The sum of the 3 r d and the 4 t h terms of a G . P . is 60 and the product of its first three terms is 1000…

Let the $\text{'}\mathrm{G}.\mathrm{P}.\text{'}$ be

$\frac{a}{r},a,ar,a{r}^2$

Given, 

$\frac{a}{r}·a·ar=1000$

$\Rightarrow a^3=1000$

$\Rightarrow a=10$

Also, $ar+a{r}^2=60$

$\Rightarrow 10\left(r+r^2\right)=60$

$\Rightarrow r+r^2=6$

$\Rightarrow r^2+r-6=0$

$\Rightarrow \left(r-2\right) \left(r+3\right)=0$

So, $r=2$ or $r=-3$

As $1^{st}$ term is positive $\therefore r=2$

$\therefore \mathrm{ }\mathrm{G}.\mathrm{P}.$ is $\text{'}5\text{'}, \text{'}10\text{'}, \text{'}20\text{'}.......$

Hence, $7^{th}$ term is $5·{\left(2\right)}^6=320.$