$\begin{array}{1 1}t_n=160 \; and \;n=6 \\t_n=320 \; and\; n=5 \\t_n=320 \;and \;n=6 \\ t_n=160\; and \;n=5 \end{array} $

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- Sum of $n$ terms of a G.P $=a.\large\frac{r^n-1}{r-1}$ where $a=1^{st}$ term and $r=$ common ratio.
- $n^{th}$ term of a G.P. $=t_n=a.r^{n-1}$

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 $=160$

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