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 between consecutive terms:
$a_2 - a_1 = 23 - 15 = 8$
$a_3 - a_2 = 31 - 23 = 8$
$a_4 - a_3 = 39 - 31 = 8$
∴ The common difference is constant, so 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 the initial amount of air = 1 unit.
Air after 1st removal = $1 - \dfrac{1}{4} = \dfrac{3}{4}$
Air after 2nd removal = $\dfrac{3}{4} - \dfrac{1}{4} \times \dfrac{3}{4} = \dfrac{9}{16}$
Air after 3rd removal = $\dfrac{9}{16} - \dfrac{1}{4} \times \dfrac{9}{16} = \dfrac{27}{64}$
The sequence is: $1, \dfrac{3}{4}, \dfrac{9}{16}, \dfrac{27}{64}, \ldots$
Differences between consecutive terms:
$a_2 - a_1 = \dfrac{3}{4} - 1 = -\dfrac{1}{4}$
$a_3 - a_2 = \dfrac{9}{16} - \dfrac{3}{4} = -\dfrac{3}{16}$
∴ $a_2 - a_1 \neq a_3 - a_2$, so no, this is not an AP.
(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 is: $150, 200, 250, 300, \ldots$
Differences between consecutive terms:
$a_2 - a_1 = 200 - 150 = 50$
$a_3 - a_2 = 250 - 200 = 50$
$a_4 - a_3 = 300 - 250 = 50$
∴ The common difference is constant, so 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 is: $10800, 11664, 12597.12, \ldots$
Differences between consecutive terms:
$a_2 - a_1 = 11664 - 10800 = 864$
$a_3 - a_2 = 12597.12 - 11664 = 933.12$
∴ $a_2 - a_1 \neq a_3 - a_2$, so no, this is not an AP.
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) $a = -2$, $d = 0$
$a_1 = -2$
$a_2 = -2 + 0 = -2$
$a_3 = -2 + 0 = -2$
$a_4 = -2 + 0 = -2$
∴ 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 $d = \dfrac{5}{3} - \dfrac{1}{3} = \mathbf{\dfrac{4}{3}}$
(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 between consecutive terms:
$a_2 - a_1 = 4 - 2 = 2$
$a_3 - a_2 = 8 - 4 = 4$
$a_4 - a_3 = 16 - 8 = 8$
∴ $a_2 - a_1 \neq a_3 - a_2$, so it is not an AP.
(ii) $2, \dfrac{5}{2}, 3, \dfrac{7}{2}, \ldots$
Differences between consecutive terms:
$a_2 - a_1 = \dfrac{5}{2} - 2 = \dfrac{1}{2}$
$a_3 - a_2 = 3 - \dfrac{5}{2} = \dfrac{1}{2}$
$a_4 - a_3 = \dfrac{7}{2} - 3 = \dfrac{1}{2}$
∴ It is an AP with $d = \dfrac{1}{2}$.
Next three terms:
$a_5 = \dfrac{7}{2} + \dfrac{1}{2} = 4$
$a_6 = 4 + \dfrac{1}{2} = \dfrac{9}{2}$
$a_7 = \dfrac{9}{2} + \dfrac{1}{2} = 5$
∴ The next three terms are $\mathbf{4, \dfrac{9}{2}, 5}$.
(iii) $-1.2, -3.2, -5.2, -7.2, \ldots$
Differences between consecutive terms:
$a_2 - a_1 = -3.2 - (-1.2) = -2$
$a_3 - a_2 = -5.2 - (-3.2) = -2$
$a_4 - a_3 = -7.2 - (-5.2) = -2$
∴ It is an AP with $d = -2$.
Next three terms:
$a_5 = -7.2 + (-2) = -9.2$
$a_6 = -9.2 + (-2) = -11.2$
$a_7 = -11.2 + (-2) = -13.2$
∴ The next three terms are $\mathbf{-9.2, -11.2, -13.2}$.
(iv) $-10, -6, -2, 2, \ldots$
Differences between consecutive terms:
$a_2 - a_1 = -6 - (-10) = 4$
$a_3 - a_2 = -2 - (-6) = 4$
$a_4 - a_3 = 2 - (-2) = 4$
∴ It is an AP with $d = 4$.
Next three terms:
$a_5 = 2 + 4 = 6$
$a_6 = 6 + 4 = 10$
$a_7 = 10 + 4 = 14$
∴ The next three terms are $\mathbf{6, 10, 14}$.
(v) $3, 3+\sqrt{2}, 3+2\sqrt{2}, 3+3\sqrt{2}, \ldots$
Differences between consecutive terms:
$a_2 - a_1 = (3 + \sqrt{2}) - 3 = \sqrt{2}$
$a_3 - a_2 = (3 + 2\sqrt{2}) - (3 + \sqrt{2}) = \sqrt{2}$
$a_4 - a_3 = (3 + 3\sqrt{2}) - (3 + 2\sqrt{2}) = \sqrt{2}$
∴ It is an AP with $d = \sqrt{2}$.
Next three terms:
$a_5 = (3 + 3\sqrt{2}) + \sqrt{2} = 3 + 4\sqrt{2}$
$a_6 = (3 + 4\sqrt{2}) + \sqrt{2} = 3 + 5\sqrt{2}$
$a_7 = (3 + 5\sqrt{2}) + \sqrt{2} = 3 + 6\sqrt{2}$
∴ The next three terms are $\mathbf{3 + 4\sqrt{2}, 3 + 5\sqrt{2}, 3 + 6\sqrt{2}}$.
