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

$$S_n=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\cdots$$ upto $$n$$ terms.

$$S_{2025}=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+.....\ 2025\ terms$$

$$S_{2025}=\dfrac{2-1}{1\times2}+\dfrac{3-2}{3\times2}+\dfrac{4-3}{4\times3}+....\ +\dfrac{2026-2025}{2025\times2026}$$

$$S_{2025}=\left(\dfrac{1}{1}-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+.....+\left(\dfrac{1}{2025}-\dfrac{1}{2026}\right)$$

$$S_{2025}=1-\dfrac{1}{2026}$$

$$2026\times S_{2025}=2026\times\frac{2025}{2026}=2025$$

The sum of the first six terms of an A.P. with first term $$-p$$ and common difference $$p$$ is $$\sqrt{2026\, S_{2025}}$$

$$-p+0+p+2p+3p+4p=\sqrt{2025}$$

$$9p=45$$

$$p=5$$

$$a_{20}=a+19d=(-p)+19p=18p$$

$$a_{15}=a+14d=(-p)+14p=13p$$

Thus, $$a_{20}-a_{15}=18p-13p=5p=5\times5=25$$