Question

For $a\in R,|a|>1,$ let
$\underset{n\to \infty }{lim}\left(\frac{1+\sqrt[3]{32}+...+\sqrt[3]{3n}}{n^{\frac{7}{3}}(\frac{1}{(an+1{)}^2}+\frac{1}{(an+2{)}^2}+...+\frac{1}{(an+n{)}^2})}\right)=54.$

Then the possible value(s) of $a$ is/are

• A. $-9$
• B. $-6$
• C. $7$
• D. 8

Answer 1

Answer: (A), (D)

The correct options are
A $-9$
D 8

$\underset{n\to \infty }{lim}\left(\frac{1^{\frac{1}{3}}+2^{\frac{1}{3}}+...+n^{\frac{1}{3}}}{\frac{n^{\frac{7}{3}}}{n^2}(\frac{1}{{(a+\frac{1}{n})}^2}+\frac{1}{{(a+\frac{2}{n})}^2}+...+\frac{1}{{(a+\frac{n}{n})}^2})}\right)=54$

$\Rightarrow \underset{n\to \infty }{lim}\left(\frac{\frac{1}{n}\sum _{r=1}^n{\left(\frac{r}{n}\right)}^{\frac{1}{3}}}{\frac{1}{n}\sum _{r=1}^n{\left(\frac{1}{a+\frac{r}{n}}\right)}^2}\right)=54$

$\Rightarrow \frac{\underset{0}{\overset{1}{\int }}{x}^{\frac{1}{3}}dx}{\underset{0}{\overset{1}{\int }}{\left(\frac{1}{a+x}\right)}^{2}dx}=\frac{\frac{3}{4}[x^{\frac{4}{3}}{]}_0^1}{-{\left[\frac{1}{a+x}\right]}_0^1}=54$

$\Rightarrow \frac{3}{4}a(a+1)=54\Rightarrow a^2+a-72=0$
$\therefore a=-9,8$