#### Question

If a, b, c and dare in continued proportion, then prove that

(a+ d)(b+ c)-(a+ c)(b+ d)= (b-c)^{2}

#### Solution

`"a"/"b" = "b"/"c" = "c"/"d" = "k"`

⇒ c = kd

b =kc= k^{2}d

a= kb= k^{3}d

(a+ d)(b + c)-(a+ c)(b + d) = (b- c)^{2}

LHS

(a+ d)(b + c)-(a+ c)(b + d)

= ab + bd + ac + cd - ab - bc - ad - cd

= bd+ ca - bc - ad

= k^{2}d^{2} + k^{4}d^{4} - k^{3}d^{2} - k^{3}d^{2}

= k^{2}d^{2} + k^{4}d^{4} - 2k^{3}d^{2}

= k^{2}d^{2} (1+ k^{2} - 2k)

RHS

(b - c)^{2} = (b - c)(b - c)

= b^{2} - 2bc + c^{2}

= k^{4}d^{4} - 2k^{3}d^{2} + k^{2}d^{2}

= k^{2}d^{2} (k^{2} - 2k + 1)

LHS = RHS. Hence, proved.

Is there an error in this question or solution?

Solution If A, B, C and Dare in Continued Proportion, Then Prove that Concept: Proportions.