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The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2 respectively. Find the last term and the number of terms.
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Verified Answer
Let $S_n=315, a=5, r=2$
$\begin{aligned}
S_n &=\frac{a\left(r^n-1\right)}{r-1}=\frac{a r^n-a}{r-1} \\
&=\frac{r \cdot a r^{n-1}-a}{r-1}=\frac{r \ell-a}{r-1}
\end{aligned}$
where $\ell=$ last term $=T_n$
$\begin{aligned}
&\therefore 315=\frac{2 \ell-5}{2-1} \Rightarrow 315=2 \ell-5 \\
&2 \ell=315+5=320 \Rightarrow \ell=160 \\
&\text { Now, } \ell=T_n=a r^{n-1} \Rightarrow 160=5(2)^{n-1} \\
&\Rightarrow \quad 2^{n-1}=\frac{160}{5}=32 \\
&\Rightarrow \quad 2^{n-1}=(2)^5 \Rightarrow n-1=5 \text { i.e. } n=6
\end{aligned}$
Hence, the last term $=160$ and number of terms $=6$
$\begin{aligned}
S_n &=\frac{a\left(r^n-1\right)}{r-1}=\frac{a r^n-a}{r-1} \\
&=\frac{r \cdot a r^{n-1}-a}{r-1}=\frac{r \ell-a}{r-1}
\end{aligned}$
where $\ell=$ last term $=T_n$
$\begin{aligned}
&\therefore 315=\frac{2 \ell-5}{2-1} \Rightarrow 315=2 \ell-5 \\
&2 \ell=315+5=320 \Rightarrow \ell=160 \\
&\text { Now, } \ell=T_n=a r^{n-1} \Rightarrow 160=5(2)^{n-1} \\
&\Rightarrow \quad 2^{n-1}=\frac{160}{5}=32 \\
&\Rightarrow \quad 2^{n-1}=(2)^5 \Rightarrow n-1=5 \text { i.e. } n=6
\end{aligned}$
Hence, the last term $=160$ and number of terms $=6$
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