# What is x if the numbers 2*square root(x-1), 3+square root(2x-6), 2*square root (x+4) are the terms of an arithmetical progression?

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### 1 Answer

If the given numbers are the consecutive terms of an arithmetical progression, we could use the arithmetical mean theorem:

3 + sqrt(2x-6) = [2sqrt(x-1)+2sqrt(x+4)]/2

3 + sqrt(2x-6) = sqrt(x-1) + sqrt(x+4)

We'll raise to square both sides:

9 + 6sqrt(2x-6) + 2x - 6 = x - 1 + 2sqrt[(x-1)(x+4)] + x + 4

We'll combine like terms both sides:

3 + 2x + 6sqrt(2x-6) = 3 + 2x + 2sqrt(x^2 + 3x - 4)

We'll eliminate 3 + 2x both sides:

6sqrt(2x-6) = 2sqrt(x^2 + 3x - 4)

We'll divide by 2:

3sqrt(2x-6) = sqrt(x^2 + 3x - 4)

We'll raise to square both sides to eliminate the square root:

9(2x-6) = x^2 + 3x - 4

x^2 + 3x - 4 - 18x + 54 = 0

x^2 - 15x + 50 = 0

The roots of the quadratic are x1 = 5 and x2 = 10.

**The given numbers are the consecutive terms of an arithmetical progression if the values of x are: x = 5 or x = 10.**