Question

If $l=1+a+a^2+,,,$ to $\infty m=1+b+b^2+...$ to $\infty$, $n=1+c+c^2+...$ to $\infty$ where $|a|<1,|b|<1,|c|<1$ and a, b, c are in AP then $l,m,n$ are in

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

Answer 1

Answer: (C)

HP

Here , $l,m,n$ are the sum of infinite GP,

Therefore,

$l=\frac{1}{1-a},m=\frac{1}{1-b},n=\frac{1}{1-c}$

On solving this , we get

$a=\frac{l-1}{l}$, $b=\frac{m-1}{m}$ , $c=\frac{n-1}{n}$ ........(1)

As $a,b,c$ are in AP,

So, $2b=a+c$ .......(2)

Substitute values of $a,b,c$ from (1) in (2). We get,

$\frac{2(m-1)}{m}=\frac{n-1}{n}+\frac{l-1}{l}$

=> $2-\frac{2}{m}=1-\frac{1}{n}+1-\frac{1}{l}$

=> $\frac{2}{m}=\frac{1}{n}+\frac{1}{l}$

Therefore, $l,m,n$ are in HP.