If a, b, c and d are in G.P. show that (a2 + b2 + c2)(b2 + c2 + d2) = (ab + bc + cd)2.

Asked by Pragya Singh | 1 year ago |  62

##### Solution :-

Given, a, b, c, d are in G.P.

So, we have

b2 = ac  … (2)

c2 = bd  … (3)

Taking the R.H.S. we have

R.H.S.

= (ab + bc + cd)2

= (ab + ad + cd)2  [Using (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  [Using (1) and (2)]

= a2b2 + a2c2 + a2c2 + b2c+ b2c2 + d2a2 + d2b2 + d2b2 + d2c2

= a2b2 + a2c2 + a2d+ b× b2 + b2c2 + b2d2 + c2b2 + c× c2 + c2d2

[Using (2) and (3) and rearranging terms]

= a2(b2 + c2 + d2) + b2 (b2 + c2 + d2) + c2 (b2+ c2 + d2)

= (a2 + b2 + c2) (b2 + c2 + d2)

= L.H.S.

Thus, L.H.S. = R.H.S.

Therefore,

(a2 + b2 + c2)(b2 + c2 + d2) = (ab + bc + cd)2

Answered by Abhisek | 1 year ago

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