Question

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

Answer 1

58.4911 to the nearest ten thousandth.

Explanation:

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

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

= 39 * 1.49977138

= 58.49108382.