What is the sum of the first five terms of a geometric series with a1=6 and r=1/3?
1 answer:
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...
You might be interested in
Correct answer:
26 = -13x
X = -2
if you need a false answer, make up any number!
Your answer is 1/4, -7/3,-4
Answer:
The answer is 17.
Step-by-step explanation:
Answer:
idk
Step-by-step explanation:
idk
Substitute with y = 4 and z = -2
= 4^2*(-2)/4 + 10
= 16*(-2)/4 + 10
= -32/4 + 10
= -8 + 10
= 2
Hope This Helped! Good Luck!