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

### 3 Answers | Add Yours

You should remember the property that links each two consecutive terms of geometric progression suhc that:

`(x/(x+1))/x = q`

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

Equating the fractions yields:

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

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

`3(x+1) = x+2 =gt 3x + 3 = x+2`

Isolating the terms in x to the left yields:

`3x - x = -3 + 2 =gt 2x = -1 =gt x = -1/2`

**Hence, evaluating the value of x for the terms of a geometric progression yields `x = -1/2` .**

**Sources:**

We need to find the value of x that makes the terms x, x/(x + 1) and 3x/[(x + 1)(x + 2)] those of a geometric sequence.

For three consecutive terms of a GP, the product of the first and the third is equal to the square of the second. Using the values we have:

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

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

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

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

=> 3x + 3 = x + 2

=> 2x = -1

=> x = -1/2

**The required value of x is -1/2.**

For the terms x, x/(x + 1) and 3x/[(x + 1)(x + 2)] to be the terms of a geometric sequence, than the following identity must be valid:

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

We'll multiply by x+1 both sides: x^2/(x+1) = 3x^2/(x+2)

We'll cross multiply: 3x^2*(x+1) = x^2*(x+2) 3x^3 + 3x^2=x^3+2x^2

We'll shift all terms to the left side: 3x^3 + 3x^2-x^3-2x^2 = 0 2x^3 + x^2 = 0

We'll factorize by x^2: x^2(2x+1) = 0

We'll cancel the 1st factor: x^2=0 x1=x2=0

We'll cancel the 2nd factor: 2x+1=0 2x=-1 x=-1/2

**The values of x that makes the three terms x, x/(x + 1) and 3x/[(x + 1)(x + 2)] those of a geometric sequence are: {-1/2 , 0}.**

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes