Question

If a,b and c are in continued proportion, prove that: a^2+b^2/b(a+c)=b(a+c)/b^2+c^2

Answer 1

Step-by-step explanation:

(a2 + b2 + c2)/(a + b + c)2 = ( a- b + c)/( a + b + c) 

now we can do cross multiplication

(a2 + b2 + c2) ( a + b + c ) = (a - b + c) (a + b + c )2

a3 + ab2 + ac2 + a2b + b3 + c2b + a2c + b2c + c3= ( a - b + c) (a2 + b2 + c2 + 2ab + 2bc + 2ca )

a3 + ab2 + ac2 + a2b + b3 + c2b + a2c + b2c + c3= a3 + ab2 + ac2 + 2a2b + 2abc + 2a2c - a2b - b3- c2b -2ab2 - 2b2c + 2abc + a2c + b2c + c3 + 2abc + 2bc2 + 2c2a

a3 + ab2 + ac2 + a2b + b3 + c2b + a2c + b2c + c3- a3 - ab2 - ac2 - 2a2b - 2abc - 2a2c + a2b + b3 + c2b + 2ab2 + 2b2c + 2abc - a2c - b2c - c3 - 2abc - 2bc2 - 2c2a = 0

a3 + ab2 + ac2 + a2b + b3 + c2b + a2c + b2c + c3- a3 - ab2 - ac2 - 2a2b - 2abc - 2a2c + a2b + b3 + c2b + 2ab2 + 2b2c + 2abc - a2c - b2c - c3 - 2abc - 2bc2 - 2c2a = 0

now solve this 

b( a+ b + c) - ac ( a + b + c) = 0

b = ac

Proved