lollyypop890 lollyypop890

09-04-2020
Mathematics

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Find the sum of the geometric series for which a1= 39, r= 1/3 , n= 8 to the nearest ten-thousandth

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PollyP52 PollyP52

09-04-2020

Answer:

58.4911 to the nearest ten thousandth.

Step-by-step explanation:

Sn = a1 * (1 - r^n) / (1 - r)

S8 = 39 * (1 - (1/3)^8) / ( 1 - 1/3)

= 39 * 1.49977138

= 58.49108382.

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