Mathematics - Sequence & Series Question with Solution | TestHub

Given:

$a+b+c+d=11$

We know that,

$\mathrm{AM}\ge \mathrm{GM}$

$\Rightarrow \frac{\frac{a}{5}+\frac{a}{5}+\frac{a}{5}+\frac{a}{5}+\frac{a}{5}+{\displaystyle \frac{b}{3}}+\frac{b}{3}+\frac{b}{3}+{\displaystyle \frac{c}{2}}+{\displaystyle \frac{c}{2}}+{\displaystyle \frac{d}{1}}}{11}\ge \sqrt[11]{\frac{a^5{b}^3{c}^{2}d}{5^5{3}^3{2}^2}}$

$\Rightarrow \frac{a+b+c+d}{11}\ge \sqrt[11]{\frac{a^5{b}^3{c}^{2}d}{5^5{3}^3{2}^2}}$

$\Rightarrow \frac{11}{11}\ge \sqrt[11]{\frac{a^5{b}^3{c}^{2}d}{5^5{3}^3{2}^2}}$

$\Rightarrow \sqrt[11]{\frac{a^5{b}^3{c}^{2}d}{5^5{3}^3{2}^2}}\le 1$

$\Rightarrow a^5{b}^3{c}^{2}d\le 5^5{3}^3{2}^2$

$\Rightarrow a^5{b}^3{c}^{2}d\le 90·3750$

So,

$\beta =90$