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Grace [21]
3 years ago
14

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

Mathematics
1 answer:
Zielflug [23.3K]3 years ago
8 0

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

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