Question

The 5th term of a g.p. is 1/3 and 9th term is 16/243 find its 4th term

Answer 1

$$\begin{gathered} \mathrm{Fifth}\mathrm{term},{\mathrm{ar}}^4=\frac{1}{3}...\left(1\right) \\ \mathrm{Ninth}\mathrm{term},{\mathrm{ar}}^8=\frac{16}{243}...\left(2\right) \\ \mathrm{Dividing}\left(2\right)\mathrm{by}\left(1\right),\mathrm{we}\mathrm{get} \\ \frac{{\mathrm{ar}}^8}{{\mathrm{ar}}^4}=\frac{{\displaystyle \frac{16}{243}}}{{\displaystyle \frac{1}{3}}} \\ {\mathrm{r}}^4=\frac{16}{81} \\ \mathrm{r}=\pm \frac{2}{3} \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{r}\mathrm{in}\left(1\right) \\ \mathrm{a}{\left(\pm \frac{2}{3}\right)}^4=\frac{1}{3} \\ \mathrm{a}\left(\frac{16}{81}\right)=\frac{1}{3} \\ \mathrm{a}=\frac{27}{16} \\ \mathrm{Fourth}\mathrm{term},{\mathrm{ar}}^3=\left(\frac{27}{16}\right){\left(\frac{2}{3}\right)}^3=\frac{1}{2}\left[\mathrm{on}\mathrm{taking}\mathrm{r}=\frac{2}{3}\right] \\ \mathrm{Fourth}\mathrm{term},{\mathrm{ar}}^3=\left(\frac{27}{16}\right){\left(-\frac{2}{3}\right)}^3=-\frac{1}{2}\left[\mathrm{on}\mathrm{taking}\mathrm{r}=-\frac{2}{3}\right] \end{gathered}$$