Question

If all non-negative real numbers $a_1\le a_2\le \dots \le a_n$ satisfying $\sum _{i=1}^n{a}_i=96,$ $\sum _{i=1}^n{a}_i^2=144$ and $\sum _{i=1}^n{a}_i^3=216,$ then the value of $n$ is
​​​​​​​(correct answer + 1, wrong answer - 0.25)

• A. $32$
• B. $64$
• C. $128$
• D. It holds for all $n\in N$

Answer 1

The correct option is B $64$

We know that for positive $a_i$ ,
$(a_1+\dots +a_n)(a_1^3+\dots +a_n^3)\ge {(a_1^2+\dots +a_n^2)}^2$

As, $96\cdot 216={144}^2$, so equality holds iff
$a_i=k{a}_i^3\Rightarrow a_i^2=\frac{1}{k}$
Now,
$\sum _{i=1}^n{a}_i=96\Rightarrow k\sum _{i=1}^n{a}_i^3=96\Rightarrow k\times 216=96\Rightarrow k=\frac{96}{216}=\frac{4}{9}$
Therefore,
​​​ ​​​​$\sum _{i=1}^n{a}_i^2=144\Rightarrow \frac{n}{k}=144\Rightarrow n=144\times \frac{4}{9}=64$