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Enter the finite geometric series from its given description, and then enter its sum.

A geometric series that starts with 4, ends with -12,500, and has a common ratio of -5.
The geometric series is _____
The sum of the geometric series is_____

User Avimimoun
by
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1 Answer

9 votes

Answer:


a_n= 4(-5)^((n-1))

-10416

Explanation:

a = 4

r = -5

geometric series:
a_n=ar^((n-1)) = 4(-5)^((n-1))

Determine which term = -12500:


a_n=-12500


4(-5)^((n-1))=-12500


(-5)^((n-1)) = -3125


(5)^((n-1)) = 3125


(n-1)ln(5) = ln(3125)


n-1 = ln(3125)/ln(5) = 5


n = 5 + 1 = 6


a_6=-12500

Using geometric sum series formula:
S_n=(a(1-r^n))/(1-r)

Therefore,
S_6=(4(1-(-5)^6))/(1-(-5)) =(-62496)/(6) =-10416

User Wayne Molina
by
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