Question

Find which term in the geometric sequence 1,3,9,27,... is the first to exceed 7,000.

Answer 1

The common ratio between terms is 3, so the sequence has general $n$-th term

$a_n=3^{n-1}$

for $n\ge1$. The term exceeds 7000 when

$3^{n-1}>7000\implies n-1>\log_37000\implies n>1+\log_37000\approx9.06$

which means the first time $a_n$ exceeds 7000 occurs when $n=10$. Indeed,

$a_{10}=3^{10-1}=19,683$

while the previous term would have been

$a_9=3^{9-1}=6561$