If a, b, c are in G.P., prove that are also in G.P a2 + b2, ab + bc, b2 + c2

Asked by Sakshi | 1 year ago |  57

##### Solution :-

a2 + b2, ab + bc, b2 + c2

Given that a, b, c are in GP.

By using the property of geometric mean,

b2 = ac

a2 + b2, ab + bc, b2 + c2 or (ab + bc)2 = (a2 + b2) (b2 + c2) [by using the property of GM]

Let us consider LHS: (ab + bc)2

Upon expansion we get,

(ab + bc)2 = a2b2 + 2ab2c + b2c2

= a2b2 + 2b2(b2) + b2c2 [Since, ac = b2]

= a2b2 + 2b4 + b2c2

= a2b2 + b4 + a2c2 + b2c2 {again using b2 = ac }

= b2(b2 + a2) + c2(a2 + b2)

= (a2 + b2)(b2 + c2)

= RHS

LHS = RHS

Hence a2 + b2, ab + bc, b2 + c2 are in GP.

Answered by Aaryan | 1 year ago

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