Question

If $a,b,c,d,e$ are in continued proportion, prove that
$(ab+bc+cd+de{)}^2=(a^2+b^2+c^2+d^2\left)\right(b^2+c^2+d^2+e^2)$.

Answer 1

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

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

We need to prove ${(ab+bc+cd+de)}^2=(a^2+b^2+c^2+d^2)(b^2+c^2+d^2+e^2)$ ....(1)

Substituting the values of $a,b,c,d$ in (1)

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

As LHS $=$ RHS

Hence, proved.