Question

The $4^{\text{th }}$ term of $GP$ is $S00$ 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

Answer: 12

$T_4=500$ 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 $S_7-S_6<\frac{1}{2}$
$S_6-S_5>1\frac{a}{m^6}<\frac{1}{2}$
$a{r}^5>1m^3>10^3$
$\frac{500}{m^2}>1m>10$...(2)
$m^2<500$.........(1)
From (1) and (2)
$m=11,12,13\dots \dots \dots \dots .,22$
So number of possible values of $m$ is $12$