How do you find #S_n# for the geometric series #a_1=12#, #a_5=972#, r=-3?

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How do you find #S_n# for the geometric series #a_1=12#, #a_5=972#, r=-3?

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Answer 1 · 2017-08-30T12:02:33

#S_n = (12(1-(-3)^n))/4#

Explanation:

The sum to #n# of a geometric series, denoted as #S_n#, is given by the formula #S_n = (a(1-r^n))/(1-r# where #a# is the first term of the series and #r# is the common ratio.

#S_n = (12(1-(-3)^n))/(1--3 )= (12(1-(-3)^n))/4#

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How do you find #S_n# for the geometric series #a_1=12#, #a_5=972#, r=-3?

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