CAT 2023

Slot 3

QA

Question & Solution

The given sequence can be written as:

$1\left(1+\ \frac{1}{4}+\frac{1}{16}+\frac{1}{64}+...\right)+\frac{1}{3}\left(\frac{1}{4}+\frac{1}{16}+...\right)+\frac{1}{9}\left(\frac{1}{16}+\frac{1}{64}+...\right)+..$

We know that the sum of an infinite G.P. is $\dfrac{a}{1-r}$, where a is the first term and r is the common ratio.

=> The first term = $\frac{1}{1-\frac{1}{4}}=\dfrac{4}{3}$

=> The second term = $\frac{1}{3}\left(\frac{\left(\frac{1}{4}\right)}{1-\left(\frac{1}{4}\right)}\right)=\dfrac{1}{9}$

=> The third term = $\frac{1}{9}\left(\frac{\left(\frac{1}{16}\right)}{1-\left(\frac{1}{4}\right)}\right)=\dfrac{1}{108}$

Observing these three terms, we see that they are in G.P. with a common ratio of $\dfrac{1}{12}$

=> Sum of this infinite G.P. = $\dfrac{\left(\dfrac{4}{3}\right)}{1-\left(\dfrac{1}{12}\right)}=\dfrac{16}{11}$