The 4 th term of GP is 500 and its common ratio is 1 m ,   m ∈ N . Let S n denote the sum of the…

Given:

$T_4=500$

Let $a=$ first term

$r=$common ratio $=\frac{1}{m}, m\in N$

So,

$a{r}^3=500$

$\Rightarrow \frac{a}{m^3}=500$

Now,

$S_n-S_{n-1}=a{r}^{n-1}$

$S_6>S_5+1$

$\Rightarrow S_6-S_5>1$ 

$\Rightarrow \frac{a}{m^5}>1$

$\Rightarrow \frac{a}{m^3}\times \frac{1}{m^2}>1$

$\Rightarrow \frac{500}{m^2}>1$

$\Rightarrow m^2<500 ...\left(1\right)$

And,

$S_7-S_6<\frac{1}{2}$

$\Rightarrow \frac{a}{m^6}<\frac{1}{2}$

$\Rightarrow \frac{a}{m^3}\times \frac{2}{m^3}<1$

$\Rightarrow 500\times \frac{2}{m^3}<1$

$\Rightarrow m^3>{10}^3$

$\Rightarrow m>10 ....\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$, we have

$m=11, 12, 13\dots \dots \dots \dots .., 22$

So, number of possible values of $m$ is $12$.