a, 6, b, 24 are terms of a G.P

Let r be the common difference such that:

6= a*r .........(1)

b= 6*r ..........(2)

24= b*r .........(3)

Let us use the substitution methos:

substitue (2) in (3):

==> 24 = b*r = (6r)*6 = 6r^2

==> 24= 6r^2

Now divide by 6:

==> r^2 = 24/6 = 4

==> r= 2

Now substitue in (2):

b= 6*r = 6*2 = 12

**==> b= 12**

Now let ussubstitue in (1):

6= a*r

==> a= 6/2 = 6/2 = 3

**==> a= 3**

**==> 3, 6, 12, 24 are terms of a G.P with r= 2**

a,6,b,24 are in GP. To determine a and b:

Since the terms are in GP, the successive terms should have common ratio. So

6/a = b/6 = 24/b.

From the last pair, b/6 = 24/b we get: b^2 = 24*6 =144.

So b = sqrt(144) = 12or -12.

Therefore r = 24/12 = 2. or 24/-12 = -2.

From the first pair, 36 = ab Or 36 = a*12. So a = 36/12 = 3.

Or so a =3 or -3. and b = 12 or -12 Therefore the series is

3 , 6, 12 , 24. Or

-3 , 6 , -12 , 24 , if r = -2.

We are given that a, 6, b, 24 are the terms of a G.P.

Therefore

a*b = 6^2 = 36

6* 24 = b^2

=> b= sqrt [ 6* 24]

=> b = sqrt 144

=> b = 12

Now using b=12 in a*b = 36,

we get 12a = 36 or a = 36/ 12 = 3.

**Therefore a = 3 and b is 12**