1. **State the problem:** We need to find the possible values of $k$ such that the constant term in the expansion of $\left(x^2 + x^k\right)^{12}$ has a coefficient of 495. 2. **Recall the binomial expansion formula:** $$\left(a + b\right)^n = \sum_{r=0}^n \binom{n}{r} a^{n-r} b^r$$ 3. **Apply the formula to our expression:** $$\left(x^2 + x^k\right)^{12} = \sum_{r=0}^{12} \binom{12}{r} (x^2)^{12-r} (x^k)^r = \sum_{r=0}^{12} \binom{12}{r} x^{2(12-r)} x^{kr} = \sum_{r=0}^{12} \binom{12}{r} x^{24 - 2r + kr}$$ 4. **Find the constant term:** The constant term corresponds to the power of $x$ being zero: $$24 - 2r + kr = 0$$ 5. **Solve for $r$:** $$24 - 2r + kr = 0 \implies 24 = r(2 - k) \implies r = \frac{24}{2 - k}$$ 6. **Conditions for $r$:** Since $r$ is an integer between 0 and 12 inclusive, $r$ must satisfy: $$0 \leq r \leq 12$$ 7. **Coefficient of the constant term:** $$\binom{12}{r}$$ must equal 495. 8. **Find $r$ such that $\binom{12}{r} = 495$:** Check binomial coefficients for $n=12$: - $\binom{12}{2} = 66$ - $\binom{12}{3} = 220$ - $\binom{12}{4} = 495$ - $\binom{12}{8} = 495$ (since $\binom{12}{r} = \binom{12}{12-r}$) So $r = 4$ or $r = 8$. 9. **Find $k$ for each $r$:** For $r=4$: $$4 = \frac{24}{2 - k} \implies 2 - k = \frac{24}{4} = 6 \implies k = 2 - 6 = -4$$ For $r=8$: $$8 = \frac{24}{2 - k} \implies 2 - k = \frac{24}{8} = 3 \implies k = 2 - 3 = -1$$ 10. **Final answer:** The possible values of $k$ are $\boxed{-4 \text{ and } -1}$.