Question 37

What is the maximum sum of the terms in the arithmetic progression $$25, 24\frac{1}{2}, 24, ..............?$$

Solution

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It is a decreasing AP 25,24.5,24,23.5,...

The sum will be largest, if nth term is zero then

25 + (n-1)*(-.5) = 0

=> n = 51

after 51 terms AP contains -ve terms

so, max sum will be obtained upto 50/51th term

S = $$\frac{51}{2}$$ *(25+0) = 637.5

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