Same thing as last time. Pull out 15 from both - a1 = 1, a3 = 9.
Ask: What can I multiply 1 by twice to get 9?
\(1 * x * x = 9\)
\(9 = x^2, x = 3\)
Okay so if a geometric equation is of form:
\(y=ab^x\)
Then a = 15 (what we pulled out to make the problem easier to solve) and b = 3
\(y = 15(3^x)\)
+36915 a1 * r * r = a3
15 r^2 = 135
r^2 = 9
r = 3
an = a1 (3n-1)
an = 15 * 3n-1
