In an infinite geometric series, the third term is 108 and the sixth term is -4. Determine the sum of this series

Question

Hi there, would love for any help on this topic. Thanks in advance!

Answer 1

a₃ = a₁r² = 108

 

a₆ = a₁r⁵ = -4

 

So, (108/r²)r⁵ = -4   r³ = -4/108 = -1/27

 

                              r = -1/3

 

                             a₁ = 108/r² = 972

 

Since r lies in the interval (-1,1), the geometric series has a finite sum.

 

S = sum of the series = a₁/(1-r) = 972/(4/3) = 729

 

Answer 2

a₆ = a₃(r³)

-4 = 108(r³)

-0.0373737 = r³

 

Please check your nunbers.