Verified Solution • 2025–26 Session moderate

QJEEM25Ma9521: JEE Main 2025 Mathematics - Class 11 JEEM25Ma9521 Solution: Let upto terms. If the sum of the first six terms of an A.P.…

Let Sn=12+16+112+120+…S_n=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\ldots upto nn terms. If the sum of the first six terms of an A.P. with first term -p and common difference p is 2026 S2025\sqrt{2026 \mathrm{~S}_{2025}} , then the absolute difference betwen 20th 20^{\text {th }} and 15th 15^{\text {th }} terms of the A.P. is

Multiple Choice Options

  • A 20
  • B 45
  • C 90
  • D 25

Step-by-Step Verified Solution

amp;Sn=12+16+112+120….N terms amp;S2025=∑n=120251n(n+1)=∑n=12025(1n−1n+1)amp;=(11−12)+(12−13)⋯⋅(12025−12026)\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 \cdots \cdot\left(\frac{1}{2025}-\frac{1}{2026}\right)\end{aligned}

amp;=20252026amp;2026.S2025=2025=45amp; Given : 62[−2p+(6−1)p]=45amp;9p=45amp;p=5amp;∣ A20−A15∣=∣−5+19×5∣−[−5+14×5]amp;=∣90−65∣amp;=25\begin{aligned}& =\frac{2025}{2026} \\& \sqrt{2026 . \mathrm{S}_{2025}}=\sqrt{2025}=45 \\& \text { Given : } \frac{6}{2}[-2 \mathrm{p}+(6-1) \mathrm{p}]=45 \\& 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}

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