Asked by Naledi obotseng
The first term of the geometric sequance is 27,the last term is 8 and the sum of the series is 65.how many terms are there in the series?
Answers
Answered by
Steve
T1 = a = 27
Tn = ar^(n-1) = 8
Sn = 65 = a*(1-r^n)/(1-r)
27r^(n-1) = 8
r^(n-1) = 8/27
Just as a guess, I'd say r=2/3 and n=4, since 8/27 = (2/3)^3
However, let's go on.
27(1-r^n)/(1-r) = 65
27(1-r*8/27) / (1-r) = 65
27(1-8r/27) = 65 - 65r
27 - 8r = 65 - 65r
57r = 38
r = 38/57 = 2/3
So, now we have
27*(2/3)^(n-1) = 8
(2/3)^(n-1) = 8/27 = (2/3)^3
n-1 = 3
n=4
Sequence:
27 18 12 8
Sum: 65
Tn = ar^(n-1) = 8
Sn = 65 = a*(1-r^n)/(1-r)
27r^(n-1) = 8
r^(n-1) = 8/27
Just as a guess, I'd say r=2/3 and n=4, since 8/27 = (2/3)^3
However, let's go on.
27(1-r^n)/(1-r) = 65
27(1-r*8/27) / (1-r) = 65
27(1-8r/27) = 65 - 65r
27 - 8r = 65 - 65r
57r = 38
r = 38/57 = 2/3
So, now we have
27*(2/3)^(n-1) = 8
(2/3)^(n-1) = 8/27 = (2/3)^3
n-1 = 3
n=4
Sequence:
27 18 12 8
Sum: 65
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