Question

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

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

Answer 1

The correct option is A $\frac{n{.2}^{n+1}}{2^n-1}$

Let the first term =a
$\Rightarrow (a,a+2,a+4,...),(0+2\times 2n)(a+4n),(a+4n\left(5\right))...(a+4n){\left(5\right)}^{n-1}$
Middle term $={(n+1)}^{th}$
$\Rightarrow a+(n+1-1)4=(a+4n){\left(5\right)}^n\Rightarrow a=\frac{4n{\left(5\right)}^n-2n}{1-{\left(5\right)}^n}$
Therefore middle term $=a+4n=2n\left[\frac{2\times {\left(5\right)}^n-1}{1-{\left(5\right)}^n}\right]=\frac{n.2^{n+1}}{2^n-1}$