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Q.
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 $a^2+b^2+c^2+d^2-e^2$.
Sequences and Series
Solution:
$\Theta \frac{ a +\frac{ b }{2}+\frac{ b }{2}+\frac{ c }{3}+\frac{ c }{3}+\frac{ c }{3}+\frac{ d }{4}+\frac{ d }{4}+\frac{ d }{4}+\frac{ d }{4}+\frac{ e }{5}+\frac{ e }{5}+\frac{ e }{5}+\frac{ e }{5}+\frac{ e }{5}}{15}=1$
$\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. } $
$\therefore a =\frac{ b }{2}=\frac{ c }{3}=\frac{ d }{4}=\frac{ e }{5} \Rightarrow a =1, b =2, c =3, d =4, e =5$
$\therefore a ^2+ b ^2+ c ^2+ d ^2- e ^2=5$