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.
$$3 + \frac{3}{2} - \frac{3}{4} + \frac{3}{8} - .....$$
B.
$$3 + \frac{3}{2} + \frac{3}{4} + \frac{3}{8} + .....$$
C.
$$3 - \frac{3}{2} + \frac{3}{4} - \frac{3}{8} + .....$$
D.
None of these
Answer :
$$3 - \frac{3}{2} + \frac{3}{4} - \frac{3}{8} + .....$$
Solution :
Let first term = $$a,$$ common ratio = $$r,$$ where $$ – 1 < r < 1$$
Then, $$\frac{a}{{1 - r}} = 2\,\,{\text{and }}\frac{{{a^3}}}{{1 - {r^3}}} = 24$$
$$\eqalign{
& \therefore \frac{{1 - {r^3}}}{{{{\left( {1 - r} \right)}^3}}} = \frac{1}{3} \cr
& {\text{i}}{\text{.e}}{\text{., }}1 - 2r + {r^2} = 3\left( {1 + r + {r^2}} \right) \cr
& {\text{or, }}2{r^2} + 5r + 2 = 0 \cr
& \therefore r = - 2{\text{ or }}\frac{{ - 1}}{2}{\text{ As }} - 1 < r < 1 \cr
& \therefore {\text{we have }}r = - \frac{1}{2} \cr
& \therefore {\text{The series is }}3 - \frac{3}{2} + \frac{3}{4} - \frac{3}{8} + ..... \cr} $$