In an arithmetic progression, if S _{40}=1030 and S _{12}=57 , then S _{30}- S

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8. Sequences and Series

medium

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

A$510$

B$515$

C$525$

D$505$

(JEE MAIN-2025)

Solution

Let $a \& d$ are first term and common diff of an $AP .$
$S_{40}=\frac{40}{2}[2 a+39 d]=1030$
$S_{12}=\frac{12}{2}[2 a+11 d]=57$
$\text { by }(1) \ (2)$
$a=-\frac{7}{2} d=\frac{3}{2}$
$\therefore S _{30}- S _{10}=\frac{30}{2}[2 a +29 d]-\frac{10}{2}[2 a +9 d]$
$=20 a -390 d$
$=515$

Standard 11

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Solution

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