Question

Find the $4^{\mathrm{th}}$ term from the end of the GP: $\frac{2}{27},\frac{2}{9},\frac{2}{3},\dots ,162$

Answer 1

Find the $4^{\mathrm{th}}$ term from the end of the GP:

Formula:

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

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

Here, last term, $l=162$

$\begin{array}{rcl}r& =& \frac{a_2}{a_1}\\ & =& \frac{\left({\displaystyle \frac{2}{9}}\right)}{\left({\displaystyle \frac{2}{27}}\right)}\\ & =& \frac{2}{9}\times \frac{27}{2}\\ & =& 3\\ n& =& 4\end{array}$

$\begin{array}{rcl}\Rightarrow a_4& =& 162{\left(\frac{1}{3}\right)}^{4-1}\\ & =& 162{\left(\frac{1}{3}\right)}^{4-1}\\ & =& 162{\left(\frac{1}{3}\right)}^3\\ & =& 162\times \frac{1}{27}\\ & =& 6\end{array}$

Hence, the $4^{th}$ term from last is $6$.