Question

Let $a,b,c$ be in AP and $\left|a\right|<1,\left|b\right|<1$, $\left|c\right|<1$. If $x=1+a+a^2+...+\infty ,y=1+b+b^2+...+\infty ,z=1+c+c^2+...+\infty$, then $x,y\&z$ are in:

• A. AP
• B. GP
• C. HP
• D. None of these

Answer 1

The correct option is C

HP

Explanation For The Correct Option:

Step 1: Determining the sum of series $x$

The given series,

$x=1+a+a^2+\dots +\infty$

This is an infinite GP with common $r=a$

Since the sum of geometric series with common ratio $r$ $=\frac{1}{(1-a)}$

Therefore,

$x=\frac{1}{(1-a)}..\left(i\right)$

Step 2: Determining the sum of series $y$

Given that

$y=1+b+b^2+\dots +\infty$

This is an infinite GP with common $r=b$

Therefore, the sum of series

$y=\frac{1}{(1-b)}..\left(ii\right)$

Step 3: Determining the sum of series $z$

Given that

$z=1+c+c^2+\dots +\infty$

Similarly,

$z=\frac{1}{(1-c)}..\left(iii\right)$

Step 4: Finding the nature of $x,y,z$

Since $a,b,c$ in AP

Then, $\left(1-a\right),\left(1-b\right),\left(1-c\right)$be also in AP

Therefore, $\frac{1}{(1-a)},\frac{1}{(1-b)},\frac{1}{(1-c)}$are in HP.

Thus, from $\left(i\right).\left(ii\right)\&\left(iii\right)$, $x,y,z$ are in HP.

Therefore, option (C) is the correct answer.