The first two terms of a geometric progression are 27 and 9. How many terms are added up to get a sum of 364/9?

We are trying to solve

27*∑ (1/3)^(n-1)   from 1 to n =  364/9

WolframAlpha shows that 6 terms are needed

Here are the sums from n = 1 to n = 6

27 + 9 + 3 + 1 + 1/3 + 1/9 = 364/9

a=27     r=1/3

27+9+3+1+(1/3)+(1/9)=364/9

$$\boxed{S_n=\frac{a(1-r^n)}{1-r}}\\\\\\
\frac{364}{9}=\frac{27(1-(1/3)^n)}{1-(1/3)}}\\\\
\frac{364}{9*27}=\frac{1-(1/3)^n}{2/3}}\\\\
\frac{364*2}{9*27*3}=1-(1/3)^n\\\\
\frac{728}{729}-1=-(1/3)^n\\\\
\frac{-1}{729}=-\frac{1}{3^n}\\\\
729=3^n\\\\
log_3(729)=log_3 3^n\\\\
n=log_3(729)\\\\
n=log729/log3\\\\
n=6$$

Thank you Chris for helping me spot my previous error