Question

If a, b, c, d are in proportion, then prove that 

i. (11a² + 9ac)/(11b² + 9bd) = (a² + 3ac)/(b² + 3bd)

ii. √((a² + 5c²)/(b² + 5d²)) = a/b

iii. (a² + ab + b²)/(a² - ab + b²) = (c² + cd + d²)/(c² - cd + d²)

Answer 1

a, b, c are in continued proportion. … [Given]

Thus, (11a² + 9ac)/(11b² + 9bd) = (a² + 3ac)/(b² + 3bd)

= b²(k² + k + 1)/b²(k² - k + 1)