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Q.
The fourth, seventh and the last term of a $G.P$. are $10,80$and $2560,$ respectively. Then the first term and number of terms in the G.P. are, respectively,
Sequences and Series
Solution:
Let $a$ be the first term and $r$ be the common ratio of the given $G.P$. Then
$ a_{4}=10, a_{7}=80$
$\Rightarrow a r^{3}=10$ and $a r^{6}=80$
$\Rightarrow \frac{a r^{6}}{a r^{3}}=\frac{80}{10}$
$\Rightarrow r^{3}=8$
$\Rightarrow r=2$
Putting $r=2$ in $a r^{3}=10,$ we get: $a=\frac{10}{8} = \frac{5}{4}$
Let there be $n$ terms in the given G.P. then, $a_{n}=2650$
$\Rightarrow a r^{n-1}=2560$
$\Rightarrow \frac{10}{8}\left(2^{n-1}\right)=2560 \Rightarrow 2^{n-4}=256 \Rightarrow 2^{n-4}=2^{8} \Rightarrow n-4=8$
$\Rightarrow n= 12$