Mathematics - Sequence & Series Question with Solution | TestHub

${\mathrm{a}}_2+{\mathrm{a}}_4+\dots +{\mathrm{a}}_{\mathrm{n}}=30$ ...(1)
${\mathrm{a}}_1+{\mathrm{a}}_3+\dots +{\mathrm{a}}_{\mathrm{n}-1}=24$ ...(2)
$(1)-(2)$
$\left(a_2-a_1\right)+\left(a_4-a_3\right)\dots \left(a_n-a_{n-1}\right)=6$
$\begin{array}{rl}& \Rightarrow \frac{\mathrm{n}}{2}\mathrm{d}=6\Rightarrow \mathrm{nd}=12\\ & {\mathrm{a}}_{\mathrm{n}}-{\mathrm{a}}_1=(\mathrm{n}-1)\mathrm{d}=\frac{21}{2}\\ & \Rightarrow \mathrm{nd}-\mathrm{d}=\frac{21}{2}\Rightarrow 12-\frac{21}{2}=\mathrm{d}\\ & \Rightarrow \mathrm{d}=\frac{3}{2},\mathrm{n}=8\end{array}$
$\begin{array}{rl}& \text{ Sum of odd terms }=\frac{4}{2}[2\mathrm{a}+(4-1)3]=24\\ & \Rightarrow \mathrm{a}=\frac{3}{2}\\ & \text{ A.P. }\Rightarrow \frac{3}{2},3,\frac{9}{2},6,\frac{15}{2},9,\frac{21}{2},12\end{array}$
no. of integer terms $=4$