# Which is the value of a, for the terms a-2,a-8, 6a to be the consecutive terms of a geometric progression?

hala718 | Certified Educator

a-2, a-8, 6a     are terms of a geometric progression

Then :

a-8 = (a-2)r.....(1)

6a = (a-8) r ....(2)

(r) is a constant ratio between the terms.

From (1):

a= ar-2r+8

ar-a= 2r-8

a(r-1)= 2r-8

a= (2r-8)/(r-1)

Now substitute

a=2r-8 in (2)

6a= (a-8)r 6(2r-8)/(r-1) = (2r-8)/(r-1)  -8) r

After simplifying:

6r^2 +12r -48= 0

Divide by 6:

r^2 + 2r -8=0

Factorize:

(r+4)(r-2)=0

r1= 2    ==> a1= (2r1-8)/(r1-1)= -4/1= -4

To check:

a-2, a-8, 6a   = -6, -12, -24

r2= -4   ==> a2= (2r2-8)/(r2-1)= -16/-5 = 16/5

To check:

a-2, a-8, 6a   =   6/5 , -24/5,  96/5

giorgiana1976 | Student

If a-2, a-8, 6a are consecutive terms of the geometric progression, that means that:

a-8 = q*(a-2), where q is the ratio

We'll divide by (a-2)

q = (a-8)/(a-2) (1)

6a = q*(a-8)

We'll divide by (a-8)

q = 6a/(a-8) (2)

From (1) and (2) it results:

(a-8)/(a-2) = 6a/(a-8)

We'll cross multiplying:

(a-8)^2 = 6a(a-2)

We'll expand the square:

a^2 - 16a + 64 = 6a^2 - 12a

We'll move all terms to one side:

a^2 - 16a + 64 - 6a^2 + 12a = 0

-5a^2 - 4a + 64 = 0

We'll multiply by -1:

5a^2 + 4a - 64 = 0

a1 = [-4+sqrt(16+1280)]/10

a1 = (-4+6)/10

a1 = 2/10

a1 = 1/5

a2 = (-4-6)/10

a2 = -10/10

a2 = -1

krishna-agrawala | Student

In a geometric progression the ratio of any two consecutive terms is constant.

Three consecutive terms of the given geometric series are:

a-2, a-8, and 6a

As ratio of first and second term is same as the ration of second and third term:

(a-2)/(a-8) = (a-8)/6a

Cross multiplying terms in denominators of left and right side of above equation:

6a(a - 2) = (a - 8)^2

6a^2 - 12 a = a^2 -16a + 64

6a^2 - a^2 - 12a + 16a - 64 = 0

5a^2 + 4a - 64 = 0

5a^2 + 20a - 16a - 64 = 0

5a(a + 4) - 16(a + 4) = 0

(a + 4)(5a - 16) = 0

5(a + 4)(a - 16/5) = 0

5(a + 4)(a - 3.2) = 0

Therefore:

a = -4, 3.2

a = -4, and a = 3.2

neela | Student

The terms a-2, a-8 and 6a are the consecutive terms of a geometric progression.To determine a.

Solution:

As the  successive terms should bear the same common ratio,

(a-8)/(a-2) = 6a/(a-8). Cross multiplying,

(a-8)^2 = 6a(a-2).

a^2-16a+64 = 6a^2-12a

0 = (6-1)a^2+16a-12a-64. Or

0 =5a^2+4a-64

a1 = {-4+sqrt4^2+4*5*64)}/(2*5) = (-4+36)/10 = -3.2. Or

a2 = (-4-36}/10 = -4.

So the values of a are a = 3.2 or a = -4.

Therefore  -4-2 , (-4-8) and 6*(-4) are in GP. Or

-4, -12 and -24 or

3.2-2 , 3.2-8 and 6*3.2 Or

1.2 , -4.8 and 19.2 are in GP