Mathematics - Sequence & Series Question with Solution | TestHub

Let the first term is a , then first $(2\mathrm{n}+1)$ terms are $a,a+2,a+4,\dots .a+2.2n$. Clearly the middle term of the sequence of $4n+1$ term is $(2n+1{)}^{\text{th }}$ term, i.e . $a+4n$ also the middle term of the A.P of $(2n+1)$ term is $(n+1{)}^{\text{th }}$ term i.e., $a+2n$. Again for the last $(2n+1)$ terms the first term will be $(2n+1{)}^{\text{th }}$ term of the A.P i.e. $a+4n$

$\therefore$ G.P. is (a + 4n), (1 + 4n) (0.5)ⁿ

Its middle term is $(a+4n)(0.5{)}^n$

According to the given condition,

$a+2n=(1+4n)(0.5{)}^n$

$\therefore a=\frac{2n-4n(0.5{)}^n}{(0.5{)}^n-1}$

∴ Required middle term $=a+4n=$

$\frac{2n-4n(0.5{)}^n}{(0.5{)}^n-1}+4n=\frac{2n}{1-{\left(\frac{1}{2}\right)}^2}=\frac{n.2^{n-1}}{2^n-1}$