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


Question and Answers Forum

All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
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 byTawa11 last updated on 06/Jan/22

	$Great sir$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum