Question:

Step 1: For three terms in A.P., the middle term equals the average of the first and third: $2\beta = -6\alpha + 3\alpha^2 + 3\beta$. Step 2: Rearranging: $2\beta - 3\beta = -6\alpha + 3\alpha^2$, so $-\beta = 3\alpha^2 - 6\alpha$, giving $\beta = 6\alpha - 3\alpha^2 = 3\alpha(2 - \alpha)$. Step 3: Since $\alpha$ and $\beta$ are natural numbers, $\beta > 0$, so $3\alpha(2-\alpha) > 0$, which requires $0 < \alpha < 2$. Thus $\alpha = 1$. Step 4: With $\alpha = 1$: $\beta = 3(1)(2-1) = 3$. So $\alpha + \beta = 1 + 3 = 4$. Step 5: The three terms are: $-6(1) = -6$, $\beta = 3$, $3(1)^2 + 3(3) = 3 + 9 = 12$. Common difference $d = 3 - (-6) = 9$. Step 6: Sum of first 9 terms: $S_9 = \frac{9}{2}[2(-6) + 8(9)] = \frac{9}{2}[-12 + 72] = \frac{9}{2}(60) = 270$.