If (a+2b+c),(a-c) and (a-2b+c) are in continued proportion, prove that b is the mean proportion between a and c

For b to be the mean proportion of a and c, the following
relation has to b verified:

b =
sqrt(a*c)

We'll replace b by (a-c) and the product a*c by
the product (a+2b+c)*(a-2b+c)

We notice that the product
(a+2b+c)*(a-2b+c) returns the difference of two
squares:

(a+2b+c)*(a-2b+c) = (a+c)^2 -
(2b)^2

The relation that has to be verified
is:

(a-c) = sqrt[(a+c)^2 -
(2b)^2]

We'll raise to square both
sides:

(a-c)^2 =(a+c)^2 -
(2b)^2

We'll expand the
binomials:

a^2 - 2ac + c^2 = a^2 +2ac + c^2 -
4b^2

We'll eliminate a^2 +
c^2:

- 2ac = 2ac - 4b^2

But
b^2 = ac

- 2ac = 2ac -
4ac

-2ac =
-2ac

Since the LHS = RHS, therefore b is the
mean proportion between a and c, if (a+2b+c),(a-c) and (a-2b+c) are in continuous
proportion.