If as many numbers as we please be in continued proportion, and there be subtracted from the second and the last numbers equal to the first, then, as the excess of the second is to the first, so will the excess of the last be to all those before it.
Proof
For a,ar,ar2,…,arn: the excess of the last over the
first is a(rn−1), and the sum of the rest is a(1+r+r2+⋯+rn−1)=a(rn−1)/(r−1); the ratio of excess to
sum equals (r−1)=(ar−a)/a, the excess of the second over
the first divided by the first.