In a geometric progression the ratio of the sum of the first 5 terms to the sum of their reciprocals is 49 and sum of the first and the third term is 35.The fifth term of the G.P. is

Let five terms in G.P. be $\frac{a}{r^2}, \frac{a}{r}, a, ar, a{r}^2$

Then $\frac{a\left(r^{-2}+r^{-1}+1+r+r^2\right)}{(1/a)\left(r^2+r+1+r^{-1}+r^{-2}\right)}=49$

$\Rightarrow a^2=49\Rightarrow a=\pm 7.$

Also, $\frac{a}{r^2}+a=35$

Therefore $a = – 7$ is not possible

Thus, $a=7$ and $a/r^2=28.$

Now, fifth term $=a{r}^2=a\left(\frac{a}{28}\right)=\frac{7}{4}$