Question

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

• A. $3+\frac{3}{2}-\frac{3}{4}+\frac{3}{8}-\dots$
• B. $3+\frac{3}{2}+\frac{3}{4}+\frac{3}{8}+\dots$
• C. $3-\frac{3}{2}+\frac{3}{4}-\frac{3}{8}+\dots$
• D. None of these

Answer 1

Answer: (C)

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 $1-2r+r^2=3\left(1+r+r^2\right)$ or $2r^2+5r+2=0$
$\therefore r=-2$ or $\frac{-1}{2}$ As $-1<r<1$
$\therefore$ we have $r=-\frac{1}{2}$
$\therefore$ The series is $3-\frac{3}{2}+\frac{3}{4}-\frac{3}{8}+\dots$