Question

Middle term of the given sequence, i.e., $T_m$ is equal to -

• A. $\frac{16}{7}$
• B. $\frac{32}{7}$
• C. $\frac{48}{7}$
• D. $\frac{16}{9}$

Answer 1

The correct option is C $\frac{48}{7}$

Given that, there are $4n+1$ terms in a sequence of which first $2n+1$ are in Arithmetic Progression and last $2n+1$ are in Geometric Progression the common difference of Arithmetic Progression is $2$ and common ratio of Geometric Progression is $\frac{1}{2}$.
Let, $a$ be the first term of AP.
$\Rightarrow$ first term of GP is $a+(2n+1-1)d=a+4n$
Middle terms of AP and GP are equal.
$\Rightarrow a+2n=\frac{a+4n}{2^n}$ -----(1)
But, Middle term of the whole sequence is $T_m$ which is sum of infinite GeometricProgression whose sum of first Two terms is ${\left(\frac{5}{4}\right)}^{2}n$ and ratio of these terms is $\frac{9}{16}$
let $A$ be the first term of infinite geometric series.
$\Rightarrow A(1+r)=\frac{25}{16}n$
$\Rightarrow A=n$
$\Rightarrow T_m=a+4n=\frac{A}{1-r}=\frac{16n}{7}$ -----(2)
from (1) and (2)
$\frac{16n}{7}-2n=\frac{16n}{{7.2}^n}$
$\Rightarrow n=3$
$\therefore T_m=\frac{16n}{7}=\frac{48}{7}$.
Hence, option C.