Geometric sequences, please help! 10 points

first term = 54
4th term = 2

I think the common ratio is a third?
and the sum to infinity is 81?

This is the bit I'm not really sure about:

After how many terms is the sum of the series greater than 99% of the sum to infinity?

Please show working! thanks

a=54
ar^3=2

divide the second equation on the first one

r^3=2/54
r^3=1/27
r=1/3

sum to infinity = a/1-r
=54/1-1/3
sum to infinity = 54/2/3 = 54*3/2 = 81

both are correct

sum = 99% sum to infinity
a(r^n -1)/r-1 = 99/100 * 81
54(1/3^n -1)/-2/3 = 99/100 *81
-81 (1/3^n -1) = 99/100*81
1/3^n -1 = -99/100
1/3^n = 1/100
n log 1/3 = log 1/100
n= log 1/100 / log 1/3
n= -2/log 1/3
n= 4.19
therefore starting the fifth term the sum of the series would be greater than 99% of the sum to infinity.

Hope This Helps :D

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Un=n-th term
r=ratio
Sn=sum to Un

geometric:
Un=U1*r^(n-1)
U1=54.
U4=U1*r^3
2=54*r^3
r^3=2/54=1/27
r=cubic root (1/27)
r=1/3

Sum to infinity=U1/(1-r)
=54/(1-1/3)
=81

99%*81=80.19
Sn=U1*(1-r^n)/(1-r)
80.19=54*(1-(1/3)^n)/(1-(1/3))
1-(1/3)^n=0.99
(1/3)^n=0.01
n=log 0.01 (base 1/3)=4

so,the sum to 5th term is greater than 99% of the sum to infinity.

a * r = 54
a * r^4 = 2

(a * r^4) / (a * r) = 2 / 54
r^3 = 1/27
r = (1/3)


Now let's find the sum to infinity

S = ar + ar^2 + ar^3 + ... + ar^n
S * r = ar^2 + ar^3 + ar^4 + ... + ar^n + ar * r^n

Sr - S = ar * r^n - ar
S * (r - 1) = ar * (r^n - 1)
S = ar * (r^n - 1) / (r - 1)

We know that ar = 54, r = 1/3, and (1/3)^inf = 0

S = 54 * (0 - 1) / (1/3 - 1)
S = -54 / (-2/3)
S = 54 * (3/2)