(vi) $0.2, 0.22, 0.222, 0.2222, \ldots$
Differences between consecutive terms:
$a_2 - a_1 = 0.22 - 0.2 = 0.02$
$a_3 - a_2 = 0.222 - 0.22 = 0.002$
∴ $a_2 - a_1 \neq a_3 - a_2$, so it is not an AP.
(vii) $0, -4, -8, -12, \ldots$
Differences between consecutive terms:
$a_2 - a_1 = -4 - 0 = -4$
$a_3 - a_2 = -8 - (-4) = -4$
$a_4 - a_3 = -12 - (-8) = -4$
∴ It is an AP with $d = -4$.
Next three terms:
$a_5 = -12 + (-4) = -16$
$a_6 = -16 + (-4) = -20$
$a_7 = -20 + (-4) = -24$
∴ The next three terms are $\mathbf{-16, -20, -24}$.
(viii) $-\dfrac{1}{2}, -\dfrac{1}{2}, -\dfrac{1}{2}, -\dfrac{1}{2}, \ldots$
Differences between consecutive terms:
$a_2 - a_1 = -\dfrac{1}{2} - \left(-\dfrac{1}{2}\right) = 0$
$a_3 - a_2 = -\dfrac{1}{2} - \left(-\dfrac{1}{2}\right) = 0$
$a_4 - a_3 = -\dfrac{1}{2} - \left(-\dfrac{1}{2}\right) = 0$
∴ It is an AP with $d = 0$.
∴ The next three terms are $\mathbf{-\dfrac{1}{2}, -\dfrac{1}{2}, -\dfrac{1}{2}}$.
(ix) $1, 3, 9, 27, \ldots$
Differences between consecutive terms:
$a_2 - a_1 = 3 - 1 = 2$
$a_3 - a_2 = 9 - 3 = 6$
$a_4 - a_3 = 27 - 9 = 18$
∴ $a_2 - a_1 \neq a_3 - a_2$, so it is not an AP.
(x) $a, 2a, 3a, 4a, \ldots$
Differences between consecutive terms:
$a_2 - a_1 = 2a - a = a$
$a_3 - a_2 = 3a - 2a = a$
$a_4 - a_3 = 4a - 3a = a$
∴ It is an AP with $d = a$.
Next three terms:
$a_5 = 4a + a = 5a$
$a_6 = 5a + a = 6a$
$a_7 = 6a + a = 7a$
∴ The next three terms are $\mathbf{5a, 6a, 7a}$.
(xi) $a, a^2, a^3, a^4, \ldots$
Differences between consecutive terms:
$a_2 - a_1 = a^2 - a = a(a - 1)$
$a_3 - a_2 = a^3 - a^2 = a^2(a - 1)$
(The two are equal only when $a = 0$ or $a = 1$.)
∴ $a_2 - a_1 \neq a_3 - a_2$, so in general it is not an AP.
(xii) $\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \ldots$
Simplifying the terms:
$\sqrt{8} = 2\sqrt{2}$
$\sqrt{18} = 3\sqrt{2}$
$\sqrt{32} = 4\sqrt{2}$
So the list is $\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}, \ldots$
Differences between consecutive terms:
$a_2 - a_1 = 2\sqrt{2} - \sqrt{2} = \sqrt{2}$
$a_3 - a_2 = 3\sqrt{2} - 2\sqrt{2} = \sqrt{2}$
$a_4 - a_3 = 4\sqrt{2} - 3\sqrt{2} = \sqrt{2}$
∴ It is an AP with $d = \sqrt{2}$.
Next three terms:
$a_5 = 4\sqrt{2} + \sqrt{2} = 5\sqrt{2} = \sqrt{50}$
$a_6 = 5\sqrt{2} + \sqrt{2} = 6\sqrt{2} = \sqrt{72}$
$a_7 = 6\sqrt{2} + \sqrt{2} = 7\sqrt{2} = \sqrt{98}$
∴ The next three terms are $\mathbf{\sqrt{50}, \sqrt{72}, \sqrt{98}}$.
(xiii) $\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \ldots$
Differences between consecutive terms:
$a_2 - a_1 = \sqrt{6} - \sqrt{3} = \sqrt{3}(\sqrt{2} - 1)$
$a_3 - a_2 = \sqrt{9} - \sqrt{6} = \sqrt{3}(\sqrt{3} - \sqrt{2})$
∴ $a_2 - a_1 \neq a_3 - a_2$, so it is not an AP.
(xiv) $1^2, 3^2, 5^2, 7^2, \ldots$
The terms are $1, 9, 25, 49, \ldots$
Differences between consecutive terms:
$a_2 - a_1 = 9 - 1 = 8$
$a_3 - a_2 = 25 - 9 = 16$
$a_4 - a_3 = 49 - 25 = 24$
∴ $a_2 - a_1 \neq a_3 - a_2$, so it is not an AP.
(xv) $1^2, 5^2, 7^2, 73, \ldots$
The terms are $1, 25, 49, 73, \ldots$
Differences between consecutive terms:
$a_2 - a_1 = 25 - 1 = 24$
$a_3 - a_2 = 49 - 25 = 24$
$a_4 - a_3 = 73 - 49 = 24$
∴ It is an AP with $d = 24$.
Next three terms:
$a_5 = 73 + 24 = 97$
$a_6 = 97 + 24 = 121$
$a_7 = 121 + 24 = 145$
∴ The next three terms are $\mathbf{97, 121, 145}$.