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MathsMediumMCQ2022 · 29 Jul Shift 1

Q62.If (20−a)(40−a) 1 + (40−a)(60−a)1 + … … + (180−a)(200−a)1 = 2561 , then the maximum value of a is (1) 198 (2) 202 (3) 212 (4) 218

What This Question Tests

This question requires evaluating a sum of terms that can be resolved using partial fraction decomposition, leading to a telescoping series, and then solving for an unknown variable.

Concepts Tested

Partial fraction decompositionTelescoping seriesArithmetic Progression

Formulas Used

1/(k-a)(k+d-a) = 1/d * [1/(k-a) - 1/(k+d-a)]

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