If a, b, c are in A.P. and a, x, b and b, y, c are in G.P., show that x2, b2, y2 are in A.P.

If a,b,c are in AP it follows that


a + c = 2b……..(1)


and a,x,b and b,y,c are in individual GPs which follows


x2 = ab …….(2)


y2 = bc ……..(3)


Adding eqn 2 and 3 we get,


x2 + y2 = ab + bc


= b(a + c)


= b.2b ( from eqn 1)


= 2b2


So we get x2 + y2 = 2b2 which shows that they are in AP.


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