Question

In a sequence of $4 n+1$ terms, the first $2 n+1$ terms are in A.P. having common difference $2$ and the last $2 n+1$ terms are in G.P. having common ratio $\frac{1}{2}$. If the middle term of the A.P. is equal to the middle term of the G.P. then the middle term of the sequence is

• A. $\frac{n \cdot 2^{n+1}}{2^{n}+1}$
• B. $\frac{n \cdot 2^{n+1}}{2^{n}-1}$
• C. $\frac{n \cdot 2^{n}}{2^{n}-1}$
• D. None of these

Answer 1

Answer: (B)

The middle term of the $4 n+1$ terms is the $(2 n+1)$ th term. Let it be $m$.
The middle term of $(2 n+1)$ terms is the $(n+1)$ th term.
Thus, the middle term of the A.P. is
$=m-(n+1-1) 2=m-2 n$
and the middle term of the G.P is
$=m\left(\frac{1}{2}\right)^{n+1-1}=\frac{m}{2^{n}}$
According to the given condition, we have
$m-2 n=\frac{m}{2^{n}} $
$\Rightarrow m=\frac{n \cdot 2^{n+1}}{2^{n}-1}$