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Which statement is correct for (8,8+3q,8+6q,8+9q,\ldots)?

Answer and explanation

Correct answer: Arithmetic progression for every (q)

The sequence is an arithmetic progression for every value of q, so A is correct. Subtract consecutive terms: (8+3q)−8=3q, (8+6q)−(8+3q)=3q, and (8+9q)−(8+6q)=3q. Since the same algebraic difference occurs each time, the sequence is an AP, with common difference d=3q. This remains true even when q=0, when all terms become 8; a constant sequence is an AP with d=0. B is wrong because q need not equal 1. C is wrong because q=0 is only one special case, not the only case. D is wrong because equal successive differences prove that it is an AP. Memory cue: compare consecutive subtractions symbolically, not by choosing one value of q.

Related tags

ApAlgebraic Common DifferenceConditionHard

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