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Question 37
What is the maximum sum of the terms in the arithmetic progression $$25, 24\frac{1}{2}, 24, ..............?$$
Solution
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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