# What value of x makes the three terms x, x/(x + 1) , 3x/(x + 1)(x + 2) those of a geometric sequence?

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For the terms x , x/(x + 1) and 3x/(x + 1)(x + 2) to be those of a geometric progression we need the following to be satisfied:

[3x/(x + 1)(x + 2)]/[x/(x + 1)] = [x/(x + 1)]/ x

=> 3 / (x+2) = 1/ (x+1)

Multiply each side by (x+1)(x+2)

=> 3*(x+1) = 1*(x+2)

=> 3x + 3 = x +2

Subtract x from both the sides

=> 2x + 3 = 2

Subtract 3 from both the sides

=> 2x = -1

Therefore x = -1/2.

**The required value of x makes the three terms x, x/(x + 1) , 3x/(x + 1)(x + 2) those of a geometric sequence is -1/2.**

If a, b and c are three consecutive terms of a gemetric series, then

ac = b^2.

Similarly if x, x/x+1 and 3x/(x+1)(x+2) are the three consecutive terms of a geometric series, then

x* 3x/(x+1)(x+2) = {x/(x+1)}^2.

Now we multiply both sides by the LCM ,(x+1)^2*(x+2) of denominators.

3x^2(x+1) = x^2*(x+2)

Divide by x^2 both sides:

3(x+1) = x+2.

3x+3 = x+2.

3x-x = 2-1 = -1.

2x = -1.

x = -1/2.

Therefore x = -1/2 makes the the three terms in GP :

First term x = -1/2 ,

2nd term x/(x+1) = -0.5/(-0.5+1) = -1 and

3rd term 3x/(x+1)(x+2) = -1.5/(0,5)(1.5) = -2 which is in geometrac ratio with common ratio 1/(-1/2) = -2/(-1) = 2.

We'll use the theorem of the geometric mean of 3 consecutive terms:x, x/(x + 1) , 3x/(x + 1)(x + 2)

[x/(x + 1)]^2 = x*3x/(x + 1)(x + 2)

We'll raise to square to the left side:

x^2/(x+1)^2 = 3x^2/(x + 1)(x + 2)

We'll simplify by (x+1) both sides:

x^2/(x+1) = 3x^2/(x+2)

We'll simplify by x^2 both sides:

1/(x+1) = 3/(x+2)

We'll cross multiply and we'll get:

3x + 3 = x + 2

We'll subtract x + 2 both sides:

2x + 1 = 0

We'll subtract 1:

2x = -1

**x = -1/2**

**For x = -1/2, the 3 given terms are the terms of a geometric progression.**