Asked by shantel
The 3rd term of a geometric series is 18 , and the 5the term is 162 .Determine the sum of the of the first 7 terms ,where r < 0
Answers
Answered by
Steve
162 = 18r^2
r = 3
18 = 9a, so a = 2
S7 = 2(1-3^7)/(1-3) = 2186
check:
2+6+18+54+162+486+1458 = 2186
r = 3
18 = 9a, so a = 2
S7 = 2(1-3^7)/(1-3) = 2186
check:
2+6+18+54+162+486+1458 = 2186
Answered by
Mamosa
Thanks for helping me
Answered by
Emmanuel
Yes
Answered by
Petit Coke
Everything is right, but
r = -3
not r = 3
because the problem stated that r < 0
r = -3
not r = 3
because the problem stated that r < 0
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