The coefficient of the third term in the expansion of $\left(x^2-\dfrac{1}{4}\right)^n$, when expanded in the descending power of $x$ is 31, then $n$ is

The general term in the expansion of $(x^2 - \dfrac{1}{4})^n$ is given by $T_{r+1} = ^{n}C_{r} (x^2)^{n-r} (-\dfrac{1}{4})^r$.

The third term corresponds to $r = 2$.

$T_{3} = ^{n}C_{2} (x^2)^{n-2} (-\dfrac{1}{4})^2 = ^{n}C_{2} x^{2n-4} (\dfrac{1}{16})$.

The coefficient of the third term is given as 31. Since the expansion is in descending powers of $x$, the coefficient is independent of $x$ if $2n-4 = 0$, but here the coefficient is simply the constant part of the term.

The coefficient is $\dfrac{^{n}C_{2}}{16} = 31$.

Therefore, $^{n}C_{2} = 31 \times 16 = 496$.

Using the formula $^{n}C_{2} = \dfrac{n(n-1)}{2}$, we have $\dfrac{n(n-1)}{2} = 496$.

$n(n-1) = 992$.

Solving $n^2 - n - 992 = 0$ using the quadratic formula $n = \dfrac{1 \pm \sqrt{1 - 4(1)(-992)}}{2} = \dfrac{1 \pm \sqrt{1 + 3968}}{2} = \dfrac{1 \pm \sqrt{3969}}{2}$.

Since $\sqrt{3969} = 63$, $n = \dfrac{1 + 63}{2} = 32$.

Answer: 32