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In a sequence of $21$ terms, the first $11$ terms are in $AP$ with common difference $2$ and the last $11$ terms are in $GP$ with common ratio $2$. If the middle term of $AP$ be equal to the middle term of the $GP$, then the middle term of the entire sequence is
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The correct answer is:
$-\frac{10}{31}$
$-\frac{10}{31}$
Since, the first $11$ terms are in $AP$, $d=2$
$\therefore \quad a_{11}=a+10 d=a+20$
The middle term of $AP$ is
$T_6=a+5 d=a+10$
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,
$\begin{aligned}
\Rightarrow \quad a+10 & =(a+20) 32 \\
\Rightarrow \quad 31 a & =10-640 \\
a & =-\frac{630}{31}
\end{aligned}$
$\therefore$ Middle term of entire sequence is $11th$ term
$\begin{aligned} \therefore \quad T_{11} & =\frac{-630}{31}+10 \times d \\ & =\frac{-630}{31}+10 \times 2=\frac{-10}{31}\end{aligned}$
$\therefore \quad a_{11}=a+10 d=a+20$
The middle term of $AP$ is
$T_6=a+5 d=a+10$
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,
$\begin{aligned}
\Rightarrow \quad a+10 & =(a+20) 32 \\
\Rightarrow \quad 31 a & =10-640 \\
a & =-\frac{630}{31}
\end{aligned}$
$\therefore$ Middle term of entire sequence is $11th$ term
$\begin{aligned} \therefore \quad T_{11} & =\frac{-630}{31}+10 \times d \\ & =\frac{-630}{31}+10 \times 2=\frac{-10}{31}\end{aligned}$
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