# 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 $$left$n+2 n^2$right)$. If the $n^{$text {th }}$ term of the progression is divisible by 9 , then the smallest possible value of $n$ is 1. 4 2. 8 3. 7 4. 9 ## Best CAT Online Coaching Try upto 40 hours for free Learn from the best! #### 2IIM : Best Online CAT Coaching. ### Video Explanation ## Best CAT Coaching in Chennai #### CAT Coaching in Chennai - CAT 2022Limited Seats Available - Register Now! ### Explanatory Answer Let $S _ { n }$stand for the sum of the first n terms of an arithmetic progression. Given, $S_n=\left$n+2 n^2$right$$ 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

Choice C is the correct answer.

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