Ageometric 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)

The answer would be choice b

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