Question

The $4^{\text {th }}$ term of $G P$ is $S 00$ and its common ratio is $\frac{1}{m}, m \in N$. Let $S_n$ denote the sum of the first $n$ terms of this GP. If $S_6>S_5+1$ and $S_7< S_6+\frac{1}{2}$, then the number of possible values of $m$ is

Answer 1

$T _4=500 $ where $a =$ first term,
$r =\text { common ratio }=\frac{1}{ m }, m \in N$
$ a r^3=500 $
$ \frac{a}{m^3}=500 $
$ S_n-S_{n-1}=a r^{n-1}$
$ S_6>S_5+1$
and $S _7- S _6< \frac{1}{2}$
$S _6- S _5>1 \,\,\,\,\,\frac{ a }{ m ^6}<\frac{1}{2}$
$ar ^5>1 \,\,\,\, m ^3>10^3$
$\frac{500}{ m ^2}>1 \,\,\,\,\, m >10$...(2)
$m ^2< 500$.........(1)
From (1) and (2)
$m =11,12,13 \ldots \ldots \ldots \ldots ., 22$
So number of possible values of $m$ is $12$