Question

In a GP the 3^rd term is 24 and the 6^th term is 192. Find the 10^th term.

Answer 1

$$\begin{gathered} \mathrm{Let}a\mathrm{be}\mathrm{the}\mathrm{first}\mathrm{term}\mathrm{and}r\mathrm{be}\mathrm{the}\mathrm{common}\mathrm{ratio}. \\ \therefore a_3=24\mathrm{and}\mathit{}{a}_6=192 \\ \mathit{\Rightarrow }a{r}^2=24\mathrm{and}a{r}^5=192 \\ \Rightarrow \frac{a{r}^5}{a{r}^2}=\frac{192}{24} \\ \Rightarrow r^3=8 \\ \Rightarrow r^3=2^3 \\ \Rightarrow r=2 \\ \mathrm{Putting}\mathit{}r=2\mathrm{in}\mathit{}a{r}^2=24 \\ a{\left(2\right)}^2=24 \\ \Rightarrow a\mathit{}=6 \\ \mathrm{Now},{10}^{\mathrm{th}}\mathrm{term}=a_{10}=a{r}^9 \\ \mathrm{Putting}a=6\mathrm{and}r=2\mathrm{in}{a}_{10}=a{r}^9 \\ \Rightarrow a_{10}=\left(6\right){\left(2\right)}^9=3072 \\ \mathrm{Thus},\mathrm{the}{10}^{\mathrm{th}}\mathrm{term}\mathrm{of}\mathrm{the}\mathrm{G}.\mathrm{P}.\mathrm{is}3072. \end{gathered}$$