CAT 2023 Slot 3 Solved Paper

The given sequence can be written as:
We know that the sum of an infinite G.P. is $\frac{\mathit{a}}{1-\mathit{r}}$, where $\mathrm{a}$ is the first term and $\mathrm{r}$ is the common ratio.
$\Rightarrow$ The first term $=\frac{1}{1-\frac{1}{4}}=\frac{4}{3}$
⇒ The second term $=\frac{1}{3}\left(\frac{\left(\frac{1}{4}\right)}{1-\left(\frac{1}{4}\right)}\right)=\frac{1}{9}$
⇒ The third term $=\frac{1}{9}\left(\frac{\left(\frac{1}{16}\right)}{1-\left(\frac{1}{4}\right)}\right)=\frac{1}{108}$
Observing these three terms, we see that they are in G.P. with a common ratio of $\frac{1}{12}$
⇒ Sum of this infinite G.P. $=\frac{\left(\frac{4}{3}\right)}{1-\left(\frac{1}{12}\right)}=\frac{16}{11}$