Arithmetic Progressions - Exercise 5.1

Fare for $1^{\text{st}}\text{ km} = a_1 = 15$.
Fare for $2^{\text{nd}}\text{ km} = a_2 = 15 + 8 = 23$.
Fare for $3^{\text{rd}}\text{ km} = a_3 = 23 + 8 = 31$.
Fare for $4^{\text{th}}\text{ km} = a_4 = 31 + 8 = 39$.
The sequence is $15, 23, 31, 39, \dots$
Since $a_{k+1} - a_k = 8$ (constant difference throughout):
$\therefore$ Yes, it forms an Arithmetic Progression (AP) with common difference $d = 8$.

Let the initial volume of air present in the cylinder be $V$.
After $1^{\text{st}}$ stroke: Volume left $= V - \dfrac{1}{4}V = \dfrac{3}{4}V$.
After $2^{\text{nd}}$ stroke: Volume left $= \dfrac{3}{4}V - \dfrac{1}{4}\left(\dfrac{3}{4}V\right) = \dfrac{3}{4}V\left(1 - \dfrac{1}{4}\right) = \left(\dfrac{3}{4}\right)^2 V = \dfrac{9}{16}V$.
Differences between consecutive terms:
$a_2 - a_1 = \dfrac{3}{4}V - V = -\dfrac{1}{4}V$
$a_3 - a_2 = \dfrac{9}{16}V - \dfrac{3}{4}V = -\dfrac{3}{16}V$
Since $a_2 - a_1 \neq a_3 - a_2$, the difference is not constant.
$\therefore$ No, this does not form an AP (it forms a Geometric Progression).

Cost for digging $1^{\text{st}}\text{ metre} = a_1 = ₹150$.
Cost for digging $2^{\text{nd}}\text{ metre} = a_2 = 150 + 50 = ₹200$.
Cost for digging $3^{\text{rd}}\text{ metre} = a_3 = 200 + 50 = ₹250$.
The sequence of costs is $150, 200, 250, 300, \dots$
Here $a_2 - a_1 = a_3 - a_2 = 50$, which is constant.
$\therefore$ Yes, it forms an AP with first term $a = 150$ and common difference $d = 50$.

Initial principal $P = ₹10,000$, rate $R = 8\%$ p.a.
Amount after $n$ years is given by $A_n = P\left(1 + \dfrac{R}{100}\right)^n$.
At end of year 1: $A_1 = 10000\left(1 + \dfrac{8}{100}\right) = 10000(1.08) = ₹10,800$.
At end of year 2: $A_2 = 10000(1.08)^2 = ₹11,664$.
At end of year 3: $A_3 = 10000(1.08)^3 = ₹12,597.12$.
Differences:
$A_2 - A_1 = 11664 - 10800 = 864$
$A_3 - A_2 = 12597.12 - 11664 = 933.12 \neq 864$
$\therefore$ No, it does not form an AP.

$a_1 = a = 10$
$a_2 = a + d = 10 + 10 = 20$
$a_3 = a_2 + d = 20 + 10 = 30$
$a_4 = a_3 + d = 30 + 10 = 40$
$\therefore$ The first four terms are $10, 20, 30, 40$.

$a_1 = -2$
$a_2 = -2 + 0 = -2$
$a_3 = -2 + 0 = -2$
$a_4 = -2 + 0 = -2$
$\therefore$ The first four terms are $-2, -2, -2, -2$.

$a_1 = 4$
$a_2 = 4 + (-3) = 1$
$a_3 = 1 + (-3) = -2$
$a_4 = -2 + (-3) = -5$
$\therefore$ The first four terms are $4, 1, -2, -5$.

$a_1 = -1$
$a_2 = -1 + \dfrac{1}{2} = -\dfrac{1}{2}$
$a_3 = -\dfrac{1}{2} + \dfrac{1}{2} = 0$
$a_4 = 0 + \dfrac{1}{2} = \dfrac{1}{2}$
$\therefore$ The first four terms are $-1, -\dfrac{1}{2}, 0, \dfrac{1}{2}$.

$a_1 = -1.25$
$a_2 = -1.25 + (-0.25) = -1.50$
$a_3 = -1.50 + (-0.25) = -1.75$
$a_4 = -1.75 + (-0.25) = -2.00$
$\therefore$ The first four terms are $-1.25, -1.50, -1.75, -2.00$.

First term $a = 3$.
Common difference $d = a_2 - a_1 = 1 - 3 = -2$.
$\therefore a = 3, d = -2$.

First term $a = -5$.
Common difference $d = a_2 - a_1 = -1 - (-5) = -1 + 5 = 4$.
$\therefore a = -5, d = 4$.

First term $a = \dfrac{1}{3}$.
Common difference $d = a_2 - a_1 = \dfrac{5}{3} - \dfrac{1}{3} = \dfrac{4}{3}$.
$\therefore a = \dfrac{1}{3}, d = \dfrac{4}{3}$.

First term $a = 0.6$.
Common difference $d = a_2 - a_1 = 1.7 - 0.6 = 1.1$.
$\therefore a = 0.6, d = 1.1$.

