if a, b, c are in G.P., prove that (a + 2b + 2c) (a – 2b + 2c) = a2 + 4c2

Asked by Sakshi | 1 year ago |  73

##### Solution :-

(a + 2b + 2c) (a – 2b + 2c) = a2 + 4c2

Given that a, b, c are in GP.

By using the property of geometric mean,

b2 = ac

Let us consider LHS: (a + 2b + 2c) (a – 2b + 2c)

Upon expansion we get,

(a + 2b + 2c) (a – 2b + 2c) = a2 – 2ab + 2ac + 2ab – 4b2 + 4bc + 2ac – 4bc + 4c2

= a2 + 4ac – 4b2 + 4c2

= a2 + 4ac – 4(ac) + 4c2 [Since, b2 = ac]

= a2 + 4c2

= RHS

LHS = RHS

Hence proved.

Answered by Aaryan | 1 year ago

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