The knowledge of the basic concept of A.P. is further tested in this exercise. Problems related to finding the common difference and the nth term of an A.P. is thoroughly practised here. The RD Sharma Solutions Class 10 prepared by subject experts at BYJU’S is a very useful resource for students for reference and preparation for exams. Also, download RD Sharma Solutions for Class 10 Maths Chapter 9 Arithmetic Progressions Exercise 9.3 PDF provided below for detailed solutions to problems.
RD Sharma Solutions for Class 10 Maths Chapter 9 Arithmetic Progressions Exercise 9.3
Access RD Sharma Solutions for Class 10 Maths Chapter 9 Arithmetic Progressions Exercise 9.3
1. For the following arithmetic progressions write the first term a and the common difference d:
(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:
We know that if a is the first term and d is the common difference, the arithmetic progression is a, a + d, a + 2d + a + 3d,….
(i) – 5, –1, 3, 7,…
Given arithmetic series is – 5, –1, 3, 7,…
c a, a + d, a + 2d + a + 3d,….
Thus, by comparing these two we get, a = – 5, a + d = 1, a + 2d = 3, a + 3d = 7
First term (a) = – 5
By subtracting the second and first terms, we get
(a + d) – (a) = d
-1 – (- 5) = d
4 = d
⇒ Common difference (d) = 4.
(ii) 1/5, 3/5, 5/5, 7/5, ………….
Given arithmetic series is 1/5, 3/5, 5/5, 7/5, ……………
It is seen that it’s of the form of 1/5, 2/5, 5/5, 7/5, ……….. a, a + d, a + 2d, a + 3d,
Thus, by comparing these two, we get
a = 1/5, a + d = 3/5, a + 2d = 5/5, a + 3d = 7/5
First term (a) = 1/5
By subtracting the first term from the second term, we get
d = (a + d)-(a)
d = 3/5 – 1/5
d = 2/5
⇒ common difference (d) = 2/5
(iii) 0.3, 0.55, 0.80, 1.05, …………
Given arithmetic series 0.3, 0.55, 0.80, 1.05, ……….
It is seen that it’s of the form of a, a + d, a + 2d, a + 3d,
Thus, by comparing, we get,
a = 0.3, a + d = 0.55, a + 2d = 0.80, a + 3d = 1.05
First term (a) = 0.3.
By subtracting the first term from the 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 –1.1, – 3.1, – 5.1, –7.1, ……..
It is seen that it’s of the form of a, a + d, a + 2d, a + 3d, ………..
Thus, by comparing these two, we get
a = –1.1, a + d = –3.1, a + 2d = –5.1, a + 3d = –7.1
First term (a) = –1.1
Common difference (d) = (a + d) – (a)
= -3.1 – ( – 1.1)
⇒ Common difference (d) = – 2
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 = 1/2
(iii) a = –1.5, d = – 0.5
Solution:
We know that, 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, the first term (a) = 4
Common difference (d) = -3
Then arithmetic progression is: a, a + d, a + 2d, a + 3d, ……
⇒ 4, 4 – 3, 4 + 2(-3), 4 + 3(-3), ……
⇒ 4, 1, – 2, – 5, – 8 ……..
(ii) a = -1, d = 1/2
Given, the first term (a) = -1
Common difference (d) = 1/2
Then arithmetic progression is: a, a + d, a + 2d, a + 3d,
⇒ -1, -1 + 1/2, -1 + 2½, -1 + 3½, …
⇒ -1, -1/2, 0, 1/2
(iii) a = –1.5, d = – 0.5
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, …….
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.
(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:
(i) Given,
Cost of digging a well for the first meter (c1) = Rs.150.
And the cost rises by Rs.20 for each succeeding meter
Then,
Cost of digging for the second meter (c2) = Rs.150 + Rs 20 = Rs 170
Cost of digging for the third meter (c3) = Rs.170 + Rs 20 = Rs 210
Hence, its clearly seen that the costs of digging a well for different lengths are 150, 170, 190, 210, ….
Evidently, this series is in A∙P.
With first term (a) = 150, common difference (d) = 20
(ii) Given,
Let the initial volume of air in a cylinder be V liters each time 3th/4 of air in a remaining i.e
1 -1/4
First time, the air in cylinder is V.
Second time, the air in cylinder is 3/4 V.
Third time, the air in cylinder is (3/4)2 V.
Thus, series is V, 3/4 V, (3/4)2 V,(3/4)3 V, ….
Hence, the above series is not a A.P.
(iii) 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
Cleary, these amounts 1100, 1210 and 1331 are not in an A.P. since the difference between them is not the same.
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