SOLUTION: Find the sum of the geometric series for which a1 = 125, r = ⅖, and n = 4.

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Question 1041852: Find the sum of the geometric series for which a1 = 125, r = ⅖, and n
= 4.
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
It is a1[(1-r^n)/(1-r)]
125*(1-(2/5)^4)/3/5;
first deal with 1-(2/5)^4=625/625-16/625=609/625
The numerator is 125*609/625, with 125 being 1/5 of 625, so the numerator is 609/5
That is divided by3/5, so invert and multiply: 609/5*5/3=203
The terms are:
125
50
20
8
They add to 203. Check