JEE Advanced 2019 · Paper 2, Mathematics question 6

Numerator.
$$\sum_{k=1}^nk^{1/3}\sim\frac34n^{4/3}.$$

Denominator (as a Riemann sum):
$$n^{7/3}\sum_{k=1}^n\frac1{(an+k)^2}=n^{4/3}\cdot\frac1n\sum\frac1{(a+k/n)^2}\to n^{4/3}\int_0^1\frac{dx}{(a+x)^2}=\frac{n^{4/3}}{a(a+1)}.$$

The limit.
$$\frac34a(a+1)=54\implies a^2+a-72=0\implies a=8\text{ or }a=-9.$$
Both satisfy $|a|>1$.

(A), (D)