Question

If $\frac{3}{5}+\frac{5}{15}+\frac{7}{45}+\frac{9}{135}+\dots =\frac{a}{b},$ then
(where $a$ and $b$ are coprime)

• A. $a=6$
• B. $b=5$
• C. $a=5$
• D. $b=6$

Answer 1

Answer: (A), (B)

The correct options are
A $a=6$
B $b=5$

Let $S=\frac{3}{5}+\frac{5}{15}+\frac{7}{45}+\frac{9}{135}+\dots$
$\Rightarrow S=\frac{1}{5}[\frac{3}{1}+\frac{5}{3}+\frac{7}{9}+\frac{9}{27}+\dots \infty ]$
Since, $\frac{3}{1}+\frac{5}{3}+\frac{7}{9}+\frac{9}{27}+\dots \infty$ is AGP with $a=3,d=2,r=\frac{1}{3}$
$\therefore S=\frac{1}{5}[\frac{3}{1-\frac{1}{3}}+\frac{2\times \frac{1}{3}}{{(1-\frac{1}{3})}^2}]$
$=\frac{1}{5}[\frac{9}{2}+\frac{3}{2}]$
$\Rightarrow S=\frac{6}{5}$

Alternate Solution:
Let $S=\frac{3}{5}+\frac{5}{15}+\frac{7}{45}+\frac{9}{135}+\dots$
$\frac{1}{3}S=\frac{3}{5\cdot 3}+\frac{5}{5\cdot 3^2}+\frac{7}{5\cdot 3^3}+\dots$
Now,
$S-\frac{1}{3}S=\frac{3}{5}+\frac{2}{5\cdot 3}+\frac{2}{5\cdot 3^2}+\frac{2}{5\cdot 3^3}+\dots$
$\Rightarrow \frac{2}{3}S=\frac{3}{5}+\frac{2}{5\cdot 3}(1+\frac{1}{3}+\frac{1}{3^2}+\dots )$
$\Rightarrow \frac{2}{3}S=\frac{3}{5}+\frac{2}{5\cdot 3}\times \frac{1}{1-\frac{1}{3}}$
$\Rightarrow \frac{2}{3}S=\frac{3}{5}+\frac{2}{5\cdot 3}\times \frac{3}{2}$
$\Rightarrow \frac{2}{3}S=\frac{3}{5}+\frac{1}{5}=\frac{4}{5}$
$\Rightarrow S=\frac{6}{5}$