Question

If $A=1+r^a+r^{2a}+r^{3a}+\cdots \infty ,a>0$ and $B=1+r^{2b}+r^{4b}+r^{6b}+\cdots \infty ,b>0,$ for $|r|<1,$ then $\frac{a}{b}$ is equal to

• A. ${\log }_{B}A$
• B. ${\log }_{1-B}(1-A)$
• C. ${\log }_{\frac{B-1}{B}}{\left(\frac{A-1}{A}\right)}^2$
• D. $2{\log }_{A}B$

Answer 1

The correct option is C ${\log }_{\frac{B-1}{B}}{\left(\frac{A-1}{A}\right)}^2$

$A=1+r^a+r^{2a}+r^{3a}+\cdots \infty ,a>0$
which is an infinite G.P. with common ratio $r^a$
$A=\frac{1}{1-r^a}$
$\Rightarrow 1-r^a=\frac{1}{A}$
$\Rightarrow r^a=1-\frac{1}{A}=\frac{A-1}{A}\cdots \left(1\right)$

$B=1+r^{2b}+r^{4b}+r^{6b}+\cdots \infty ,b>0$
which is an infinite G.P. with common ratio $r^{2b}$
$B=\frac{1}{1-r^{2b}}$
$\Rightarrow 1-r^{2b}=\frac{1}{B}$
$\Rightarrow r^{2b}=1-\frac{1}{B}=\frac{B-1}{B}\cdots \left(2\right)$

$a\log r=\log \left(\frac{A-1}{A}\right)\left[\text{From (1)}\right]$
and $2b\log r=\log \left(\frac{B-1}{B}\right)\left[\text{From (2)}\right]$
$\therefore \frac{a}{b}={\log }_{\frac{B-1}{B}}{\left(\frac{A-1}{A}\right)}^2$