$a_2 - a_1 = 4 - 2 = 2$; $a_3 - a_2 = 8 - 4 = 4$.
Since $a_2 - a_1 \neq a_3 - a_2$, it is not an AP.

$a_2 - a_1 = \dfrac{5}{2} - 2 = \dfrac{1}{2}$
$a_3 - a_2 = 3 - \dfrac{5}{2} = \dfrac{1}{2}$
$a_4 - a_3 = \dfrac{7}{2} - 3 = \dfrac{1}{2}$
Difference is constant $\left(d = \dfrac{1}{2}\right)$. Yes, it is an AP.
Next three terms:
$a_5 = \dfrac{7}{2} + \dfrac{1}{2} = 4$
$a_6 = 4 + \dfrac{1}{2} = \dfrac{9}{2}$
$a_7 = \dfrac{9}{2} + \dfrac{1}{2} = 5$
$d = \dfrac{1}{2}$; next terms: $4, \dfrac{9}{2}, 5$.

$a_2 - a_1 = -3.2 - (-1.2) = -2$
$a_3 - a_2 = -5.2 - (-3.2) = -2$
Difference is constant ($d = -2$). Yes, it is an AP.
Next three terms: $-9.2, -11.2, -13.2$.

$a_2 - a_1 = -6 - (-10) = 4$
$a_3 - a_2 = -2 - (-6) = 4$
Difference is constant ($d = 4$). Yes, it is an AP.
Next three terms: $2 + 4 = 6$, $6 + 4 = 10$, $10 + 4 = 14$.

$a_2 - a_1 = (3+\sqrt{2}) - 3 = \sqrt{2}$
$a_3 - a_2 = (3+2\sqrt{2}) - (3+\sqrt{2}) = \sqrt{2}$
Difference is constant ($d = \sqrt{2}$). Yes, it is an AP.
Next three terms: $3+4\sqrt{2}, 3+5\sqrt{2}, 3+6\sqrt{2}$.

$a_2 - a_1 = 0.22 - 0.2 = 0.02$
$a_3 - a_2 = 0.222 - 0.22 = 0.002 \neq 0.02$
Not an AP.

$a_2 - a_1 = -4 - 0 = -4$
$a_3 - a_2 = -8 - (-4) = -4$
Difference is constant ($d = -4$). Yes, it is an AP.
Next three terms: $-16, -20, -24$.

$a_2 - a_1 = -\dfrac{1}{2} - \left(-\dfrac{1}{2}\right) = 0$.
Difference is constant ($d = 0$). Yes, it is an AP.
Next three terms: $-\dfrac{1}{2}, -\dfrac{1}{2}, -\dfrac{1}{2}$.

$a_2 - a_1 = 3 - 1 = 2$; $a_3 - a_2 = 9 - 3 = 6 \neq 2$.
Not an AP (Geometric Progression with common ratio $r = 3$).

$a_2 - a_1 = 2a - a = a$; $a_3 - a_2 = 3a - 2a = a$.
Difference is constant ($d = a$). Yes, it is an AP.
Next three terms: $5a, 6a, 7a$.

$a_2 - a_1 = a^2 - a = a(a - 1)$
$a_3 - a_2 = a^3 - a^2 = a^2(a - 1) \neq a(a - 1)$ (unless $a = 0$ or $1$).
Not an AP.

Simplifying radicals:
$\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}, \dots$
$a_2 - a_1 = 2\sqrt{2} - \sqrt{2} = \sqrt{2}$
$a_3 - a_2 = 3\sqrt{2} - 2\sqrt{2} = \sqrt{2}$
Difference is constant ($d = \sqrt{2}$). Yes, it is an AP.
Next three terms:
$a_5 = 4\sqrt{2} + \sqrt{2} = 5\sqrt{2} = \sqrt{50}$
$a_6 = 5\sqrt{2} + \sqrt{2} = 6\sqrt{2} = \sqrt{72}$
$a_7 = 6\sqrt{2} + \sqrt{2} = 7\sqrt{2} = \sqrt{98}$
$d = \sqrt{2}$; next terms: $\sqrt{50}, \sqrt{72}, \sqrt{98}$.

$a_2 - a_1 = \sqrt{6} - \sqrt{3} = \sqrt{3}(\sqrt{2} - 1)$
$a_3 - a_2 = \sqrt{9} - \sqrt{6} = 3 - \sqrt{6} \neq \sqrt{6} - \sqrt{3}$.
Not an AP.

Terms are $1, 9, 25, 49, \dots$
$a_2 - a_1 = 9 - 1 = 8$; $a_3 - a_2 = 25 - 9 = 16 \neq 8$.
Not an AP.

Terms are $1, 25, 49, 73, \dots$
$a_2 - a_1 = 25 - 1 = 24$
$a_3 - a_2 = 49 - 25 = 24$
$a_4 - a_3 = 73 - 49 = 24$
Difference is constant ($d = 24$). Yes, it is an AP.
Next three terms:
$a_5 = 73 + 24 = 97$
$a_6 = 97 + 24 = 121 = 11^2$
$a_7 = 121 + 24 = 145$
$d = 24$; next terms: $97, 121, 145$.