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Question Number 163324 by ZiYangLee last updated on 06/Jan/22

$$\mathrm{Given}\mathrm{that}\left\{a_n\right\}\mathrm{is}\mathrm{a}\mathrm{geometric}\mathrm{sequence}$$$$\mathrm{where}\mathrm{the}\mathrm{first}\mathrm{term},a_1>1\mathrm{and}\mathrm{the}\mathrm{common}$$$$\mathrm{ratio},r>0.$$$$\mathrm{If}{b}_n={\log }_2{a}_n\mathrm{where}n\in \mathbb{N},b_1+b_3+b_5=6,$$$$\mathrm{and}{b}_1\cdot b_3\cdot b_5=0,\mathrm{find}\mathrm{the}\mathrm{general}\mathrm{term}\mathrm{of}\left\{a_n\right\}.$$

Answered by mr W last updated on 06/Jan/22

$$a_n=a_1{r}^{n-1}$$$$b_n={\log }_2{a}_n={\log }_2\left(a_1{r}^{n-1}\right)={\log }_2{a}_1+(n-1){\log }_{2}r$$$$let\alpha ={\log }_2{a}_1\ne 0,\beta ={\log }_{2}r$$$$\Rightarrow b_n=\alpha +(n-1)\beta$$$$b_1+b_3+b_5=3\alpha +(0+2+4)\beta =6$$$$\Rightarrow \alpha +2\beta =2\dots \left(i\right)$$$$b_1{b}_3{b}_5=\alpha (\alpha +2\beta )(\alpha +4\beta )=0$$$$\alpha (\alpha +2\beta )(\alpha +4\beta )=0$$$$\Rightarrow \alpha +4\beta =0\dots \left(ii\right)$$$$from\left(i\right)and\left(ii\right)weget$$$$\alpha =4,\beta =-1$$$${\log }_2{a}_1=4\Rightarrow a_1=2^4=16$$$${\log }_{2}r=-1\Rightarrow r=2^{-1}=\frac{1}{2}$$$$\Rightarrow a_n=2^4\times {\left(\frac{1}{2}\right)}^{n-1}=\frac{1}{2^{n-5}}$$

Commented by Tawa11 last updated on 06/Jan/22

$$\mathrm{Great}\mathrm{sir}$$

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