Mathematics - Sequence & Series Question with Solution | TestHub

Let the common difference of $t_i$ be $d$, so $t_2=t_1+d$. Then $s_1=t_1,s_2=2\left(t_1+d\right)$, and the common difference of $s_i$ is $s_2-s_1=t_1+2d$. $\begin{array}{rlr}& \frac{10}{2}\left[2t_1+9\left(t_1+2d\right)\right]=\frac{15}{2}\left[2t_1+14d\right]\Rightarrow 2\left(11t_1+18d\right)=3\left(2t_1+14d\right)\Rightarrow 16t_1=6d\Rightarrow 8t_1=3d.& \frac{s_2-s_1}{t_2-t_1}=\frac{t_1+2d}{d}=\frac{\frac{3d}{8}+2d}{d}=\frac{19}{8}.\end{array}$