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Mixed Progression Reasoning

Hardmathematics

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?

Select the correct option:

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About This Question

Subject
mathematics
Chapter
sequence and series
Topic
mixed progression reasoning
Difficulty
Hard
Year
2025
Tags
advanced-calculus-drillap-gp-combinationdiscriminantconstant-sequencenecessary-condition

Solution

Correct 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.

This hard difficulty mathematics question is from the chapter sequence and series, covering the topic of mixed progression reasoning. It appeared in the 2025 exam.

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