Question

Find the $6^{\mathrm{th}}$ term from the end of the GP $8,4,2$$.....\frac{1}{1024}$

Answer 1

Find the $6^{\mathrm{th}}$ term from the end of the GP:

The $n^{th}$term from the end is given by:

$a_n=l{\left(\frac{l}{r}\right)}^{n-1}$where, $l$ is the last term, $r$ is the common ratio, $n$ is the $n^{th}$ term

Here, last term, $\mathrm{l}=\frac{1}{1024}$

First term is $\mathrm{a}=8$

Common ratio $r=\frac{a_2}{a_1}=\frac{4}{8}=\frac{1}{2}$

$6^{\mathrm{th}}$ term from the end of the GP will be

$\begin{array}{rcl}& \Rightarrow & a_6=\frac{1}{1024}\times (2{)}^{6-1}\\ & =& \frac{1}{1024}\times (2{)}^5\\ & =& \frac{1}{32}\end{array}$

Hence, the $6^{th}$value is $\frac{1}{32}$.