If $\sum _{\mathrm{j}=1}^{21} {\mathrm{a}}_{\mathrm{j}}=693$ where a₁, a₂, ... , a₂₁ are in A.P. and  $\sum _{\mathrm{i}=0}^{10} {\mathrm{a}}_{2\mathrm{i}+1}=$

${\mathrm{a}}_1+{\mathrm{a}}_2+\dots \dots +{\mathrm{a}}_{21}=693\Rightarrow {\mathrm{a}}_1+\left({\mathrm{a}}_1+\mathrm{d}\right)+\dots \dots +\left({\mathrm{a}}_1+20\mathrm{d}\right)=693$
$\begin{array}{l}\Rightarrow 21{\mathrm{a}}_1+\mathrm{d}(1+2+3+\dots \dots +20)=693\\ \Rightarrow 21{\mathrm{a}}_1+21\times 10\mathrm{d}=693\Rightarrow {\mathrm{a}}_1+10\mathrm{d}=33\\ \Rightarrow {\mathrm{a}}_1+{\mathrm{a}}_3+\dots \dots ..{\mathrm{a}}_{21}=11\left({\mathrm{a}}_1+10\mathrm{d}\right)=11\times 33=363\end{array}$