Mathematics - Sequence & Series Question with Solution | TestHub

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