Question Number 194878 by oMwarimu last updated on 18/Jul/23
Answered by Rasheed.Sindhi last updated on 18/Jul/23
$${a}+{d}={ar}^{\mathrm{3}} \:\:\wedge\:\:{a}+\mathrm{9}{d}={ar}^{\mathrm{6}} \\ $$$${a}\left({r}^{\mathrm{3}} −\mathrm{1}\right)={d}\:\wedge\:{a}\left({r}^{\mathrm{6}} −\mathrm{1}\right)=\mathrm{9}{d} \\ $$$${a}=\frac{{d}}{{r}^{\mathrm{3}} −\mathrm{1}}\:\wedge\:{a}=\frac{\mathrm{9}{d}}{{r}^{\mathrm{6}} −\mathrm{1}} \\ $$$$\frac{{d}}{{r}^{\mathrm{3}} −\mathrm{1}}=\frac{\mathrm{9}{d}}{{r}^{\mathrm{6}} −\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{{r}^{\mathrm{3}} −\mathrm{1}}=\frac{\mathrm{9}}{{r}^{\mathrm{6}} −\mathrm{1}} \\ $$$${r}^{\mathrm{6}} −\mathrm{1}=\mathrm{9}{r}^{\mathrm{3}} −\mathrm{9} \\ $$$${r}^{\mathrm{6}} −\mathrm{9}{r}^{\mathrm{3}} +\mathrm{8}=\mathrm{0} \\ $$$$\left({r}^{\mathrm{3}} −\mathrm{1}\right)\left({r}^{\mathrm{3}} −\mathrm{8}\right)=\mathrm{0} \\ $$$${r}^{\mathrm{3}} =\mathrm{1},\mathrm{8}\Rightarrow{r}=\mathrm{1},\mathrm{2} \\ $$
Answered by Rasheed.Sindhi last updated on 18/Jul/23
$${a}+{d}={ar}^{\mathrm{3}} \:\wedge\:{a}+\mathrm{9}{d}={ar}^{\mathrm{6}} \\ $$$$\begin{cases}{\mathrm{9}{a}+\mathrm{9}{d}=\mathrm{9}{ar}^{\mathrm{3}} }\\{{a}+\mathrm{9}{d}={ar}^{\mathrm{6}} }\end{cases}\Rightarrow\mathrm{8}{a}=\mathrm{9}{ar}^{\mathrm{3}} −{ar}^{\mathrm{6}} \\ $$$$\Rightarrow\mathrm{8}=\mathrm{9}{r}^{\mathrm{3}} −{r}^{\mathrm{6}} \Rightarrow{r}^{\mathrm{6}} −\mathrm{9}{r}^{\mathrm{3}} +\mathrm{8}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\left({r}^{\mathrm{3}} −\mathrm{1}\right)\left({r}^{\mathrm{3}} −\mathrm{8}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:{r}^{\mathrm{3}} =\mathrm{1},\mathrm{8}\Rightarrow{r}=\mathrm{1},\mathrm{2} \\ $$