Mixed Progression Reasoning

Question

If the numbers a, b, c are simultaneously in arithmetic progression and in geometric progression with all terms nonzero, what must necessarily hold among them?

RankGuru · Accepted Answer

This problem combines the defining conditions of two progressions, a synthesis JEE Advanced favours for testing conceptual depth. Being in AP requires 2b = a + c, and being in GP requires b^2 = ac. Substitute a + c = 2b into the GP condition by treating a and c as roots of t^2 - 2bt + b^2 = 0, which factors as (t - b)^2 = 0, forcing a = c = b. Thus all three terms coincide. Option a + c = 2b only captures the AP condition alone and is incomplete. Option b^2 = ac only states the GP condition alone, again partial. Option claiming HP is false, since constant equal terms are trivially in every progression but the strict HP-only claim is not what is forced. Hence a = b = c. The structural reason is that the AP condition makes b the arithmetic mean of a and c while the GP condition makes b the geometric mean, and the arithmetic and geometric means of two numbers coincide only when those numbers are equal, instantly collapsing the triple. Plausibility check: any constant sequence like 5, 5, 5 satisfies both 2b = a + c and b^2 = ac, and no non-constant triple can satisfy both, confirming the equality is the unique necessary outcome.

Mixed Progression Reasoning

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