Question

The sum of an infinite geometric progression (G.P.) is 2 and the sum of the G.P. made from the cubes of the terms of this infinite series is 24. The values a and r respectively (where a is the first term and r denotes the common ratio of the series):

• A. - 2, 3
• B. $3,-\frac{1}{2}$
• C. $0,\frac{1}{2}$
• D. can't be determined

Answer 1

The correct option is B $3,-\frac{1}{2}$

First of all choice (c) is ruled out since 'a' can not be zero. Again choice (a) is ruled out because $|r|\le 1$
Now let us check the option (b)
$S_n=3+(-\frac{3}{2})+\frac{9}{4}+(-\frac{27}{8})+...\infty \therefore S_n=\frac{3}{1-(-\frac{1}{2})}=\frac{3}{\frac{3}{2}}=2Again,27-\frac{27}{8}+\frac{729}{64}-...\infty S_{\infty }=\frac{27}{1-(-\frac{1}{8})}\frac{27}{\frac{9}{8}}=24$
Hence choice (b) is the appropriate choice.


Alternatively:
$S_1=\frac{a}{1-r}=2S_2=\frac{a^3}{1-r^3}=24$
By solving these two equations, we get
$a=3,r=-\frac{1}{2}$