Let ${\text{S}}_{\mathrm{n}}=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\dots$ 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 between 20th and 15th terms of the A.P. is

$\begin{array}{l}{S}_n=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}···\mathrm{n}\text{ terms }\end{array}$

$S_n=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}···\frac{1}{n\left(n+1\right)}$

${\mathrm{S}}_{2025}=\sum _{\mathrm{n}=1}^{2025} \frac{1}{\mathrm{n}(\mathrm{n}+1)}=\sum _{\mathrm{n}=1}^{2025} \left(\frac{1}{\mathrm{n}}-\frac{1}{\mathrm{n}+1}\right)$

$=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)\cdots \cdot \left(\frac{1}{2025}-\frac{1}{2026}\right)$
$\begin{array}{l}\begin{array}{rl} & =\frac{2025}{2026}\\ & \sqrt{2026.{\mathrm{S}}_{2025}}=\sqrt{2025}=45\end{array}\\ \text{ Given : }\frac{6}{2}[-2\mathrm{p}+(6-1\left)\mathrm{p}\right]=45\\ \begin{array}{rl} & 9\mathrm{p}=45\\ & \mathrm{p}=5\\ & \left|{\mathrm{A}}_{20}-{\mathrm{A}}_{15}\right|=|-5+19\times 5|-[-5+14\times 5]\\ & =|90-65|\\ & =25\end{array}\end{array}$