Let $S_{n}=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dots$ 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 $\sqrt{2026\cdot S_{2025}}$, then the absolute difference between $20^{\text{th}}$ and $15^{\text{th}}$ terms of the A.P.\ is:
Step-by-Step Solution
Key Concept: $\dfrac{1}{k(k+1)}=\dfrac{1}{k}-\dfrac{1}{k+1}$ telescopes to $S_{n}=\dfrac{n}{n+1}.$ Then $2026\,S_{2025}=2025$, so the right-hand side simplifies to $45$, pinning down $p$.
$S_{n}=\sum_{k=1}^{n}\dfrac{1}{k(k+1)}=\sum_{k=1}^{n}\!\left(\dfrac{1}{k}-\dfrac{1}{k+1}\right)=1-\dfrac{1}{n+1}=\dfrac{n}{n+1}.$
$S_{2025}=\dfrac{2025}{2026}\Rightarrow 2026\,S_{2025}=2025\Rightarrow \sqrt{2026\,S_{2025}}=45.$
A.P.\ first term $-p$, common difference $p$: sum of first $6$ terms $=\dfrac{6}{2}(-2p+5p)=9p=45\Rightarrow p=5.$
$A_{n}=(n-2)p$, so $|A_{20}-A_{15}|=|18p-13p|=5p=25.$
Correct Answer: 4