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17 votes
17 votes
Find the sum of the first 40 terms of a geometric sequence where the first term is 16 and the common ratio is 1.1

User Maycca
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1 Answer

20 votes
20 votes

Given:


a=16\text{ ; }r=1.1;S_{n\text{ }}=\text{?}
S_n=(a(r^n-1))/(r-1)
S_n=(16(1.1^(40)-1))/(1.1-1)
S_n=(16(45.26-1))/(0.1)
S_n=(16(44.26))/(0.1)
S_n=(708.16)/(0.1)
S_n=7081.6

User Leenah
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