Let

where we assume |r| < 1. Multiplying on both sides by r gives

and subtracting this from
gives

As n → ∞, the exponential term will converge to 0, and the partial sums
will converge to

Now, we're given


We must have |r| < 1 since both sums converge, so


Solving for r by substitution, we have


Recalling the difference of squares identity, we have

We've already confirmed r ≠ 1, so we can simplify this to

It follows that

and so the sum we want is

which doesn't appear to be either of the given answer choices. Are you sure there isn't a typo somewhere?
$25.
25/5=5
5x4=20
4/5 x 25 = 20
180, is the answer because when I had a question like that you have to add
Y=-3x+1 because in the points the equation will be y+2=-3(x-1) this converted into slope intercept form is y=-3(x-1)
Answer:
7
Step-by-step explanation:
because y is square by green and 7 is also green I guess