Sn=21+61+121+201….N terms S2025=n=1∑2025n(n+1)1=n=1∑2025(n1−n+11)=(11−21)+(21−31)…⋯(20251−20261) $\begin{aligned}
& =\frac{2025}{2026} \
& \sqrt{2026 . \mathrm{S}{2025}}=\sqrt{2025}=45
\end{aligned}Given:\frac{6}{2}[-2 p+(6-1) p]=45\begin{aligned}
& 9 \mathrm{p}=45 \
& \mathrm{p}=5 \
& \left|\mathrm{~A}{20}-\mathrm{A}_{15}\right|=|-5+19 \times 5|-[-5+14 \times 5] \
& =|90-65| \
& =25
\end{aligned}$
Let S_n= 1 2+ 1 6+ 1 12+ 1 20+… upto n terms. If the sum of the first six terms of an A.P. with first term -p and common difference p is √2026…
JEE Main 2025 — Mathematics Algebra
2025mcqhard
Let Sn=21+61+121+201+… upto n terms. If the sum of the first six terms of an A.P. with first term -p and common difference p is 2026 S2025, then the absolute difference betwen 20th and 15th terms of the A.P. is
Official previous-year question
Held on 24 Jan 2025 · Verified 6 Jul 2026.
Options
- A
20
- B
90
- C
45
- D
25
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