an infinite geometric series has all positive terms if first term=64 and 3rd term=1 determine sum of series?

Question

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Answer 1

The general formula for the sum of a geometric series is a/(1-r), where a is the first term and r = first term/second term. In this we are given a = 64 and ar^2 = 1, so r = 1/8. Plugging these into the formula gives us:

64/(1-1/8) = 512/7 which is about 73.14

Answer 2

for any geometric series nth term is written as a* r (n-1), here a=64 and 3rd term is 1 so, 1= 64 * r^(3-1)
1=64* r^2
so r^2=1/64
so r=1/8

here r<1
so sum of infinite series is = a / (1-r)= 64/(1-1/8)= 64 /(7/8)= 512/7

Answer 3

S = 64 + 8 + 1 + 1/8 + 1/64 + ...
S/8 = 8 + 1 + 1/8 + 1/64 + ...
S - S/8 = 64
S(1 - 1/8) = 64
S(7/8) = 64
S = 64 (8/7) = 512 / 7 Answer

Answer 4

A geometric series can be expressed as

a ar ar² ar³...

ar²/a = 1/64
r² = 1/64
r = 1/8

The sum is

S = 64 + 8 + 1 + 1/8...
(1 - 1/8)S = 64
S = 64/(7/8) = 512/7

Answer 5

64 * r^2 = 1 so r=1/8

Sum = 64 +8 + 1/(1-1/8) = 72+8/7 = 512/7