question_answer

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 the value of \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}+\,{{d}^{2}}+{{e}^{2}}\] is

A)  40                              

B)  45               

C)  50                              

D)  55

Correct Answer: D

Solution :

 Given, \[a+b+c+d+e=15\] \[\Rightarrow \]            \[AM=1\] Now,    \[GM={{\left( a\times \frac{{{b}^{2}}}{{{2}^{2}}}\times \frac{{{c}^{3}}}{{{3}^{3}}}\times \frac{{{d}^{4}}}{{{4}^{4}}}\times \frac{{{e}^{5}}}{{{5}^{5}}} \right)}^{1/15}}\] \[={{\left[ \frac{{{(120)}^{3}}\cdot 50}{{{2}^{2}}\cdot {{3}^{3}}\cdot {{4}^{4}}\cdot {{5}^{5}}} \right]}^{1/15}}=1\] \[\therefore \]    \[AM=GM\] Hence,             \[a=\frac{b}{2}=\frac{c}{3}=\frac{d}{4}=\frac{e}{5}\] \[\Rightarrow \,\,a=1,\,b=2,\,c=3,\,\,d=4,\,e=5\] \[\therefore \]    \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{d}^{2}}+{{e}^{2}}\] \[={{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}+{{5}^{2}}\] = 55