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

Question

Aicharya · Accepted Answer

$\begin{aligned}
& S_{40}=1030 \Rightarrow \frac{40}{2}[2 a+39 d]=1030 \\ & \Rightarrow \quad 2 a+39 d=\frac{103}{2}....(1) \\ & \quad S_{12}=57 \Rightarrow \frac{12}{2}[2 a+11 d]=57 \\ & \Rightarrow \quad 2 a+11 d=\frac{57}{6}...(2)
\end{aligned}$
Equation (1) - equation (2)
$\begin{aligned}
& 28 d=\frac{103}{2}-\frac{57}{6} \\ & 28 d=\frac{309-57}{6}
\end{aligned}$
$d=\frac{3}{2}$
$\Rightarrow \quad a=-\frac{7}{2}$
$S_{30}-S_{10}=\frac{30}{2}[2 a+29 d]-\frac{10}{2}[2 a+9 d]$
$=15[2 a+29 d]-5[2 a+9 d]$
$=5[6 a+87 d-2 a-9 d]$
$=5[4 a+78 d]$
$=5[-14+117]$
$=515$ *

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

Solution