If a, b, c and d are in G.P. show that (a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc + cd)2.
Given: a, b, c and d are in G.P.
Here,
a, b, c, d are in G.P.
∴ bc = ad — (1)
b2 = ac —(2)
c2 = bd –(3)
we have to be prove that,
(a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc – cd)2
From R.H.S.
(ab + bc + cd)2
= (ab + ad + cd)2 (From eq–1)
= (ab + d (a + c))2
= a2b2 + 2abd (a + c) + d2 (a + c)2
= a2b2 +2a2bd + 2acbd + d2(a2 + 2ac + c2)
= a2b2 + 2a2c2 + 2b2c2 + d2a2 + 2d2b2 + d2c2 (from eq– 1 and eq—2)
= a2b2 + a2c2 + a2c2 + b2c2 + b2c2 + d2a2 + d2b2 + d2b2 + d2c2
= a2b2 + a2c2 + a2d2 + b2 × b2 + b2c2 + b2d2 + c2b2 + c2 × c2 + c2d2
From eq – 2 and eq – 3
= a2(b2 + c2 + d2) + b2 (b2 + c2 + d2) + c2 (b2+ c2 + d2)
= (a2 + b2 + c2) (b2 + c2 + d2)
= L.H.S.
∴ L.H.S. = R.H.S.
∴(a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc + cd)2.