Question Number 207395 by hardmath last updated on 13/May/24
$$\mathrm{Geometric}\:\mathrm{series}: \\ $$$$\frac{\mathrm{b}_{\mathrm{4}} \:\centerdot\:\mathrm{b}_{\mathrm{7}} \:\centerdot\:\mathrm{b}_{\mathrm{10}} }{\mathrm{b}_{\mathrm{1}} \:\centerdot\:\mathrm{b}_{\mathrm{3}} \:\centerdot\:\mathrm{b}_{\mathrm{5}} }\:\:=\:\:\mathrm{2}^{\mathrm{12}} \:\:\:\:\:\mathrm{find}:\:\:\:\frac{\mathrm{b}_{\mathrm{5}} }{\mathrm{b}_{\mathrm{2}} }\:\:=\:\:? \\ $$
Answered by efronzo1 last updated on 13/May/24
$$\:\Leftrightarrow\:\frac{\mathrm{r}^{\mathrm{3}+\mathrm{6}+\mathrm{9}} }{\mathrm{r}^{\mathrm{2}+\mathrm{4}} }\:=\:\mathrm{2}^{\mathrm{12}} \\ $$$$\:\Leftrightarrow\mathrm{r}^{\mathrm{18}−\mathrm{6}} =\:\mathrm{r}^{\mathrm{12}} =\:\mathrm{2}^{\mathrm{12}} \Rightarrow\begin{cases}{\mathrm{r}=\mathrm{2}}\\{\mathrm{r}=−\mathrm{2}}\end{cases} \\ $$$$\:\Leftrightarrow\frac{\mathrm{r}^{\mathrm{4}} }{\mathrm{r}}\:=\:\mathrm{r}^{\mathrm{3}} =\:\begin{cases}{−\mathrm{8}}\\{\mathrm{8}}\end{cases} \\ $$
Commented by hardmath last updated on 13/May/24
$$\mathrm{answer}:\:−\mathrm{1}\:? \\ $$
Commented by MM42 last updated on 13/May/24
$${notice} \\ $$$${geometric}\:{series}\::\:{b}_{{n}} =\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{ar}^{{i}} \\ $$$${geometric}\:{sequence}\::\:{b}_{{n}} ={ar}^{{n}−\mathrm{1}} \:\: \\ $$$$ \\ $$