Mathematics - Sequence & Series Question with Solution | TestHub

Since, the first $11$ terms are in $AP$, $d=2$ $\therefore a_{11}=a+10d=a+20$ The middle term of $AP$ is $T_6=a+5d=a+10$ For the next $11$ terms in $GP$ $r=2$ $\therefore$ The middle term of $GP$ is $b(2{)}^5$ where, $b$ is the first term of a $GP$ which is the last term of $AP$ $b(2{)}^5=(a+20)32$ According to the given condition, $\begin{array}{rl}\Rightarrow a+10& =(a+20)32\\ \Rightarrow 31a& =10-640\\ a& =-\frac{630}{31}\end{array}$ $\therefore$ Middle term of entire sequence is $11th$ term $\begin{array}{rl}\therefore T_{11}& =\frac{-630}{31}+10\times d\\ & =\frac{-630}{31}+10\times 2=\frac{-10}{31}\end{array}$