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