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

Since the terms of this AP keep falling by $\frac{1}{2}$ each time, the running sum keeps growing only as long as the terms being added are positive. Once a term turns negative, adding it pulls the sum back down, so the sum is greatest right at the last non-negative term.

1. General term: $a_n = 25 - \frac{1}{2}(n-1)$.
2. Last positive term: for $a_n > 0$ we need $n - 1 < 50$, so $a_{50} = 25 - 24.5 = 0.5$ is the last strictly positive term, and $a_{51} = 0$ exactly.
3. Sum through the zero term: including $a_{51}=0$ adds nothing extra, so $S_{51} = S_{50}$, both equal to the maximum sum.
4. Apply the sum formula: $S_{51} = \frac{51}{2}\left[2(25) + (51-1)\left(-\frac{1}{2}\right)\right] = \frac{51}{2}\left[50 - 25\right] = \frac{51 \times 25}{2} = 637.5$.

So the maximum possible sum of the AP is $637\frac{1}{2}$. $$\boxed{637\tfrac{1}{2}}$$