Given that sum of $n$ terms of a G.P. $=315$, with $1^{st}$ term $=a=5$ and $r=2$
$\Rightarrow\:S_n=a.\large\frac{r^n-1}{r-1}$$=315$
$\Rightarrow\:5.\large\frac{2^n-1}{2-1}$$=315$
$\Rightarrow\:2^n-1=63$ or $2^n=64$
$\Rightarrow\:n=6$
Step 2
$\therefore\:$ No. of terms $=n=6$
We know that the last term $=t_n-a.r^n=5.2^{6-1}=5\times 32=160$
$i.e.,$ The last term $=160$