# 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

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### 2 Answers

It is given that (a+2b+c),(a-c) and (a-2b+c) are in continued proportion.

This gives (a + 2b + c)/(a - c) = (a - c)/(a - 2b + c)

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

=> (a + c)^2 - 4b^2 = (a - c)^2

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

=> 4ac = 4b^2

=> ac = b^2

=> b/a = c/b

**This proves that b is the mean proportion between a and c or b/a = c/b.**

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.**