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UPSC CSE CSAT 2019 · Paper II (CSAT) · Question 51 · Number system

अनुक्रम 4, 196, 16, 144, 36, 100, 64, X में X क्या है ?

UPSC Civil Services (Preliminary) Examination 2019, General Studies Paper II (CSAT), Question 51 — Number system.

CSAT 2019 · Q51

Number system Easy

अनुक्रम 4, 196, 16, 144, 36, 100, 64, X में X क्या है ?

Read this question in English

What is X in the sequence 4, 196, 16, 144, 36, 100, 64, X?

(a) 48 · (b) 64 · (c) 125 · (d) 256

उत्तर और व्याख्या · explanation in English

Answer: (b) 64

Every term is a square: 2², 14², 4², 12², 6², 10², 8². Bases in odd places rise 2, 4, 6, 8 and in even places fall 14, 12, 10, so the next even-place base is 8 and X = 64.

  1. 4 = 2², 196 = 14², 16 = 4², 144 = 12², 36 = 6², 100 = 10², 64 = 8².
  2. Odd positions: bases 2, 4, 6, 8 (rising by 2). Even positions: 14, 12, 10, … (falling by 2).
  3. X is in an even position, so its base is 8 and X = 8² = 64.

Remember · In mixed sequences, split the odd and even positions and look for squares or cubes in each.

Question and answer: UPSC's official GS Paper II (2019, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·

Cite this question: UPSC CSE CSAT 2019, Q51 — minimalistias.com/pyqs/csat/2019/q51

In the UPSC syllabus: Number system › Sequences, cycles & extreme values See it and tick it covered →

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