SOLUTION: find the sum of the infinite geometric series 64 + 48 + 36 + 27 + ... if there is a sum

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Question 287878: find the sum of the infinite geometric series 64 + 48 + 36 + 27 + ... if there is a sum
Answer by CharlesG2(834) About Me  (Show Source):
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find the sum of the infinite geometric series 64 + 48 + 36 + 27 + ... if there is a sum
64 * 3/4 = 48
48 * 3/4 = 36
36 * 3/4 = 27
27 * 3/4 = 20.25
64 * (3/4)^0 = 64 * 1 = 64
64 * (3/4)^1 = 64 * 3/4 = 48
64 * (3/4)^2 = 64 * 9/16 = 36
64 * (3/4)^3 = 64 * 27/64 = 27
64 * (3/4)^4 = 64 * 81/256 = 20.25
+64+%2A+sum%28%283%2F4%29%5En%2C0%2Cinfinity%29+
S(sub infinity) = a(sub 1)/(1 - r) (|r|<1)
r = 3/4, a(sub 1) = 64
S(sub infinity) = 64/(1 - 3/4) = 64/(1/4) = 256
sum is 256