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 solve this problem, we will first determine the expression for the general term of the given series and use it to find the sum \( S_{2025} \). Then, we will relate the information given about the arithmetic progression (A.P.) to compute the absolute difference between the 20th and 15th terms.

1. Let's express the general formula for the sequence \( \frac{1}{2}, \frac{1}{6}, \frac{1}{12}, \frac{1}{20}, \dots \). These terms suggest they follow the pattern of the form: \(T_n = \frac{1}{n(n+1)}\). 
2. We need to compute the sum \( S_{2025} \): \(S_{2025} = \sum_{n=1}^{2025} \frac{1}{n(n+1)}\).
3. The term \(\frac{1}{n(n+1)}\) can be decomposed using partial fraction: \(\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}\). This is a telescoping series, and it 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. Now, the sum of the first six terms of the A.P. given is: \(6a + 15d = \sqrt{2026 S_{2025}}\). Substituting the value of \(S_{2025}\): \(\sqrt{2026 \times \left(1 - \frac{1}{2026}\right)} = \sqrt{2025}\).
6. Since the first term of the A.P. is \(-p\) and the common difference is \( p \), the sum of the first six terms: \(6(-p) + 15p = \sqrt{2025}\). Simplify: \(9p = 45 \Rightarrow p = 5\).
7. Compute the absolute difference between the 20th and 15th terms of the A.P.: \(a_n = a + (n-1)d\).
   • 20th term: \(a_{20} = -p + (19)p = 19p - p = 18p\)
   • 15th term: \(a_{15} = -p + (14)p = 14p - p = 13p\)

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

Answer 2

To solve the problem, we first need to understand the sequence \( S_n = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \dots \). This is a series where each term can be expressed as \(\frac{1}{n(n+1)}\). The terms of this series can be rewritten as a telescoping series: \[ \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1} \] Thus, the sum of the series up to \( S_n \) is: \[ S_n = \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \ldots + \left(\frac{1}{n} - \frac{1}{n+1}\right) = 1 - \frac{1}{n+1} \] Therefore, \( S_{2025} = 1 - \frac{1}{2026} \). Next, we address the A.P. problem. We are given that the sum of the first six terms of an A.P. with first term \(-p\) and common difference \(p\) is \(\sqrt{2026 S_{2025}}\). The sum of the first six terms (\(S_6\)) of an A.P. is: \[ S_6 = \frac{6}{2} \times [2(-p) + 5p] = 3(-2p + 5p) = 9p \] According to the problem: \[ 9p = \sqrt{2026 \times \left(1 - \frac{1}{2026}\right)} = \sqrt{2026 - 1} = \sqrt{2025} = 45 \] Thus, \(9p = 45\) which gives us \(p = 5\). We are required to find the absolute difference between the 20th and 15th terms of the A.P.: 20th term = \(-p + 19p = 19p - p = 18p\) 15th term = \(-p + 14p = 14p - p = 13p\) The absolute difference is: \[ |18p - 13p| = |5p| = 5 \times 5 = 25 \] Therefore, the absolute difference between the 20th and 15th terms of the A.P. is 25.