find a1 for given geometric series.

Sn= 405, r=3, n=4

I don't know how to start the problem

use your formula for S(n)

S(n) = a(r^n - 1)/(r-1)

405 = a(3^4 - 1)/(3-1)
405 = a(81 - 1)/2
405 = 40a
a = 405/40 = 81/8