Question

If $a,b,c,d$ are in continued proportion, prove that
$(a^2-b^2)(c^2-d^2)={(b^2-c^2)}^2$

Answer 1

Given $a,b,c,d$ are in continued proportion

$⟹\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k($say$)$

$⟹c=dk,b=ck=k^{2}d,a=bk=k^{3}d$

$LHS=(a^2-b^2)(c^2-d^2)=(k^6{d}^2-k^4{d}^2)(k^2{d}^2-d^2)=k^4{d}^4(k^2-1{)}^2=(k^4{d}^2-k^2{d}^2{)}^2=\left(\right(k^{2}d{)}^2-(kd{)}^2{)}^2$

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

Hence Proved