# If a, b and 8 are a GP and a, 8 and b are an AP, find a and b.

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For three terms a, b and c in a GP, ac = b^2. And for three terms a, b and c which are in an AP a+c= 2b.

As a, b and 8 are in a GP 8a = b^2

And as a, 8 and b are in an AP a+b = 2*8= 16

Now we will use the two equations:

8a = b^2 => a = b^2 / 8

substitute this in a+ b = 16 => b^2/ 8 + b =16

=> b^2 + 8 b = 8*16

=> b^2 + 8b – 128 =0

=> b^2 + 16b- 8b -128 =0

=> b(b+16) – 8(b+16)=0

=> (b-8)(b+16) =0

=> b = 8 or b= -16

Now a+b=16

=> a +8 = 16

=> a = 8

This gives an AP with difference 0 and common ratio 1.

For a-16 =16

=> a = 32

Here the common difference is -24 and common ratio is -2

**The required sets of a and b are (32, -16) and (8,8)**

a,b and 8 are in GP.

Therefore a*8 = b^2 .........(1)

a, 8 , b are inAP. So a+b = 2*8 = 16........(2).

From (2) b = 16-a. Substitute b = 16-a in (1):

a*8 = (16-a)^2.

8a = 256 -32a+a^2.

a^2-32a-8a +256 = 0

a^2-40a+256 = 0

(a-32)(a-8) = 0

a-32 = 0 . Or a-8 = 0.

a =32 . Or a = 8.

a = 8 gives b= 16-a = 8

a = 32 gives b = 16-32 = -16.

So the required (a, b, 8) = (8 , 8, 8). Or with common ratio 1, and common difference 0

(a,b, 8) = (32 , -16 , 8 ) with comoon ratio ,-2 and common diff = -24