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UPSC CSE CSAT 2021 · Paper II (CSAT) · Question 50 · Puzzles, arrangements & general mental ability

What is the value of ‘X’ in the sequence 2, 7, 22, 67, 202, X, 1822?

UPSC Civil Services (Preliminary) Examination 2021, General Studies Paper II (CSAT), Question 50 — Puzzles, arrangements & general mental ability.

What is the value of ‘X’ in the sequence 2, 7, 22, 67, 202, X, 1822?

Answer & explanation

Answer: (c) 607

Every term is three times the previous one plus 1. So X = 202 × 3 + 1 = 607, and 607 × 3 + 1 = 1822 confirms it.

  1. 2 × 3 + 1 = 7, 7 × 3 + 1 = 22, 22 × 3 + 1 = 67, 67 × 3 + 1 = 202 — each term is 3 × previous + 1.
  2. X = 202 × 3 + 1 = 607.
  3. Check: 607 × 3 + 1 = 1822, the next term.

Remember · When terms roughly triple, test '× 3 ± k', and always verify with the term after the blank.

Question and answer: UPSC's official GS Paper II (2021, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·

Cite this question: UPSC CSE CSAT 2021, Q50 — minimalistias.com/pyqs/csat/2021/q50

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