A geometric sequence has a first term of x-y and a common ratio of x+y. what if the third term of the sequence?

First term t (1) = (x-y)

common ratio = (x+y) (don't omit the parentheses)

then the nth term:

t (n) = (x-y) (x+y) ^ (n-1)

check:

t (1) = (x-y) (x+y) ^ (1-1) = (x-y) (x+y) ^0 = (x-y) (1) = (x-y)

t (3) = (x-y) (x+y) ^ (3-1) = (x-y) (x+y) ^2

When n is small, t (n) can also be found by repeated multiplication of the common ratio (x+y)