# 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.

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

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