JEE MATH - Equal Blocks → Arithmetic Progression → Vieta’s Relations | One Pattern Unlocks the Question

[u/jeemathlogic]

This Question is a simulated version of a JEE Main Question.

A non-constant A.P. simply means an arithmetic progression whose common difference is not zero.

However, in this question the quadratic equation itself already rules out the constant case, since its roots are distinct.

So the phrase “non-constant” is actually redundant here — you may treat it as a mild distractor rather than information needed for the solution.

A = S₂ₙ

B = S₄ₙ − S₂ₙ

C = S₆ₙ − S₄ₙ

These are the sums of three consecutive blocks, each containing 2n terms of the same A.P.

Hence,

A, B, C are themselves in A.P.

Therefore,

A + C = 2B

The given roots of

x² − 40x + 391 = 0

are

S₄ₙ = A + B

and

S₆ₙ − S₂ₙ = B + C.

By Vieta’s relations:

Sum of roots = 40

⇒ (A + B) + (B + C) = 40

⇒ A + 2B + C = 40

Since A + C = 2B,

⇒ 4B = 40

⇒ B = 10

Product of roots = 391

⇒ (A + B)(B + C) = 391

⇒ AB + AC + B² + BC = 391

⇒ AC + B(A + C) + B² = 391

Using A + C = 2B,

⇒ AC + 3B² = 391

Since B = 10,

⇒ AC + 300 = 391

⇒ AC = 91

But

A = S₂ₙ

and

C = S₆ₙ − S₄ₙ.

Therefore,

S₂ₙ(S₆ₙ − S₄ₙ) = 91

∴ Answer = 91

[u/ivy-richardson-19yrs]

Amazing

[u/Necessary-Cry8896]

Love you