CAT 2022 Slot 1 QA Q18 Progressions & Series

Question 18 of the Quantitative Ability (QA) section, CAT 2022 Slot 1, with CAT's official answer and a worked solution.

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. A
    4
  2. B
    8
  3. C
    7
  4. D
    9

Answer: (C) 7 Progressions & Series MCQ

Solution

  1. \(a_n = S_n - S_{n-1} = (n + 2n^2) - (n - 1 + 2(n-1)^2) = 4n - 1\).
  2. Need \(9 \mid 4n - 1\): \(n = 1,2,\dots\) give \(3, 7, 11, 15, 19, 23, 27\).
  3. First multiple of 9 is \(27\) at \(n = 7\).

Same concept in other years

See every question on nth term = S_n − S_(n−1).

More Progressions & Series questions from CAT

All 38 Progressions & Series questions asked since 2017.