CAT 2023 Slot 3 · Quant · Question 20

The n-th term is a geometric sum in \frac{1}{3} multiplied by a power of \frac{1}{4}.

That inner sum has n terms, so it equals \frac{1-{\left( \frac{1}{3}\right)}^{n}}{1-\frac{1}{3}}=\frac{3}{2}\left( 1-\frac{1}{{3}^{n}}\right).

The outer factor on the n-th term is {\left( \frac{1}{4}\right)}^{n-1}.

So the whole series is \frac{3}{2}\sum_{n\ge 1}{{\left( \frac{1}{4}\right)}^{n-1}}-\frac{3}{2}\sum_{n\ge 1}{\frac{1}{{3}^{n}}}{\left( \frac{1}{4}\right)}^{n-1}.

The first sum is geometric with first term 1 and ratio \frac{1}{4}, so it equals \frac{1}{1-\frac{1}{4}}=\frac{4}{3}.

That contributes \frac{3}{2}\times \frac{4}{3}=2.

For the second sum, write \frac{1}{{3}^{n}}{\left( \frac{1}{4}\right)}^{n-1}=4\times{\left( \frac{1}{12}\right)}^{n}.

Summing the geometric series, 4\times \frac{\frac{1}{12}}{1-\frac{1}{12}}=4\times \frac{1}{11}=\frac{4}{11}.

That contributes \frac{3}{2}\times \frac{4}{11}=\frac{6}{11}.

The total is 2-\frac{6}{11}=\frac{22-6}{11}=\frac{16}{11} — the second option.

Shortcut. A rough estimate settles it fast: the first two terms alone give 1+\frac{4}{3}\times \frac{1}{4}=1.33, and the remaining terms are small, so the answer is a little under 1.5. Only \frac{16}{11}\approx 1.45 fits.