CAT 2023 Slot 2 QA Q20 Progressions & Series

Question 20 of the Quantitative Ability (QA) section, CAT 2023 Slot 2, with CAT's official answer and a worked solution.

Let both the series \(a_1, a_2, a_3, \ldots\) and \(b_1, b_2, b_3 \ldots\) be in arithmetic progression such that the common differences of both the series are prime numbers. If \(a_5=b_9, a_{19}=b_{19}\) and \(b_2=0\), then \(a_{11}\) equals
  1. A
    79
  2. B
    83
  3. C
    86
  4. D
    84

Answer: (A) 79 Progressions & Series MCQ

Solution

  1. Let the common differences be primes \(p\) (for \(a\)) and \(q\) (for \(b\)). From \(b_2 = 0\), \(b_n = (n-2)q\).
  2. \(a_5 = b_9\): \(a_1 + 4p = 7q\). \(a_{19} = b_{19}\): \(a_1 + 18p = 17q\).
  3. Subtracting: \(14p = 10q \Rightarrow 7p = 5q\); with both prime, \(p = 5, q = 7\).
  4. \(a_1 = 49 - 20 = 29\), so \(a_{11} = 29 + 10\times5 = 79\).

Same concept in other years

See every question on AP problems: write terms as a + (n − 1)d and solve.

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