If the sum of the second, fourth and sixth terms of a G.P. of positive terms is 21 and the sum of its eighth, tenth and twelfth terms is 15309 , then…

$$\begin{aligned} & \mathrm{ar}+\mathrm{ar}^3+\mathrm{ar}^5=21, \quad \mathrm{ar}^7+\mathrm{ar}^9+\mathrm{ar}^{11}=15309 \\ & \Rightarrow \operatorname{ar}\left(1+\mathrm{r}^2+\mathrm{r}^4\right)=21, \quad \operatorname{ar}^7\left(1+\mathrm{r}^2+\mathrm{r}^4\right)=15309\end{aligned}$$​ar+ar3+ar5=21,ar7+ar9+ar11=15309⇒ar(1+r2+r4)=21,ar7(1+r2+r4)=15309​

$$\mathrm{Eq}^{\mathrm{n}}\left({\overline{2)} \div \mathrm{eq}^{\mathrm{n}}(1) \quad \quad \quad \text { (2) }}_{(1)}^{(2)}\right.$$Eqn(2)​÷eqn(1) (2) (1)(2)​

$$\begin{aligned} & \Rightarrow \frac{\mathrm{a} \cdot \mathrm{r}^7}{\mathrm{ar}}=\frac{15309}{21} \Rightarrow \mathrm{r}^6=729 \\ & \Rightarrow \frac{\mathrm{a} \cdot\left(\mathrm{r}^9-1\right)}{\mathrm{r}-1}=\frac{\frac{7}{91}(19683-1)}{2}=\frac{7 \times 19682}{91 \times 2} \\ & =\frac{9841}{13}=757\end{aligned}$$​⇒ara⋅r7​=2115309​⇒r6=729⇒r−1a⋅(r9−1)​=2917​(19683−1)​=91×27×19682​=139841​=757​