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}\times2^{6-1} \\ 6th\text{ term}=11\times2^5 \\ 6th\text{ term}=11\times32 \\ 6th\text{ term}=352\text{ } \end{gathered}$