I need help with question 8 this is non graded

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Bonner asked Dec 11, 2023

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I need help with question 8 this is non graded

Mathematics
college

Bonner asked Dec 11, 2023

by Bonner

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Given

Step-by-step explanation

From the above,

$\begin{gathered} a_6=4a_(6-1)\Rightarrow4a_5 \\ From\text{ the definition of a common ratio} \\ r=(a_6)/(a_5)=4 \\ Also; \\ a_6=ar^5 \\ 8192=a4^5 \\ a=(8192)/(4^5)=(8192)/(1024)=8 \\ \end{gathered}$

Therefore, we can have the first four terms as

$\begin{gathered} a=8 \\ ar=8*4=32 \\ ar^2=8*4^2=128 \\ ar^3=8*4^3=512 \end{gathered}$

Answer: Option C

Frederik Deweerdt answered Dec 17, 2023

by Frederik Deweerdt

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