An arithmetic series has a first term of -4 and a common difference of 1. A geometric series has a first term of 8 and a common ratio of 0.5. After how many terms does the sum of the arithmetic series exceed the sum of the geometric series?
4 answers
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can anyone help?
for the AS
Sn = n/2(-8 + n-1)
= n/2(n-9)
for a GS
Sn = 8(1-.5^n)/(1-.5)
so n/2(n-9) > 8(1-.5^n)/(1-.5)
n/2(n-9) > 16(1-.5^n)
n^2 - 9n > 32 - 32(.5^n)
32(.5)^n + n^2 - 9n - 32 > 0
set it equal to zero and attempt to solve it.
This would be a very nasty equation to solve, except you know that n has to be a whole number, so do some trial and error calculations.
eg. n = 2 we get -38 > 0 which is false
if n = 5 we get -44 > 0 which is false
if n = 20 we get 188 > 0 which is true
So somewhere between n=5 and n=20 there should be a solution
(n=12 ---> 4.0078 > 0 mmmmhhh?)
Sn = n/2(-8 + n-1)
= n/2(n-9)
for a GS
Sn = 8(1-.5^n)/(1-.5)
so n/2(n-9) > 8(1-.5^n)/(1-.5)
n/2(n-9) > 16(1-.5^n)
n^2 - 9n > 32 - 32(.5^n)
32(.5)^n + n^2 - 9n - 32 > 0
set it equal to zero and attempt to solve it.
This would be a very nasty equation to solve, except you know that n has to be a whole number, so do some trial and error calculations.
eg. n = 2 we get -38 > 0 which is false
if n = 5 we get -44 > 0 which is false
if n = 20 we get 188 > 0 which is true
So somewhere between n=5 and n=20 there should be a solution
(n=12 ---> 4.0078 > 0 mmmmhhh?)
1 answer