CSAT Solved Papers/ 2021/Q50

2021 CSAT — Q50

Quant Logical & quantitative reasoning 2.5 marks Easy

What is the value of 'XX' in the sequence 2, 7, 22, 67, 202, X, 18222,\ 7,\ 22,\ 67,\ 202,\ X,\ 1822?

  1. A 603
  2. B 605
  3. C 607 Answer
  4. D 608

Worked rationale

Test the term-to-term rule "×3+1\times 3 + 1":

2×3+1=7,7×3+1=22,22×3+1=67,67×3+1=202. ✓2\times3+1 = 7,\quad 7\times3+1 = 22,\quad 22\times3+1 = 67,\quad 67\times3+1 = 202.\ \checkmark

So the next term is

X=202×3+1=607.X = 202\times 3 + 1 = 607.

Verify it propagates: 607×3+1=1822607\times 3 + 1 = 1822, matching the given final term. ✓

Answer: (c) 607.

Why the other options miss

  • A
    an arithmetic slip: computes 202×3=606202\times 3 = 606 then subtracts/mis-adds, or uses ×3−3\times 3 - 3.
  • B
    the wrong rule: applies a slightly wrong rule (e.g. ×3−1\times 3 - 1) giving 605605.
  • D
    adds one too many: adds 22 instead of 11 after tripling (202×3+2202\times3+2).

Specialist insight

When successive ratios are roughly 33 but not clean, test the affine rule an+1=3an+ca_{n+1}=3a_n + c and read off cc from one step (7=3⋅2+17 = 3\cdot2 + 1). Here c=1c=1, and the real safety net is the given last term: 607×3+1=1822607\times3+1 = 1822 confirms the rule both forward and across the gap, so there’s no ambiguity. Always validate a recursive rule on two steps and, when a later term is supplied, close the loop with it.

The trap, in one line

Rule ×3+1\times3+1: X=202⋅3+1=607X = 202\cdot3+1 = 607, and 607⋅3+1=1822607\cdot3+1 = 1822 checks out ⇒\Rightarrow (c).

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