Question

If $a,b,c,d$ are in continued proportion, prove that
${(\frac{a-b}{c}+\frac{a-c}{b})}^2-{(\frac{d-b}{c}+\frac{d-c}{b})}^2={(a-d)}^2(\frac{1}{c^2}+\frac{1}{b^2})$

Answer 1

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

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

$⟹c=dk,b=ck=k^{2}d,a=bk=k^{3}d$

$LHS=(\frac{a-b}{c}+\frac{a-c}{b}{)}^2-(\frac{d-b}{c}+\frac{d-c}{b}{)}^2=(\frac{k^{3}d-k^{2}d}{kd}+\frac{k^{3}d-kd}{k^{2}d}{)}^2-(\frac{d-k^{2}d}{kd}+\frac{d-kd}{k^{2}d}{)}^2$

$=(k^2-\frac{1}{k}{)}^2-(\frac{1}{k^2}-k{)}^2=k^4-\frac{1}{k^4}+\frac{1}{k^2}-k^2+\frac{2}{k}-2k$

$RHS=(a-d{)}^2(\frac{1}{c^2}+\frac{1}{b^2})=(k^3-1{)}^2(\frac{1}{k^2}+\frac{1}{k^4})=k^4-\frac{1}{k^4}+\frac{1}{k^2}-k^2+\frac{2}{k}-2k$

$LHS=RHS$

Hence Proved