Question

If $a,b,c,d$ are in continued proportion, prove that

$(a^2+b^2+c^2)(b^2+c^2+d^2)=(ab+bc+cd{)}^2$.

Answer 1

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

$\Rightarrow \frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k\Rightarrow a=bk,b=ck,c=dk,$

To prove $a^2+b^2+c^2\left)\right(b^2+c^2+d^2)={(ab+bc+cd+)}^2$
substituting $a,b,c$

$\Rightarrow ({\left(bk\right)}^2+{\left(ck\right)}^2+{\left(dk\right)}^2)(b^2+c^2+d^2)={(b^{2}k+c^{2}k+d^{2}k)}^2\Rightarrow k^2(b^2+c^2+d^2)(b^2+c^2+d^2)=k^2{(b^2+c^2+d^2)}^2\Rightarrow {(b^2+c^2+d^2)}^2={(b^2+c^2+d^2)}^2$

Hence proved.