Menu Close

Given-that-a-n-is-a-geometric-sequence-where-the-first-term-a-1-gt-1-and-the-common-ratio-r-gt-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




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 by Tawa11 last updated on 06/Jan/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *