Q37 / 75 · JEE Main 2025 (24 Jan Shift 1) · Mathematics
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
- A$20$
- B$90$
- C$45$
- Correct answerD$25$
Solution
$\begin{aligned} & \mathrm{Sn}=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20} \ldots . \mathrm{N} \text { terms } \\ & \mathrm{S}_{2025}=\sum_{\mathrm{n}=1}^{2025} \frac{1}{\mathrm{n}(\mathrm{n}+1)}=\sum_{\mathrm{n}=1}^{2025}\left(\frac{1}{\mathrm{n}}-\frac{1}{\mathrm{n}+1}\right) \\ & =\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right) \quad \ldots \cdots\left(\frac{1}{2025}-\frac{1}{2026}\right)\end{aligned}$
$\begin{aligned}
& =\frac{2025}{2026} \\
& \sqrt{2026 . \mathrm{S}_{2025}}=\sqrt{2025}=45
\end{aligned}$
Given : $\frac{6}{2}[-2 p+(6-1) p]=45$
$\begin{aligned}
& 9 \mathrm{p}=45 \\
& \mathrm{p}=5 \\
& \left|\mathrm{~A}_{20}-\mathrm{A}_{15}\right|=|-5+19 \times 5|-[-5+14 \times 5] \\
& =|90-65| \\
& =25
\end{aligned}$
Hence, the correct answer is option: 'D'.