# Determine the numbers a, b if 3 ,a ,b , 24 is a geometric sequence.

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We are given that 3 ,a ,b , 24 is a geometric sequence.

So we have a/ 3 = b/a = 24 / b = k

So 24 / b = k => b = 24 / k

a/3 = k => a = 3k

Now b/a = 24 / b

=> (24/k)/ 3k = 24/(24/k)

=> 8/k^2 = k

=> 8 = k^3

=> k = 2

Therefore b = 24 / k = 24/2 = 12

a = 3k => a = 3*2 = 6

**Therefore a = 6 and b = 12.**

check : We see that 24/12 = 12/6 = 6/3 = 2

Since 3,a,b, 24 are in geometric progression, the consecutive terms should be in the same ratio.

So a/3 = b/a = 24/b = r.

Fromt the 1st equality, a/3 = b/a. So a^2= 3b....(1).

From b/a = 24/b, we get: b^2= 24a.....(2).

From (1), we get: b = a^2/3. We substitite b = a^2/3 in (2):

(a^2/3)^2 = 24a.

(a^4)/9 = 24a.

a^4 = 24*9a = 0.

a^4-216a = 0.

a(a^3-216) = 0.

a(a^3-6^3) = 0.

a(a-6) (a^2+6a+36) = 0.

a = 0, or a= 6.

Therefore a = 6 is the practical solution. r = a/3 = 2.

So b = 6*2 = 12.

Therfore a =6 , b = 12.

We'll use the theorem of geometric mean of a geometric sequence.:

a^2 = 3b (1)

b^2 = 24a (2)

We'll raise to square (1):

a^4 = 9b^2

We'll divide by 9 both sides:

b^2 = a^4/9 (3)

We'll substitute (3) in (2):

a^4/9 = 24a

We'll divide by a:

a^3/9 = 24

We'll cross multiply and we'll get:

a^3 = 24*9

a^3 = 2^3*3^3

a = 2*3

**a = 6**

For a = 6, we'll get b:

b^2 = 24a

b = sqrt 24*6

b = sqrt144

**b = 12**

**So, for a=6 and b=12, the terms of the geometric series, whose common ratio is r =2, are: 3 , 6 , 12 , 24, ....**