Question

Find the 9th term of the geometric sequence 8,32,128,

Answer 1

$\text{First term,}~ a = 8\\\\\text{Common ratio,}~ r= \dfrac{32}8 = 4\\\\\text{nth term} = ar^{n-1} \\\\\text{9th term} = 8\cdot 4^{9-1}\\\\\\~~~~~~~~~~~~~=8 \cdot 4^8\\\\\\~~~~~~~~~~~~~=524288\\\\\text{The 9th of the geometric sequence is 525288.}$