Three numbers are in AP. Their sum is $24$ and their product is $440$. Find the smallest of the three.

1. Let the three terms be $(a-d), a, (a+d)$ where $a$ is the middle term and $d$ is the common difference. 2. Sum: $(a-d) + a + (a+d) = 3a = 24$, so $a = 8$. 3. Product: $(a-d)\,a\,(a+d) = a(a^2 - d^2) = 8(64 - d^2) = 440$. 4. Solve: $64 - d^2 = 55$, so $d^2 = 9$, giving $d = \pm 3$. 5. With $d = 3$: the terms are $5, 8, 11$. Smallest is $5$. 6. Verify: sum $5+8+11 = 24$ ✓, product $5 \cdot 8 \cdot 11 = 440$ ✓. 7. Other options pick the middle (8) or largest (11) terms instead of smallest. _Source: NCERT Class 11 Mathematics, Ch 8, §8.2 + worked examples (AP problem-solving), p. 4–5._