The 4 th term of GP is 500 and its common ratio is (1)/(m), m ∈ N ....

Given:

${T}_{4} = 500$

Let $a =$ first term

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

So,

$a{r}^{3} = 500$

$⇒ \dfrac{a}{{m}^{3}} = 500$

Now,

${S}_{n} - {S}_{n - 1} = a{r}^{n - 1}$

${S}_{6} > {S}_{5} + 1$

$⇒ {S}_{6} - {S}_{5} > 1$

$⇒ \dfrac{a}{{m}^{5}} > 1$

$⇒ \dfrac{a}{{m}^{3}} \times \dfrac{1}{{m}^{2}} > 1$

$⇒ \dfrac{500}{{m}^{2}} > 1$

$⇒ {m}^{2} < 500 . . . \left(1\right)$

And,

${S}_{7} - {S}_{6} < \dfrac{1}{2}$

$⇒ \dfrac{a}{{m}^{6}} < \dfrac{1}{2}$

$⇒ \dfrac{a}{{m}^{3}} \times \dfrac{2}{{m}^{3}} < 1$

$⇒ 500 \times \dfrac{2}{{m}^{3}} < 1$

$⇒ {m}^{3} > {10}^{3}$

$⇒ m > 10 . . . . \left(2\right)$

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

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

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