Let $S_n=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\ldots$ upto $n$ terms. If the sum of the first six terms of an A.P. with first term -p and common difference p is $\sqrt{2026 \mathrm{~S}_{2025}}$, then the absolute difference betwen $20^{\text {th }}$ and $15^{\text {th }}$ terms of the A.P. is

<p>To find the value of $ S_{2025} $, calculate the partial sum</p> <p>$ S_n = \sum\limits_{n=1}^{2025} \frac{1}{n(n+1)} = \sum\limits_{n=1}^{2025} \left( \frac{1}{n} - \frac{1}{n+1} \right) $</p> <p>This series is telescopic, meaning most terms cancel out with each other:</p> <p>$ = \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) $</p> <p>Simplifying the expression, we find:</p> <p>$ = \frac{1}{1} - \frac{1}{2026} = 1 - \frac{1}{2026} = \frac{2025}{2026} $</p> <p>Now, evaluate $\sqrt{2026 \cdot S_{2025}}$:</p> <p>$ \sqrt{2026 \cdot \frac{2025}{2026}} = \sqrt{2025} = 45 $</p> <p>For the arithmetic progression with the first term $-p$ and common difference $p$, the sum of the first six terms is given by:</p> <p>$ \frac{6}{2} [-2p + (6-1)p] = 3(5p) = 15p $</p> <p>Given that:</p> <p>$ 15p = 45 $</p> <p>We find:</p> <p>$ p = 3 $</p> <p>To determine the absolute difference between the $20^{\text{th}}$ and $15^{\text{th}}$ terms of the A.P., compute:</p> <p>$ |A_{20} - A_{15}| = |(-p + 19p) - (-p + 14p)| $</p> <p>Substituting the value of $p$:</p> <p>$ = |18p - 13p| = |5p| = 5 \times 5 = 25 $</p> <p>Thus, the absolute difference is $ \boxed{25} $.</p>