Let $a$, $b$, $c$ be positive integers in arithmetic progression such that the equation
$$ax^2+bx+c=0$$
has only integer solutions.
Then which of the following statements is (are) TRUE?
$c-b$ is an integer multiple of $a$
Both the roots of the equation $ax^2+bx+c=0$ are odd integers
If $c=15$, then $ab=8$
If $b=8$, then $x=3$ is a root of the equation $ax^2+bx+c=0$
Step-by-Step Solution
Key Concept: The constraint that all three of $a,b,c$ are in AP AND the roots are integers severely restricts possible values. The key is that $a$ must divide $b$ (and hence $d$), leading to essentially one solution family $(a,b,c)=(a,8a,15a)$ with roots $-3,-5$.
**Step 1: Set up AP structure**
Let $b=a+d$, $c=a+2d$ (common difference $d\geq0$). For integer roots $p,q$: $p+q=-b/a$ and $pq=c/a$ must give integers, so $a|b$. Write $d=ka$, giving $b=a(1+k)$, $c=a(1+2k)$.
**Step 2: Find valid $k$**
$(p-q)^2=(p+q)^2-4pq=(1+k)^2-4(1+2k)=k^2-6k-3$. Need $k^2-6k-3=m^2\geq0$. Factoring: $(k-3-m)(k-3+m)=12$. Only valid factor pair (same parity, $c>0$): $k=7$, giving $p+q=-8$, $pq=15$, roots $=-3,-5$.
**Step 3: Verify options**
(A) $c-b=(1+2k)a-(1+k)a=ka=7a$, a multiple of $a$. TRUE ✓
(B) Roots $-3,-5$ are both odd integers. TRUE ✓
(C) $c=15a=15 \Rightarrow a=1$, $b=8$. $ab=8$. TRUE ✓
(D) $b=8 \Rightarrow b=8a \Rightarrow a=1$, $c=15$. Check $x=3$: $9+24+15=48\neq0$. FALSE ✗
Correct Answer: A, B, C