Q5. If a_n = $\frac{n(n - 3)}{x + 4}$, then find 18th term of this sequence.

Dear Student,

$$\begin{gathered} a_n= \frac{n\left(n-3\right)}{x+4} \\ {a}_{18} = \frac{18\left(18-3\right)}{x+4} \\ {a}_{18} =\frac{18\times 15}{x+4} = \frac{270}{x+4} \\ This can\text{'}t be solved further. \end{gathered}$$

$If the question would be a_n= \frac{n\left(n-3\right)}{n+4}$

Thus, it appears that the provided question is wrong, there should be n in place of x in the denominator.

Regards