AP with reciprocal targets: In an arithmetic progression, the m-th term is 1/n and the n-th term is 1/m. What is the sum of the first mn terms?

Introduction / Context:
We are given two AP terms at positions m and n that are reciprocals in a symmetric way. Solving for the common difference and first term lets us compute the sum of the first mn terms succinctly.

Given Data / Assumptions:

• t_m = a + (m − 1)d = 1/n.
• t_n = a + (n − 1)d = 1/m.

Concept / Approach:
Subtract equations to find d, then back-solve for a. Finally, use S_k = k/2[2a + (k − 1)d] with k = mn, simplifying carefully.

Step-by-Step Solution:

Verification / Alternative check:
Symmetry suggests both a and d equal 1/(mn), which our algebra confirms, making the final sum compact and elegant.

Why Other Options Are Wrong:

• (mn−1)/4, (mn+1)/4, (mn−1)/2: Do not follow from the derived a and d; algebra leads to (mn+1)/2 specifically.

Common Pitfalls:
Sign errors when subtracting the two equations, or forgetting to divide by mn consistently across terms.

Final Answer: