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

Given that {a_n } is a geometric sequence where the first term, a_1 >1 and the common ratio, r>0. If b_n =log_2 a_n where n∈N, b_1 +b_3 +b_5 =6, and b_1 ∙b_3 ∙b_5 =0, find the general term of {a_n }.

$$\mathrm{Given}\:\mathrm{that}\:\left\{{a}_{{n}} \right\}\:\mathrm{is}\:\mathrm{a}\:\mathrm{geometric}\:\mathrm{sequence} \\ $$ $$\mathrm{where}\:\mathrm{the}\:\mathrm{first}\:\mathrm{term},\:{a}_{\mathrm{1}} >\mathrm{1}\:\mathrm{and}\:\mathrm{the}\:\mathrm{common} \\ $$ $$\mathrm{ratio},\:{r}>\mathrm{0}.\: \\ $$ $$\mathrm{If}\:{b}_{{n}} =\mathrm{log}_{\mathrm{2}} \:{a}_{{n}} \:\mathrm{where}\:{n}\in\mathbb{N},\:{b}_{\mathrm{1}} +{b}_{\mathrm{3}} +{b}_{\mathrm{5}} =\mathrm{6}, \\ $$ $$\mathrm{and}\:{b}_{\mathrm{1}} \centerdot{b}_{\mathrm{3}} \centerdot{b}_{\mathrm{5}} =\mathrm{0},\:\mathrm{find}\:\mathrm{the}\:\mathrm{general}\:\mathrm{term}\:\mathrm{of}\:\left\{{a}_{{n}} \right\}. \\ $$

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 (a_1 r^(n−1) )=log_2 a_1 +(n−1)log_2 r let α=log_2 a_1 ≠0, β=log_2 r ⇒b_n =α+(n−1)β b_1 +b_3 +b_5 =3α+(0+2+4)β=6 ⇒α+2β=2 ...(i) b_1 b_3 b_5 =α(α+2β)(α+4β)=0 α(α+2β)(α+4β)=0 ⇒ α+4β=0 ...(ii) from(i) and (ii) we get α=4, β=−1 log_2 a_1 =4 ⇒a_1 =2^4 =16 log_2 r=−1 ⇒r=2^(−1) =(1/2) ⇒a_n =2^4 ×((1/2))^(n−1) =(1/2^(n−5) )

$${a}_{{n}} ={a}_{\mathrm{1}} {r}^{{n}−\mathrm{1}} \\ $$ $${b}_{{n}} =\mathrm{log}_{\mathrm{2}} \:{a}_{{n}} =\mathrm{log}_{\mathrm{2}} \:\left({a}_{\mathrm{1}} {r}^{{n}−\mathrm{1}} \right)=\mathrm{log}_{\mathrm{2}} \:{a}_{\mathrm{1}} +\left({n}−\mathrm{1}\right)\mathrm{log}_{\mathrm{2}} \:{r} \\ $$ $${let}\:\alpha=\mathrm{log}_{\mathrm{2}} \:{a}_{\mathrm{1}} \neq\mathrm{0},\:\beta=\mathrm{log}_{\mathrm{2}} \:{r} \\ $$ $$\Rightarrow{b}_{{n}} =\alpha+\left({n}−\mathrm{1}\right)\beta \\ $$ $${b}_{\mathrm{1}} +{b}_{\mathrm{3}} +{b}_{\mathrm{5}} =\mathrm{3}\alpha+\left(\mathrm{0}+\mathrm{2}+\mathrm{4}\right)\beta=\mathrm{6} \\ $$ $$\Rightarrow\alpha+\mathrm{2}\beta=\mathrm{2}\:\:\:...\left({i}\right) \\ $$ $${b}_{\mathrm{1}} {b}_{\mathrm{3}} {b}_{\mathrm{5}} =\alpha\left(\alpha+\mathrm{2}\beta\right)\left(\alpha+\mathrm{4}\beta\right)=\mathrm{0} \\ $$ $$\alpha\left(\alpha+\mathrm{2}\beta\right)\left(\alpha+\mathrm{4}\beta\right)=\mathrm{0} \\ $$ $$\Rightarrow\:\alpha+\mathrm{4}\beta=\mathrm{0}\:...\left({ii}\right) \\ $$ $${from}\left({i}\right)\:{and}\:\left({ii}\right)\:{we}\:{get} \\ $$ $$\alpha=\mathrm{4},\:\beta=−\mathrm{1} \\ $$ $$\mathrm{log}_{\mathrm{2}} \:{a}_{\mathrm{1}} =\mathrm{4}\:\Rightarrow{a}_{\mathrm{1}} =\mathrm{2}^{\mathrm{4}} =\mathrm{16} \\ $$ $$\mathrm{log}_{\mathrm{2}} \:{r}=−\mathrm{1}\:\Rightarrow{r}=\mathrm{2}^{−\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$ $$\Rightarrow{a}_{{n}} =\mathrm{2}^{\mathrm{4}} ×\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}−\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{5}} } \\ $$

Commented byTawa11 last updated on 06/Jan/22

Great sir

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

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