9 Arithmetic Progressions
Arithmetic Progressions Class 10 RD Sharma

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.


(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
RD Sharma Class 10 Solutions Arithmetic Progressions
RD Sharma Class 10 Solutions Arithmetic Progressions Exercise 9.3
(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:
RD Sharma Class 10 Solutions Chapter 9 Arithmetic Progressions
RD Sharma Class 10 Pdf Chapter 9 Arithmetic Progressions
RD Sharma Solutions Class 10 Chapter 9 Arithmetic Progressions
Learncbse.In Class 10 Chapter 9 Arithmetic Progressions
Class 10 RD Sharma Solutions Chapter 9 Arithmetic Progressions

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.
RD Sharma Class 10 Pdf Ebook Chapter 9 Arithmetic Progressions
Solution:
RD Sharma Class 10 Pdf Free Download Full Book Chapter 9 Arithmetic Progressions
RD Sharma Class 10 Solution Chapter 9 Arithmetic Progressions
RD Sharma Maths Class 10 Solutions Pdf Free Download Chapter 9 Arithmetic Progressions
RD Sharma Class 10 Book Pdf Free Download Chapter 9 Arithmetic Progressions
RD Sharma Maths Class 10 Solutions Pdf Free Download Chapter 9 Arithmetic Progressions
RD Sharma Class 10 Book Pdf Free Download Chapter 9 Arithmetic Progressions

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:
Class 10 RD Sharma Chapter 9 Arithmetic Progressions
RD Sharma Maths Class 10 Solutions Chapter 9 Arithmetic Progressions
Class 10 RD Sharma Chapter 9 Arithmetic Progressions
RD Sharma Maths Class 10 Solutions Chapter 9 Arithmetic Progressions

We have provided complete details of RD Sharma Class 10 Solutions Chapter 9 Exercise 9.3. If you have any queries related to CBSE Class 10, feel free to ask us in the comment section below.

FAQs on RD Sharma Class 10 Solutions Chapter 9 Exercise 9.3

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What are the key benefits of learning RD Sharma Solutions for Class 10 Maths Chapter 9 Exercise 9.3?

The RD Sharma Class 10 Solutions Chapter 9 Exercise 9.3 answers were quite helpful. This makes it simple to clear any questions about arithmetic progression. Easily answer all of the questions that the Class 10 students have given for exercise.

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