Question

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

• A. \(637\frac{1}{2}\)
• B. 625
• C. \(662\frac{1}{2}\)
• D. 650

Hint:

Find the term that first touches zero, the sum stops growing once terms turn negative.

Answer 1

The Correct Option is A

Step 1: Identify the AP and its common difference.
The series is \(25, 24\frac{1}{2}, 24, 23\frac{1}{2}, \ldots\), so first term \(a = 25\) and common difference \(d = 24\frac{1}{2} - 25 = -\frac{1}{2}\). Since \(d\) is negative, the terms keep decreasing and will eventually turn negative, so the sum is maximum exactly at the point where the terms are still zero or positive.

Step 2: Find where the terms become zero.
The \(n\)th term is \(a_n = a + (n-1)d = 25 + (n-1)\left(-\frac{1}{2}\right)\).
Setting \(a_n = 0\): \(25 - \frac{1}{2}(n-1) = 0 \Rightarrow n - 1 = 50 \Rightarrow n = 51\).
So the 51st term is exactly 0, and every term after that turns negative.

Step 3: Sum up to the 51st term.
\(S_{51} = \frac{51}{2}(a_1 + a_{51}) = \frac{51}{2}(25 + 0) = \frac{51 \times 25}{2} = \frac{1275}{2} = 637.5\).
Adding the 51st term, which is 0, does not change the sum from the 50th term, so this is the maximum possible sum.

Final Answer:
The maximum sum is \(637\frac{1}{2}\). \[ \boxed{637\frac{1}{2}} \]