Harmonic progression — find the 3rd term using an AP condition: The 1st term of a harmonic progression (HP) is 1/17. The product of the 2nd and 4th terms equals the product of the 5th and 6th terms. Find the 3rd term of the HP.

Introduction / Context:In an HP, reciprocals of the terms form an AP. Use that correspondence to translate the given product condition into a simple relation in the AP's first term A and common difference d.

Given Data / Assumptions:

• HP: H_k = 1 / (A + (k−1)d).
• H_1 = 1/17 ⇒ A = 17.
• H_2 * H_4 = H_5 * H_6.

Concept / Approach:Rewrite the products using AP denominators and cancel numerators to get an equation in A and d. Then compute H_3 = 1/(A + 2d).

Step-by-Step Solution:H_2 H_4 = 1/[(A+d)(A+3d)], H_5 H_6 = 1/[(A+4d)(A+5d)]Equality ⇒ (A+4d)(A+5d) = (A+d)(A+3d)Expand: A^2 + 9Ad + 20d^2 = A^2 + 4Ad + 3d^2 ⇒ 5Ad + 17d^2 = 0With A = 17 ⇒ d = −5H_3 = 1/(A+2d) = 1/(17 − 10) = 1/7

Verification / Alternative check:Compute H_2, H_4, H_5, H_6 explicitly with A = 17, d = −5; the equality holds.

Why Other Options Are Wrong:1/14 and 1/35 correspond to wrong d values; they do not satisfy the given product constraint.

Common Pitfalls:Forgetting to invert to an AP; mishandling the negative common difference.

Final Answer:1/7