# Find x if the numbers x, 6, x-5 are the terms of a geometric series.

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The numbers x, 6 and x-5 are terms of a geometric series.

Therefore we have 6/x = (x - 5)/6

=> x(x - 5) = 36

=> x^2 - 5x - 36 = 0

=> x^2 - 9x + 4x - 36 = 0

=> x(x- 9) + 4(x- 9) =0

=> (x + 4)(x - 9) =0

From x + 4 = 0, we get x = -4

From x - 9 = 0 , we get x = 9

**Therefore x can be 9 and -4.**

The given geometric progression is x, 6 and x-5.

We assume the first term is a1 = x, then the 2nd term a2 = ar and the 3rd term is a2 = a1 *r^2 = x-5.

Therefore a2 /a1 = a3/a2 = r.

=> 6/x = (x-5)/6 .

We multiply both sides by 6x and get:

6^2 = x(x-5)

36 = x^2-5x. Or x^2-5x-36 = 0

=> (x-9)(x+4) = 0.

So x -9 = 0 or x+4 = 0.

=**> x= 9 or x= -4.**

If x= 9, then 6 and 9-5 = 4 are the terms.

If x= - 4, then 6 and -4-5 = -9 are the terms.

So (9, 6, 4 with common ratio 2/3) or ( -4, 6, -9 with common ratio -3/2) are the terms of GP.

The terms x,6,x-5 are the consecutive terms of a geometric progression (series) if and only if the middle term is the geometric mean of the neighbor terms:

6 = sqrt[x*(x-5)]

We'll raise to square both sides:

36 = x(x-5)

We'll use the symmetric property and we'll remove the brackets:

x^2 - 5x = 36

We'll subtract 36:

x^2 - 5x - 36 = 0

We'll apply the quadratic formula:

x1 = [5 + sqrt(25 + 144)]/2

x1 = (5+13)/2

**x1 = 9**

x2 = (5-13)/2

**x2 = -4**

**We'll check the solution:**

For x = -4:

-4 , 6 , -9

6/-4 = -3/2

-9/6 = -3/2

So, the common ratio of the g.p. whose terms are -4 , 6 , -9, is r = -3/2.

For x = 9

9 , 6 , 4

6/9 = 2/3

4/6 = 2/3

So, the common ratio of the g.p. whose terms are 9 , 6 , 4 is r = 2/3.