CAT Sequences and Series — practice questions

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: syllabus Mathematics, Ch 8, §8.2 + worked examples (AP problem-solving), p. 4–5._

1. Let the two numbers be $a$ and $b$. 2. AM = $(a+b)/2 = 5$, so $a + b = 10$. 3. GM = $\sqrt{ab} = 4$, so $ab = 16$. 4. These two equations describe the SUM and PRODUCT of two numbers — they are roots of the quadratic $x^2 - 10x + 16 = 0$. 5. Use quadratic formula: $x = (10 \pm \sqrt{100 - 64})/2 = (10 \pm 6)/2$, giving $x = 8$ or $x = 2$. 6. So the numbers are $\boxed{2 \text{ and } 8}$. 7. Sanity: AM = $(2+8)/2 = 5$ ✓; GM = $\sqrt{16} = 4$ ✓. 8. Other options don't satisfy both conditions. _Source: syllabus Mathematics, Ch 8, §8.4 (AM-GM relationship + Examples), p. 10–11._

1. Your syllabus: for any two POSITIVE numbers $a$ and $b$: 2. $\dfrac{a+b}{2} \geq \sqrt{ab}$ — i.e., AM $\geq$ GM. 3. EQUALITY holds if and only if $a = b$. 4. Proof sketch: $(a - b)^2 \geq 0$ (always, since square is non-negative) $\Rightarrow a^2 + b^2 \geq 2ab \Rightarrow a^2 + 2ab + b^2 \geq 4ab \Rightarrow (a+b)^2 \geq 4ab$ so $\sqrt{(a+b)^2/4} \geq \sqrt{ab}$, i.e. AM $\geq$ GM. 5. Example check: $a=2, b=8$: AM=5, GM=4. $5 \geq 4$ ✓ (strict inequality since $a \neq b$). 6. Option A is wrong unless $a=b$. Option B is the OPPOSITE inequality. Option D is much too strong. _Source: syllabus Mathematics, Ch 8, §8.4 (AM-GM relationship), p. 10–11._