Question

Which term of the progression 18, −12, 8, ... is $\frac{512}{729}$?

Answer 1

$$\begin{gathered} \mathrm{Here},\mathrm{first}\mathrm{term},a=18 \\ \mathrm{and}\mathrm{common}\mathrm{ratio},r=\frac{-2}{3} \\ \mathrm{Let}\mathrm{the}{n}^{\mathrm{th}}\mathrm{term}\mathrm{be}\frac{512}{729}. \\ \therefore a{r}^{n-1}=\frac{512}{729} \\ \Rightarrow \left(18\right){\left(\frac{-2}{3}\right)}^{n-1}=\frac{512}{729} \\ \Rightarrow {\left(\frac{-2}{3}\right)}^{n-1}=\frac{512}{729}\times \frac{1}{18}=\frac{256}{6561} \\ \Rightarrow {\left(\frac{-2}{3}\right)}^{n-1}={\left(\frac{-2}{3}\right)}^8 \\ \Rightarrow n-1=8 \\ \Rightarrow n=9 \\ \mathrm{Thus},\mathrm{the}{9}^{\mathrm{th}}\mathrm{term}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{G}.\mathrm{P}.\mathrm{is}\frac{512}{729}. \end{gathered}$$