Question

Whats the sixth term of the geometric sequence for which a1=11 and r=2

Answer 1

`\begin{gathered} a_1=first\text{ term=11} \\ r=\text{common ratio=}2 \end{gathered}`

Therefore,

`n_{th\text{ }}term=ar^(n-1)`
`\begin{gathered} 6th\text{ term=11}*2^(6-1) \\ 6th\text{ term}=11*2^5 \\ 6th\text{ term}=11*32 \\ 6th\text{ term}=352\text{ } \end{gathered}`