Question

In a sequence of $(4n+1)$ terms, the first $(2n+1)$ terms are in A.P. whose common difference is 2 and the last $(2n+1)$ terms are in G.P. with common ratio 0.5. If the middle terms of the A P. and G.P. are equal, then the middle term of the sequence is

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

Answer 1

Answer: (B)

$\frac{n.2^{n+1}}{2^n-1}$

To find the middle term of the sequence
Total number of terms are $(4n+1)$
First $(2n+1)$ terms are in AP whose common difference is $2$
and the last $(2n+1)$ terms are in GP with common ratio $0.5$
Given,
If the middle terms of the AP and GP are equal
Let the middle term be $a{}^{\prime }$
Middle term of AP $=a+4n$
Middle term of GP $=a^{\prime }$
Now, According to Question
$a+4n=a^{\prime }$
and , $(2n+1)$ terms of both the series are also equal
$a+2n={\left(\frac{1}{2}\right)}^n{a}^{\prime }$
On solving both equation , we get
$2n=a^{\prime }\left(\frac{2^n-1}{2^n}\right)$
$2^{n+1}n=a^{\prime }(2^n-1)$
$a{}^{\prime }=\frac{2^{n+1}n}{2^n-1}$

This is the middle term of the sequence

Hence , option $B$ is correct