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.
No comments:
Post a Comment