Question

Let \( S_n = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \dots \) up to \( n \) terms. If the sum of the first six terms of an A.P. with first term \( -p \) and common difference \( p \) is \( \sqrt{2026 S_{2025}} \), then the absolute difference between the 20th and 15th terms of the A.P. is:

• A. 25
• B. 90
• C. 20
• D. 45

Hint:

For problems involving the sum of terms in an arithmetic progression, use the formula for the sum of terms and the relationship between the first term, common difference, and the sum of terms. This can help simplify the problem and allow you to solve for the desired quantities.

Answer 1

The Correct Option is A

To address this problem, we will first establish the general term expression for the provided series and utilize it to calculate the sum \( S_{2025} \). Subsequently, we will correlate the given information concerning the arithmetic progression (A.P.) to determine the absolute difference between its 20th and 15th terms.

1. The general formula for the sequence \( \frac{1}{2}, \frac{1}{6}, \frac{1}{12}, \frac{1}{20}, \dots \) is \(T_n = \frac{1}{n(n+1)}\).
2. We must compute the sum \( S_{2025} = \sum_{n=1}^{2025} \frac{1}{n(n+1)} \).
3. The term \( \frac{1}{n(n+1)} \) can be resolved via partial fraction decomposition into \( \frac{1}{n} - \frac{1}{n+1} \). This constitutes a telescoping series, which simplifies to \( S_{2025} = \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{2025} - \frac{1}{2026}\right) \).
4. The series reduces to \( S_{2025} = 1 - \frac{1}{2026} \).
5. The sum of the first six terms of the A.P. is given as \( 6a + 15d = \sqrt{2026 S_{2025}} \). Substituting the value of \( S_{2025} \) yields \( \sqrt{2026 \times \left(1 - \frac{1}{2026}\right)} = \sqrt{2025} \).
6. Given that the first term of the A.P. is \( -p \) and the common difference is \( p \), the sum of the first six terms is \( 6(-p) + 15p = \sqrt{2025} \). Simplifying this equation results in \( 9p = 45 \), thus \( p = 5 \).
7. We need to calculate the absolute difference between the 20th and 15th terms of the A.P. using the formula \( a_n = a + (n-1)d \).
   • The 20th term is \( a_{20} = -p + (19)p = 19p - p = 18p \).
   • The 15th term is \( a_{15} = -p + (14)p = 14p - p = 13p \).

Consequently, the absolute difference between the 20th and 15th terms of the A.P. is 25.