CAT 2023 Question Paper | Quant Slot 3

CAT 2023 Quant was dominated by Arithmetic followed by Algebra. In Arithmetic, the questions were dominated by topics like Speed-time-distance, Mixture and Alligations. This year, there was a surprise. The questions from Geometry were relatively on the lower side as compared to the previous years. There were 8 TITA Qs this year. Overall this section was at a medium level of difficulty.

Question 20 : The value of \(1+\left(1+\frac{1}{3}\right) \frac{1}{4}+\left(1+\frac{1}{3}+\frac{1}{9}\right) \frac{1}{16}+\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right) \frac{1}{64}+\cdots\), is

Video Explanation

Explanatory Answer

\( 1 + \left( 1 + \frac { 1 } { 3 } \right) \frac { 1 } { 4 } + \left( 1 + \frac { 1 } { 3 } + \frac { 1 } { 3 ^ { 2 } } \right) \frac { 1 } { 4 ^ { 2 } } + \left( 1 + \frac { 1 } { 3 } + \frac { 1 } { 3 ^ { 2 } } + \frac { 1 } { 3 ^ { 3 } } \right) \frac { 1 } { 4 ^ { 3 } } + \ldots \)
The nth term in the series is given by, \( T _ { n } = \left( \right. \) Sum of \( n \) terms of a GP with \( \left. a = 1 , r = \frac { 1 } { 3 } \right) \left( \frac { 1 } { 4 } \right) ^ { n - 1 } \)
\( T _ { n } = \left( \frac { 1 - \left( \frac { 1 } { 3 } \right) ^ { n } } { 1 - \frac { 1 } { 3 } } \right) \frac { 1 } { 4 ^ { n - 1 } } = \frac { \left( 3 ^ { n } - 1 \right) \times 3 } { 3 ^ { n } \times 2 } \left( \frac { 1 } { 4 ^ { n - 1 } } \right) \)
\( T _ { n } = \frac { 1 } { 2 } \left( \frac { 3 ^ { n } - 1 } { 12 ^ { n - 1 } } \right) = \frac { 1 } { 2 } \left( \frac { 3 } { 4 ^ { n - 1 } } - \frac { 1 } { 12 ^ { n - 1 } } \right) = \frac { 3 } { 2 } \left( \frac { 1 } { 4 } \right) ^ { n - 1 } - \frac { 1 } { 2 } \left( \frac { 1 } { 12 } \right) ^ { n - 1 } \)
So, we have decomposed the infinite series into two infinite series by decomposing the nth term.
\( \therefore S _ { \infty } = \frac { 3 / 2 } { 1 - 1 / 4 } - \frac { 1 / 2 } { 1 - 1 / 12 } = \frac { 3 / 2 } { 3 / 4 } - \frac { 1 / 2 } { 11 / 12 } = 2 - \frac { 6 } { 11 } = \frac { 16 } { 11 } \)

The question is " The value of \(1+\left(1+\frac{1}{3}\right) \frac{1}{4}+\left(1+\frac{1}{3}+\frac{1}{9}\right) \frac{1}{16}+\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right) \frac{1}{64}+\cdots\), is "

Hence, the answer is '\(\frac{16}{11}\)'

Choice B is the correct answer.

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