Question

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 in which the common difference of Arithmetic Progression is $2$ and common ratio of Geometric Progression is $\frac{1}{2}$. The middle term of the Arithmetic Progression is equal to middle term of Geometric Progression. Let middle term of the sequence is $T_m$ and $T_m$ is the sum of infinite Geometric Progression whose sum of first Two terms is ${\left(\frac{5}{4}\right)}^{2}n$ and ratio of these terms is $\frac{9}{16}$.

First term of given sequence is equal to

• A. $-\frac{8}{7},-\frac{20}{7}$
• B. $-\frac{36}{7}$
• C. $\frac{36}{7}$
• D. $\frac{48}{7}$

Answer 1

The correct option is B $-\frac{36}{7}$

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)
$\frac{16n}{7}-2n=\frac{16n}{{7.2}^n}$
$\Rightarrow n=3$
$\Rightarrow T_m=a+4n=\frac{16n}{7}=\frac{48}{7}$.
$\therefore$ First term of the sequence $a=\frac{48}{7}-12=\frac{-36}{7}$
Hence, option B.