CAT 2022 Question Paper | Quant Slot 1

CAT Previous Year Paper | CAT Quant Questions | Question 18

CAT 2022 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 18 : For any natural number \(n\), suppose the sum of the first \(n\) terms of an arithmetic progression is \(\left(n+2 n^2\right)\). If the \(n^{\text {th }}\) term of the progression is divisible by 9 , then the smallest possible value of \(n\) is

  1. 4
  2. 8
  3. 7
  4. 9

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Explanatory Answer

Let \( S _ { n } \)stand for the sum of the first n terms of an arithmetic progression.
Given, \(S_n=\left(n+2 n^2\right)\)
Let \( T _ { n } \)stand for the \( n ^ { t h } \) term of the same arithmetic progression.
\( T _ { n } = S _ { n } - S _ { n - 1 } \)
\( T _ { n } = ( n + 2 n ^ { 2 } ) - ( n - 1 + 2 ( n - 1 ) ^ { 2 } ) \)
\( T _ { n } = ( n + 2 n ^ { 2 } ) - ( n - 1 + 2 ( n ^ { 2 } + 1 - 2 n ) ) \)
\( T _ { n } = ( n + 2 n ^ { 2 } ) - ( n - 1 + 2 n ^ { 2 } + 2 - 4 n ) \)
\( T _ { n } = ( n + 2 n ^ { 2 } - n + 1 - 2 n ^ { 2 } - 2 + 4 n ) \)
\( T _ { n } = 4 n - 1 \)
The smallest value of n for which 4n - 1 is a multiple of 9 is 7.
n = 7


The answer is '7'

Choice C is the correct answer.

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