Question

Find the 4th term form the end of the G.P. $\frac{2}{27},\frac{2}{9},\frac{2}{3},.....,162$

Answer 1

Here, first term, $a=\frac{2}{27}$

Common ratio, r = $\frac{a_2}{a_1}=\frac{\frac{2}{9}}{\frac{2}{27}}=3$

Last term, l = 162

After reversing the given G.P., we get another G.P. whose first term is l and common ratio is $\frac{1}{r}$.

4th term from the end $=l{\left(\frac{1}{r}\right)}^{4-1}=\left(162\right)$

${\left(\frac{1}{3}\right)}^3=6$