# what is x if the numbers a+x, b+x, c+x are terms of geometric progression ?

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a+x, b+x and c+x are in GP. To find x.

Solution:

Since a+x, b+x and c+x are in GP, the consecutive terms should have the same common ratio.

=> (b+x)/(a+x) = (c+x)/(b+x).

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

=> b^2+2bx+x^2 = ac+(a+c)x+x^2

=> 2bx - (a+c)x = ac-b^2

=> (2b- a-c)x = ac-b^2.

Therefore ** x= (ac-b^2/(2b-a-c).**

Since the given terms are the consecutive terms of a geometric progression, we'll write the relation between them:

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

We'll expand the square from the left side and we'll remove the brackets from the right side:

b^2 + 2bx + x^2 = ac + ax + cx + x^2

We'll eliminate x^2:

b^2 + 2bx = ac + x(a+c)

We'll move the terms in x to the left side and the terms without x, to the right side:

2bx - x(a+c) = ac - b^2

We'll factorize by x and we'll get:

x(2b - a - c) = ac - b^2

We'll divide by (2b - a - c):

**x = (b^2 - ac)/(a + c - 2b)**