In an arithmetic progression, if S _{40}=1030 | vedclass

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Question

Let $a \& d$ are first term and common diff of an $AP .$
$S_{40}=\frac{40}{2}[2 a+39 d]=1030$
$S_{12}=\frac{12}{2}[2 a+11 d]=57$
$	ext { by }(

Vedclass · Accepted Answer

$515$

Vedclass · Answer

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

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

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