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Q.
If the sum of an infinite GP is $4 / 3$ and the sum of the series obtained on squaring each term is $16 / 27$, then its common ratio is
Sequences and Series
Solution:
Let first term is a and common ratio is $r$ then
$\frac{a}{1-r}=\frac{4}{3}$......(1)
And $\frac{a^2}{1-r^2}=\frac{16}{27}$...........(2)
by $(1)^2(2)$
$\Rightarrow \frac{1+r}{1-r}=3 $
$ \Rightarrow r=1 / 2$