JEE Mains · Maths · STD 11 - 8. sequence and series
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 \(45\)
- D \(25\)
Answer & Solution
Correct Answer
(D) \(25\)
Step-by-step Solution
Detailed explanation
\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}\rig…
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