NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.1#
Welcome to our NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.1. This introductory exercise helps students identify whether a given sequence is an Arithmetic Progression (AP) and to find the common difference. A solid grasp of these basics is essential for mastering the rest of the chapter.
Arithmetic Progression (AP): A sequence $a_1, a_2, a_3, \ldots$ is an AP if the difference between consecutive terms is constant. This constant is called the Common Difference $d = a_{n+1} - a_n$.
- If $d > 0$, the AP is increasing.
- If $d < 0$, the AP is decreasing.
- If $d = 0$, all terms are equal.
1. In which of the following situations, does the list of numbers involved make an arithmetic progression, and why? #
(i) The taxi fare after each km when the fare is ₹15 for the first km and ₹8 for each additional km.
Fare for 1st km = ₹15 Fare for 2nd km = ₹15 + ₹8 = ₹23 Fare for 3rd km = ₹23 + ₹8 = ₹31 Fare for 4th km = ₹31 + ₹8 = ₹39
The sequence is: $15, 23, 31, 39, \ldots$
Differences: $23-15 = 8$, $31-23 = 8$, $39-31 = 8$.
The difference between consecutive terms is constant ($d = 8$).
∴ Yes, this is an AP with $d = 8$.
(ii) The amount of air present in a cylinder when a vacuum pump removes $\dfrac{1}{4}$ of the air remaining in the cylinder at a time.
Let initial amount of air = 1 unit.
After 1st removal: $1 - \dfrac{1}{4} = \dfrac{3}{4}$
After 2nd removal: $\dfrac{3}{4} - \dfrac{1}{4} \times \dfrac{3}{4} = \dfrac{3}{4} \times \dfrac{3}{4} = \dfrac{9}{16}$
After 3rd removal: $\dfrac{9}{16} \times \dfrac{3}{4} = \dfrac{27}{64}$
The sequence: $1, \dfrac{3}{4}, \dfrac{9}{16}, \dfrac{27}{64}, \ldots$
Differences: $\dfrac{3}{4} - 1 = -\dfrac{1}{4}$ and $\dfrac{9}{16} - \dfrac{3}{4} = -\dfrac{3}{16}$
Since $-\dfrac{1}{4} \neq -\dfrac{3}{16}$, the differences are not constant.
∴ No, this is not an AP (it is a geometric progression).
(iii) The cost of digging a well after every metre of digging, when it costs ₹150 for the first metre and rises by ₹50 for each subsequent metre.
Cost for 1st metre = ₹150 Cost for 2nd metre = ₹150 + ₹50 = ₹200 Cost for 3rd metre = ₹200 + ₹50 = ₹250 Cost for 4th metre = ₹250 + ₹50 = ₹300
The sequence: $150, 200, 250, 300, \ldots$
Difference = $50$ (constant).
∴ Yes, this is an AP with $d = 50$.
(iv) The amount of money in the account every year, when ₹10000 is deposited at compound interest at 8% per annum.
Amount after 1st year = $10000 \times 1.08 = ₹10800$ Amount after 2nd year = $10000 \times (1.08)^2 = ₹11664$ Amount after 3rd year = $10000 \times (1.08)^3 = ₹12597.12$
The sequence: $10800, 11664, 12597.12, \ldots$
Differences: $11664 - 10800 = 864$ and $12597.12 - 11664 = 933.12$
Since $864 \neq 933.12$, the difference is not constant.
∴ No, this is not an AP (it is a geometric progression).
2. Write first four terms of the AP, when the first term $a$ and the common difference $d$ are given as follows: #
(i) $a = 10$, $d = 10$
$a_1 = 10$ $a_2 = 10 + 10 = 20$ $a_3 = 20 + 10 = 30$ $a_4 = 30 + 10 = 40$
The AP is: $\mathbf{10, 20, 30, 40, \ldots}$
(ii) $a = -2$, $d = 0$
$a_1 = -2$ $a_2 = -2 + 0 = -2$ $a_3 = -2$ $a_4 = -2$
The AP is: $\mathbf{-2, -2, -2, -2, \ldots}$
(iii) $a = 4$, $d = -3$
$a_1 = 4$ $a_2 = 4 + (-3) = 1$ $a_3 = 1 + (-3) = -2$ $a_4 = -2 + (-3) = -5$
The AP is: $\mathbf{4, 1, -2, -5, \ldots}$
(iv) $a = -1$, $d = \dfrac{1}{2}$
$a_1 = -1$ $a_2 = -1 + \dfrac{1}{2} = -\dfrac{1}{2}$ $a_3 = -\dfrac{1}{2} + \dfrac{1}{2} = 0$ $a_4 = 0 + \dfrac{1}{2} = \dfrac{1}{2}$
The AP is: $\mathbf{-1, -\dfrac{1}{2}, 0, \dfrac{1}{2}, \ldots}$
(v) $a = -1.25$, $d = -0.25$
$a_1 = -1.25$ $a_2 = -1.25 + (-0.25) = -1.50$ $a_3 = -1.50 + (-0.25) = -1.75$ $a_4 = -1.75 + (-0.25) = -2.00$
The AP is: $\mathbf{-1.25, -1.50, -1.75, -2.00, \ldots}$
3. For the following APs, write the first term and the common difference: #
(i) $3, 1, -1, -3, \ldots$
First term $a = \mathbf{3}$
Common difference $d = 1 - 3 = \mathbf{-2}$
(ii) $-5, -1, 3, 7, \ldots$
First term $a = \mathbf{-5}$
Common difference $d = -1 - (-5) = \mathbf{4}$
(iii) $\dfrac{1}{3}, \dfrac{5}{3}, \dfrac{9}{3}, \dfrac{13}{3}, \ldots$
First term $a = \mathbf{\dfrac{1}{3}}$
Common difference $d = \dfrac{5}{3} - \dfrac{1}{3} = \dfrac{4}{3}$
(iv) $0.6, 1.7, 2.8, 3.9, \ldots$
First term $a = \mathbf{0.6}$
Common difference $d = 1.7 - 0.6 = \mathbf{1.1}$
4. Which of the following are APs? If they form an AP, find the common difference $d$ and write three more terms. #
(i) $2, 4, 8, 16, \ldots$
Differences: $4-2=2$, $8-4=4$, $16-8=8$
Differences are not constant.
