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Q.
The sum of an infinite geometric series is $2$ and the sum of the geometric series made from the cubes of this infinite sereis is $24$ . Then the series is
Sequences and Series
Solution:
Let first term = $a$, common ratio = $r$, where $-1<\, r <\,1$
Then, $\frac{ a }{1- r }=2$ and $\frac{ a ^{3}}{1- r ^{3}}=24 $
$\therefore \frac{1- r ^{3}}{(1- r )^{3}}=\frac{1}{3}$
i.e $1-2 r+r^{2}=3\left(1+r+r^{2}\right)$ or $2 r^{2}+5 r+2=0$
$\therefore r =-2$ or $\frac{-1}{2} $ As $-1< \,r <\,1 $
$ \therefore $ we have $r =-\frac{1}{2}$
$\therefore $ The series is $3-\frac{3}{2}+\frac{3}{4}-\frac{3}{8}+\ldots$