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If we consider a set of three numbers a, b and c which are in AP, we know that a + c =2b. Now, if they are in GP, a*c =b^2.
Now we have to see if it is possible to have three numbers satisfy both the equations.
If we express a as 2b - c and replace it in a*c = b^2, it gives us (2b-c)*c = b^2
=> 2bc – c^2 = b^2
=> b^2 + c^2 – 2bc =0
=> (b-c) ^2 =0
=> b = c
Also, as a = 2b-c = 2c-c = c
we find that all the three number have to be equal.
So a set of three numbers which are in AP as well as GP are a set of three equal numbers.
Let us see algebraically, whether this is possible.
Let a, b and c be 3 numbers which are in both AP and GP.
Since the numbers are in AP, a+c = 2b.........(1)
Since the numbers are in GP , ac = b^2 ......(2).
From (1), c = 2b-a, Substitute 2b-c for c in (2):
a(2b-a) = b^2.
Therefore b^2- 2ab + a^2 = 0
Or (b-a)^2 = 0.
Or a = b.
Or (a,b,c ) = (a,a,a,).
That implies the common diffrence = a-b = 0 and the common ratio is b/a = 1.
Under the special case, a,a and a are an AP with common diffrence d = 0 and a,a and a are in GP with common diference d =1.
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