Question

If the nth term of the Geometric Progressions $162,54,18,.....$ and $\frac{2}{81},\frac{2}{27},\frac{2}{9},.....$ are equal, then find the value of $n$.

Answer 1

Given that:

$G.P_1:162,54,18,......$ and $G.P_2:\frac{2}{81},\frac{2}{27},\frac{2}{9},.......$

$t_n$ of the both G.P are equal.

To find: $n=?$

Solution:

$G.P_1:162,54,18,......$

$a_1=162,r_1=\frac{1}{3}$

$n^{th}$ term $t_n$ of $G.P_1=162\times (\frac{1}{3}{)}^{n-1}$ $------\left(i\right)$

$G.P_2:\frac{2}{81},\frac{2}{27},\frac{2}{9},.......$

$a_2=\frac{2}{81},r_2=3$

$n^{th}$ term $t_n$ of $G.P_{=}\frac{2}{81}\times 3^{n-1}$ $------\left(ii\right)$

Equating eqn.$\left(i\right)$ and eqn.$\left(ii\right)$ we get,

$162\times (\frac{1}{3}{)}^{n-1}=\frac{2}{81}\times 3^{n-1}$

or, $3^8=3^{2n-2}$

or, $81\times 81=3^{2n-2}$

or $2n-2=8$

or, $2n=10$

or, $n=5$