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


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Question Number 163324 by ZiYangLee last updated on 06/Jan/22
$Given that {a_n } is a geometric sequence where the first term, a_1 >1 and the common ratio, r>0. If b_n =log_2 a_n where n∈N, b_1 +b_3 +b_5 =6, and b_1 ∙b_3 ∙b_5 =0, find the general term of {a_n }.$
$Given that {a_{n}} is a geometric sequence$ $where the first term, a_{1} > 1 and the common$ $ratio, r > 0 .$ $If b_{n} = \log_{2} a_{n} where n \in N, b_{1} + b_{3} + b_{5} = 6,$ $and b_{1} \cdot b_{3} \cdot b_{5} = 0, find the general term of {a_{n}} .$
	Answered by mr W last updated on 06/Jan/22

	$a_{n} = a_{1} r^{n - 1}$ $b_{n} = \log_{2} a_{n} = \log_{2} (a_{1} r^{n - 1}) = \log_{2} a_{1} + (n - 1) \log_{2} r$ $l e t α = \log_{2} a_{1} \neq 0, β = \log_{2} r$ $\Rightarrow b_{n} = α + (n - 1) β$ $b_{1} + b_{3} + b_{5} = 3 α + (0 + 2 + 4) β = 6$ $\Rightarrow α + 2 β = 2 . . . (i)$ $b_{1} b_{3} b_{5} = α (α + 2 β) (α + 4 β) = 0$ $α (α + 2 β) (α + 4 β) = 0$ $\Rightarrow α + 4 β = 0 . . . (i i)$ $f r o m (i) a n d (i i) w e g e t$ $α = 4, β = - 1$ $\log_{2} a_{1} = 4 \Rightarrow a_{1} = 2^{4} = 16$ $\log_{2} r = - 1 \Rightarrow r = 2^{- 1} = \frac{1}{2}$ $\Rightarrow a_{n} = 2^{4} \times {(\frac{1}{2})}^{n - 1} = \frac{1}{2^{n - 5}}$
	Commented by Tawa11 last updated on 06/Jan/22

	$Great sir$
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