Difficulty: Medium
Correct Answer: (mn+1)/2
Explanation:
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:
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:
Subtract: (m − n)d = 1/n − 1/m = (m − n)/(mn) ⇒ d = 1/(mn).From t_m: a = 1/n − (m − 1)d = 1/n − (m − 1)/(mn) = 1/(mn).Now S_{mn} = (mn)/2 [2a + (mn − 1)d] = (mn)/2 [2/(mn) + (mn − 1)/(mn)].Inside bracket = (mn + 1)/(mn). Multiply gives S_{mn} = (mn + 1)/2.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:
Common Pitfalls:Sign errors when subtracting the two equations, or forgetting to divide by mn consistently across terms.
Final Answer:
(mn+1)/2
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