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Q.
Find the $9^{th}$ term and the general term of the progression $\frac{1}{4}, -\frac{1}{2}, 1, -2,...$
Sequences and Series
Solution:
The given progression is clearly a $G.P$. with first term $a = 1/4$ and common ratio $r = - 2$.
$\therefore 9^{th}$ term $= a_{9} $
$= ar^{\left(9-1\right)}$
$ = \frac{1}{4}\left(-2\right)^{8} $
$= 64 $
and, General term $ = a_{n} $
$= ar^{\left(n-1\right)} $
$= \frac{1}{4}\left(-2\right)^{n-1}$
$ = \left(-1\right)^{n-1} 2^{n-3}$