Mathematics · Sequence and Series

JEE Main 2025 — 24 January, Evening Shift — Question 7

In an arithmetic progression, if $S_{40}=1030$ and $S_{12}=57$ , then $S_{30}-S_{10}$ is equal to:

Answer: B

Let a & d are first term and common diff of an AP.

$S_{40}=\frac{40}{2}[2 a+39 \mathrm{~d}]=1030$

$\mathrm{S}_{12}=\frac{12}{2}[2 \mathrm{a}+11 \mathrm{~d}]=57$

by (1) & (2)

$\mathrm{a}=-\frac{7}{2} \quad \mathrm{~d}=\frac{3}{2}$

$\therefore \mathrm{S}_{30}-\mathrm{S}_{10}=\frac{30}{2}[2 \mathrm{a}+29 \mathrm{~d}]-\frac{10}{2}[2 \mathrm{a}+9 \mathrm{~d}]$

$=20 \mathrm{a}+390 \mathrm{~d}$

$=515$

Answer key and solution verified before publishing.

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