Solve for x if x-1, x-9, 6x are consecutive terms of a geometric progression.

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krishna-agrawala's profile pic

krishna-agrawala | College Teacher | (Level 3) Valedictorian

Posted on

In a geometric progression the ration between any two consecutive terms remains same.

Therefore for the given three consecutive terms:

(x - 1)/(x - 9) = (x - 9)/6x

Cross multiplying the terms in above equation:

6x(x - 1) = (x - 9)^2

6x^2 - 6x = x^ -18x + 81

6x^2 - x^2 - 6x + 18x - 81=0

5x^2 + 12x - 81 = 0

5x^2 - 15x + 27x - 81 = 0

5x(x - 3) + 27(x - 3) = 0

(x - 3)(5x + 27) = 0

Therefore:

x = 3 and -27/5

 

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

To solve if x-1, x-9 and 6x are in the GP.

So the successive terms bear the same ratio. The  common ratio is (x-9)/(x-1) = 6x/(x-9). Multiplying by (x-1)(x-9), we get:

(x-9)^2 = 6x(x-1).

x^2-18x+81 = 6x^2-6x.

0 = 6x^2-6x -x^2+18x-81.

5x^2+12x -81 =  0.

x1 = {-12+(12^2+4*5*81)}/(2*5) = (-12+42)/10 = 3. Or

x2 = (-12-42}/10 = -5.4.

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

If x-1, x-9, 6x are consecutive terms of the geometric progression, that means that:

x-9 = q*(x-1), where q is the ratio

We'll divide by (x-1)

q = (x-9)/(x-1) (1)

6x = q*(x-9)

We'll divide by (x-9)

q = 6x/(x-9) (2)

From (1) and (2) it results:

(x-9)/(x-1) = 6x/(x-9)

We'll cross multiplying:

(x-9)^2 = 6x(x-1)

We'll expand the square:

x^2 - 18x + 81 = 6x^2 - 6x

We'll move all terms to one side:

x^2 - 18x + 81 - 6x^2 + 6x = 0

-5x^2 - 12x + 81 = 0

We'll multiply by -1:

5x^2 + 12x - 81 = 0

x1 = [-12+sqrt(144+1620)]/10

x1 = (-12+42)/10

x1 = 30/10

x1 = 3

x2 = (-12-42)/10

x2 = -54/10

x2 = -5.4

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