∴ Not an AP.
(ii) $2, \dfrac{5}{2}, 3, \dfrac{7}{2}, \ldots$
Differences: $\dfrac{5}{2}-2 = \dfrac{1}{2}$, $3-\dfrac{5}{2} = \dfrac{1}{2}$, $\dfrac{7}{2}-3 = \dfrac{1}{2}$
All differences $= \dfrac{1}{2}$ (constant).
∴ AP with $d = \dfrac{1}{2}$.
Next three terms: $4, \dfrac{9}{2}, 5$
(iii) $-1.2, -3.2, -5.2, -7.2, \ldots$
Differences: $-3.2-(-1.2) = -2$, $-5.2-(-3.2) = -2$
All differences $= -2$ (constant).
∴ AP with $d = -2$.
Next three terms: $-9.2, -11.2, -13.2$
(iv) $-10, -6, -2, 2, \ldots$
Differences: $-6-(-10)=4$, $-2-(-6)=4$, $2-(-2)=4$
∴ AP with $d = 4$.
Next three terms: $6, 10, 14$
(v) $3, 3+\sqrt{2}, 3+2\sqrt{2}, 3+3\sqrt{2}, \ldots$
Differences: $(3+\sqrt{2})-3 = \sqrt{2}$, $(3+2\sqrt{2})-(3+\sqrt{2}) = \sqrt{2}$
∴ AP with $d = \sqrt{2}$.
Next three terms: $3+4\sqrt{2}, 3+5\sqrt{2}, 3+6\sqrt{2}$
(vi) $0.2, 0.22, 0.222, 0.2222, \ldots$
Differences: $0.22-0.20=0.02$, $0.222-0.22=0.002$
Differences are not constant.
∴ Not an AP.
(vii) $0, -4, -8, -12, \ldots$
Differences: $-4-0=-4$, $-8-(-4)=-4$
∴ AP with $d = -4$.
Next three terms: $-16, -20, -24$
(viii) $-\dfrac{1}{2}, -\dfrac{1}{2}, -\dfrac{1}{2}, -\dfrac{1}{2}, \ldots$
All differences $= 0$.
∴ AP with $d = 0$.
Next three terms: $-\dfrac{1}{2}, -\dfrac{1}{2}, -\dfrac{1}{2}$
(ix) $1, 3, 9, 27, \ldots$
Differences: $3-1=2$, $9-3=6$, $27-9=18$
Differences are not constant.
∴ Not an AP (it is a GP with ratio 3).
(x) $a, 2a, 3a, 4a, \ldots$
Differences: $2a-a=a$, $3a-2a=a$, $4a-3a=a$
All differences $= a$ (constant).
∴ AP with $d = a$.
Next three terms: $5a, 6a, 7a$
(xi) $a, a^2, a^3, a^4, \ldots$
Differences: $a^2-a = a(a-1)$ and $a^3-a^2 = a^2(a-1)$
These are equal only if $a = 1$ or $a = 0$. In general, they differ.
∴ Not an AP (in general).
(xii) $\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \ldots$
Simplifying: $\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}, \ldots$
Differences: $2\sqrt{2}-\sqrt{2} = \sqrt{2}$, $3\sqrt{2}-2\sqrt{2} = \sqrt{2}$
∴ AP with $d = \sqrt{2}$.
Next three terms: $5\sqrt{2}, 6\sqrt{2}, 7\sqrt{2}$ i.e., $\sqrt{50}, \sqrt{72}, \sqrt{98}$
(xiii) $\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \ldots$
Differences: $\sqrt{6}-\sqrt{3} = \sqrt{3}(\sqrt{2}-1)$ and $\sqrt{9}-\sqrt{6} = \sqrt{3}({\sqrt{3}-\sqrt{2}})$
Since $\sqrt{2}-1 \neq \sqrt{3}-\sqrt{2}$, the differences are not constant.
∴ Not an AP.
(xiv) $1^2, 3^2, 5^2, 7^2, \ldots$
The terms are $1, 9, 25, 49, \ldots$
Differences: $9-1=8$, $25-9=16$, $49-25=24$
Differences are not constant.
∴ Not an AP.
(xv) $1^2, 5^2, 7^2, 73, \ldots$
The terms are $1, 25, 49, 73$.
Differences: $25-1=24$, $49-25=24$, $73-49=24$
All differences $= 24$ (constant).
∴ AP with $d = 24$.
Next three terms: $97, 121, 145$