Find the sum of the infinite geometric series

1/5^4, 1/5^7, 1/5^10, 1/5^13.....

It is a GP with a=1/5^4 and r =1/5^3
Sum = a/(1-r) = 1/5^4 * 1/(1-1/5^3) = 1/(5^4-5) = 1/620

the common ratio r=(1/5)^3=1/125
and the first term a=1/5^4=1/625
So we plug into formula:
S=a/(1-r)=(1/625)/(1-1/125)=(1/625)/(1…
Multiply by 625/625
=1/(124*625/125)=1/(124*5)=1/620