Question

Please elaborate all steps.

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm \: \displaystyle\sum_{n=0}^{ \infty }\rm {a}^{n} \: = \: x$

$\rm \: x = {a}^{0} + {a}^{1} + {a}^{2} + \cdots \cdots \infty$

$\rm \: x = 1 + {a}^{1} + {a}^{2} + \cdots \cdots \infty$

So, its an infinite GP series with First term 1 and common ratio 'a'.

Also, provided that |a| < 1.

So, using sum of infinite GP series, we get

$\rm \: x = \dfrac{1}{1 - a}$

$\rm \: 1 - a = \dfrac{1}{x}$

$\rm \: - a = \dfrac{1}{x}- 1$

$\rm \: a =1 - \dfrac{1}{x}$

So,

$\bf\implies \: a =1 - \dfrac{1}{x}$

Similarly,

$\bf\implies \: b =1 - \dfrac{1}{y}$

and

Similarly,

$\bf\implies \: c =1 - \dfrac{1}{z}$

Further, given that

$\bf \: a, b, c\:are\:in\:AP$

$\rm\implies \:2b = a + c$

On substituting the values of a, b, c from above, we get

$\rm \: 2\bigg(1 - \dfrac{1}{y} \bigg) = 1 - \dfrac{1}{x} + 1 - \dfrac{1}{z}$

$\rm \: 2 - \dfrac{2}{y}= 2 - \dfrac{1}{x} - \dfrac{1}{z}$

$\rm \: - \dfrac{2}{y}= - \dfrac{1}{x} - \dfrac{1}{z}$

$\rm \: \dfrac{2}{y}= \dfrac{1}{x} + \dfrac{1}{z}$

$\bf\implies \: \frac{1}{x} , \frac{1}{y} , \frac{1}{z} \:are\:in\:AP$

$\bf\implies \:x, y, z\:are\:in\:HP$

So, option 4) is correct.

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Formula Used :-

Sum of infinite GP series with first term a and common ratio r such that |r| < 1, is given by

$\boxed{ \rm{ \:S_ \infty \: = \: \dfrac{a}{1 - r} \: }}$

Answer 2

$\mathbb\red{ \tiny A \scriptsize \: N \small \:S \large \: W \Large \:E \huge \: R}$

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$\rm x=\sum_{n=0}^{\infty} a^{n} \quad\Rightarrow x=\frac{1}{1-a} \quad\Rightarrow a=\frac{x-1}{x}$

$\rm y=\sum_{n=0}^{\infty} b^{n} \quad\Rightarrow y=\frac{1}{1-b} \quad\Rightarrow b=\frac{y-1}{y}$

$\rm z=\sum_{n=0}^{\infty} c^{n} \quad\Rightarrow z=\frac{1}{1-c} \quad\Rightarrow c=\frac{z-1}{z}$

a, b, c are in A.P.

$\text{\( \rm \Rightarrow \dfrac{x-1}{x}, \dfrac{y-1}{y}, \dfrac{z-1}{z} \) are in A.P.}$

$\text{ \( \rm \Rightarrow 1-\dfrac{1}{x}, 1-\cfrac{1}{y}, 1-\dfrac{1}{z} \) are in A.P.}$

$\text{ \( \rm \Rightarrow-\dfrac{1}{x},-\cfrac{1}{y},-\dfrac{1}{z} \) are in A.P.}$

$\text{ \( \rm \Rightarrow x, y, z \) are in H.P.}$