Question

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

Answer 1

Answer: 55

We have given that,
$a+b+c+d+e=15$
$AM=\frac{\left[a+\frac{b}{2}+\frac{b}{2}+\frac{c}{3}+\frac{c}{3}+\frac{c}{3}+\frac{d}{4}+\frac{d}{4}+\frac{d}{4}+\frac{d}{4}+\frac{e}{5}+\frac{e}{5}+\frac{e}{5}+\frac{e}{5}+\frac{e}{5}\right]}{15}$
$AM=1$
$GM={\left(a\cdot \frac{b^2}{2^2}\cdot \frac{c^3}{3^3}\cdot \frac{d^4}{4^4}\cdot \frac{e^5}{5^5}\right)}^{\frac{1}{15}}$
$GM=\frac{(120{)}^3\cdot 50}{2^2\cdot 3^3\cdot 4^4\cdot 5^5}=1$
$GM=1$
$\therefore AM=GM$
Hence, $a=\frac{b}{2}=\frac{c}{3}=\frac{d}{4}=\frac{e}{5}$
$\therefore a^2+b^2+c^2+d^2+e^2$
$=1^2+2^2+3^2+4^2+5^2$
$=\frac{5\times 6\times 11}{6}=55$