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Algebra Difficulty 2.8 Junior Find the answer

Given that all terms of the geometric sequence {an}\{a_n\} are positive and a1=1a_1=1, a1+a3+a5=21a_1+a_3+a_5=21, then a2+a4+a6=a_2+a_4+a_6=

Pick one

Solution

Analysis

This problem tests the general formula of a geometric sequence and its properties, as well as reasoning and computational skills. It is considered a medium-level question.

Solution

Let's assume the common ratio of the geometric sequence {an}\{a_n\}, where all terms are positive, is q>0q > 0,

Since a1=1a_1=1 and a1+a3+a5=21a_1+a_3+a_5=21,

Therefore, 1+q2+q4=211+q^2+q^4=21. Solving this equation yields q=2q=2.

Then, a2+a4+a6=q(a1+a3+a5)=2×21=42a_2+a_4+a_6=q(a_1+a_3+a_5)=2\times21=42,

Hence, the correct choice is B\boxed{B}.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.