Question

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

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

Answer 1

The Correct Option is A

To find the maximum sum of the terms in this arithmetic progression (AP), we need to determine the point where the sum is largest before the values start getting negatively impacted by further terms. Given the AP starts at 25 and decreases by \( \frac{1}{2} \) each term, let's first determine when the sequence reaches zero.

In an AP, the n-th term \(a_n\) is given by \(a_n = a_1 + (n-1)d\), where \(a_1\) is the first term and \(d\) is the common difference.

Here, \(a_1 = 25\), \(d = -\frac{1}{2}\). We need to find n such that \(a_n = 0\).

\(0 = 25 + (n-1)\left(-\frac{1}{2}\right)\)

\((n-1)\left(-\frac{1}{2}\right) = -25\)

\(n-1 = 50\)

\(n = 51\)

The sequence becomes zero at the 51st term.

Let's calculate the sum of the 51 terms.

The sum \(S_n\) of the first n terms of an AP is given by:

\(S_n = \frac{n}{2}(a_1 + a_n)\)

Substitute \(n=51\), \(a_1=25\), \(a_n=0\):

\(S_{51} = \frac{51}{2}(25 + 0)\)

\(S_{51} = \frac{51}{2} \times 25\)

\(S_{51} = 51 \times 12.5\)

\(S_{51} = 637.5\)

Converting \(637.5\) to the form provided in options, we find:

\(637^{\frac{1}{2}}\)

Thus, the maximum sum of the terms in this arithmetic progression is \(637^{\frac{1}{2}}\).