Question 6

Since $a, b, c$ are in AP, $2b = a + c$. Let the integer roots be $\alpha$ and $\beta$.

From Vieta's formulas, $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$. Since $\alpha, \beta, a, b, c$ are integers, $\frac{b}{a}$ and $\frac{c}{a}$ must be integers.

Let $b = ka$ and $c = ma$ where $k, m \in \mathbb{Z}^+$.

Substituting into $2b = a + c$ gives $2ka = a + ma \Rightarrow m = 2k - 1$.

Now, $\alpha + \beta = -k$ and $\alpha\beta = m = 2k - 1$.

Eliminating $k$: $\alpha\beta = 2(-(\alpha + \beta)) - 1 \Rightarrow \alpha\beta + 2\alpha + 2\beta + 4 = 3 \Rightarrow (\alpha + 2)(\beta + 2) = 3$.

The possible integer factor pairs for 3 are $(1, 3)$ and $(-1, -3)$.

Case 1: $\alpha+2=1, \beta+2=3 \Rightarrow \alpha=-1, \beta=1$. Then $k = -(\alpha+\beta) = 0$. But $b$ is a positive integer, so $k > 0$.

Case 2: $\alpha+2=-1, \beta+2=-3 \Rightarrow \alpha=-3, \beta=-5$. Then $k = -(-3-5) = 8$ and $m = 2(8)-1 = 15$. Thus, $b = 8a$ and $c = 15a$.

(A) $c - b = 15a - 8a = 7a$, which is a multiple of $a$. Correct.

(B) Roots are $-3$ and $-5$, which are both odd. Correct.

(C) If $c = 15$, then $15a = 15 \Rightarrow a = 1$. Then $b = 8(1) = 8$. So $ab = 8$. Correct.

(D) If $b = 8$, then $8a = 8 \Rightarrow a = 1$. The roots are $-3$ and $-5$. $x=3$ is not a root. Incorrect.