Question

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

Answer 1

$$\begin{gathered} \mathrm{Here},\mathrm{first}\mathrm{term},a=\frac{2}{27} \\ C\mathrm{ommon}\mathrm{ratio},r=\frac{a_2}{a_1}=\frac{{\displaystyle \frac{2}{9}}}{{\displaystyle \frac{2}{27}}}=3 \\ \mathrm{Last}\mathrm{term},l=162 \\ \mathrm{After}\mathrm{reversing}\mathrm{the}\mathrm{given}\mathrm{G}.\mathrm{P}.,\mathrm{we}\mathrm{get}\mathrm{another}\mathrm{G}.\mathrm{P}.\mathrm{whose}\mathrm{first}\mathrm{term}\mathrm{is}l\mathrm{and}\mathrm{common}\mathrm{ratio}\mathrm{is}\frac{1}{r}. \\ \therefore 4\mathrm{th}\mathrm{term}\mathrm{from}\mathrm{the}\mathrm{end}=\mathit{}l{\left(\frac{1}{r}\right)}^{4-1}=\left(162\right){\left(\frac{1}{3}\right)}^3=6 \end{gathered}$$