Question

What is the sum of the first five terms of a geometric series with a1=6 and r=1/3?

Answer 1

The sum of a geometric sequence is:

s(n)=a(1-r^n)/(1-r), s(n)=nth sum, a=initial value, r=common ratio, n=term #

In this case we are told that a=6 and r=1/3 so:

s(n)=6(1-(1/3)^n)/(1-1/3)

s(n)=6(1-(1/3)^n)/(2/3)

s(n)=9(1-(1/3)^n), so the sum of the first 5 term is:

s(5)=9(1-(1/3)^5))

s(5)=9(1-(1/243))

s(5)=9(242/243)

s(5)=2178/243

s(5)=242/27

s(5)=8 26/27 ...if you wanted an approximation s(5)≈8.96 to the nearest hundredth...