Question

If $a,b,c,d,e$ are positive real numbers such that $a+b+c+d+e=15$ and $a{b}^2{c}^3{d}^4{e}^5=$ $(120{)}^3\cdot 50$, then find the value of $a^2+b^2+c^2+d^2-e^2$.

Answer 1

Answer: 5

$\Theta \frac{a+\frac{b}{2}+\frac{b}{2}+\frac{c}{3}+\frac{c}{3}+\frac{c}{3}+\frac{d}{4}+\frac{d}{4}+\frac{d}{4}+\frac{d}{4}+\frac{e}{5}+\frac{e}{5}+\frac{e}{5}+\frac{e}{5}+\frac{e}{5}}{15}=1$
$\therefore \text{ A.M. }=1$
$\text{ and }\text{ G.M. }={\left(a\cdot \frac{b}{2}\cdot \frac{b}{2}\cdot \frac{c}{3}\cdot \frac{c}{3}\cdot \frac{c}{3}\cdot \frac{d}{4}\cdot \frac{d}{4}\cdot \frac{d}{4}\cdot \frac{d}{4}\cdot \frac{e}{5}\cdot \frac{e}{5}\cdot \frac{e}{5}\cdot \frac{e}{5}\cdot \frac{e}{5}\right)}^{\frac{1}{15}}=1$
$\Theta \text{ A.M. }=\text{ G.M. }$
$\therefore a=\frac{b}{2}=\frac{c}{3}=\frac{d}{4}=\frac{e}{5}\Rightarrow a=1,b=2,c=3,d=4,e=5$
$\therefore a^2+b^2+c^2+d^2-e^2=5$