Sequence And Series Question 473
Question: The sum of an infinite geometric series is 2 and the sum of the geometric series made from the cubes of this infinite series is 24. Then the series is
Options:
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
Show Answer
Answer:
Correct Answer: C
Solution:
[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(1+r+r^{2}) $ 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}+… $
BETA
