SOLUTION: What is Sn of the geometric series with a1 = �81, r = �2/3, and n = 5? THANKS FOR THE HELP :)

Question 475149: What is Sn of the geometric series with a1 = �81, r = �2/3, and n = 5?
THANKS FOR THE HELP :)
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
Sn of a geometric series is given by the equation:
Sn = a * (1-r^n) / (1-r)
a if the first term in the sequence.
r is the common ratio
To see how this work, assume your sequence is:
2,4,8,16
The sum of this series is 30.
Using the formula, we have:
a = 2
n = 4
r = 2
Sn = a * (1-r^n) / (1-r) which becomes:
Sn = 2 * (1-2^4) / (1-2) which becomes:
Sn = 2 * (1-16)/(1-2) which becomes:
Sn = 2 * (-15 / -1) which becomes:
Sn = -30 / -1 which becomes:
Sn = 30.
The formula works and can be applied to your problem.
Your problem states:
a = �81
r = �2/3
n = 5
The formula is:
Sn = a * (1-r^n) / (1-r)
Substituting with your values in the equation gets:
Sn = -81 * (1-(-2/3)^5) / (1-(-2/3))
This becomes:
Sn = (-81 * (1 - (-32/243))) / (1 + (2/3)) which becomes:
Sn = (-81 * (1 + (32/243))) / (1 + (2/3)) which becomes:
Sn = (-81 * (275/243)) / (5/3) which becomes:
Sn = -81 * (275/243) * (3/5).
275 can be divided by 5 and 243 can be divided by 3, so your equation becomes:
Sn = -81 * (55/81) which then becomes:
Sn = -55.
The sum of the geometric series is equal to -55.
We can check this out to see if it's true because the numbers are small enough.
We start with -81 and then multiply it by the ratio of (-2/3) for 4 times.
We get:
-81
54
-36
24
-16
The sum of these 5 terms is equal to -55.
The formula works and you have your solution.

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