1. Problem 1: The sum of the first three terms of a geometric progression (G.P.) is half its sum to infinity. Find the positive common ratio. 2. Problem 2: A man bought 5 reams of duplicating paper, each supposed to contain 450 sheets. The actual sheets were 415, 410, 400, 415, and 405. --- ### Problem 1: Geometric Progression Common Ratio 1. The sum of the first three terms of a G.P. with first term $a$ and common ratio $r$ is: $$S_3 = a + ar + ar^2 = a(1 + r + r^2)$$ 2. The sum to infinity of a G.P. (for $|r| < 1$) is: $$S_\infty = \frac{a}{1-r}$$ 3. Given that $S_3$ is half of $S_\infty$, we write: $$a(1 + r + r^2) = \frac{1}{2} \times \frac{a}{1-r}$$ 4. Cancel $a$ (assuming $a \neq 0$): $$1 + r + r^2 = \frac{1}{2(1-r)}$$ 5. Multiply both sides by $2(1-r)$: $$2(1-r)(1 + r + r^2) = 1$$ 6. Expand the left side: $$2(1-r + r - r^2 + r^2 - r^3) = 1$$ Simplify inside parentheses: $$2(1 - r^3) = 1$$ 7. So: $$2 - 2r^3 = 1 \implies 2r^3 = 1 \implies r^3 = \frac{1}{2}$$ 8. Taking the positive cube root: $$r = \sqrt[3]{\frac{1}{2}} = \frac{1}{\sqrt[3]{2}} \approx 0.7937$$ --- ### Problem 2: Paper Sheets and Price Error A. Calculate the average percentage error: 1. Percentage error for each packet is: $$\text{Error} = \frac{\text{Actual} - \text{Expected}}{\text{Expected}} \times 100$$ 2. Expected sheets = 450. 3. Calculate errors: - Packet 1: $\frac{415 - 450}{450} \times 100 = -7.78\%$ - Packet 2: $\frac{410 - 450}{450} \times 100 = -8.89\%$ - Packet 3: $\frac{400 - 450}{450} \times 100 = -11.11\%$ - Packet 4: $\frac{415 - 450}{450} \times 100 = -7.78\%$ - Packet 5: $\frac{405 - 450}{450} \times 100 = -10.00\%$ 4. Average percentage error: $$\frac{-7.78 - 8.89 - 11.11 - 7.78 - 10.00}{5} = \frac{-45.56}{5} = -9.11\%$$ Rounded to nearest whole number: $-9\%$. B. Calculate amount cheated: 1. Total expected sheets = $5 \times 450 = 2250$ sheets. 2. Total actual sheets = $415 + 410 + 400 + 415 + 405 = 2045$ sheets. 3. Shortfall in sheets = $2250 - 2045 = 205$ sheets. 4. Price per sheet: $$\frac{35 \times 5}{2250} = \frac{175}{2250} \approx 0.0778$$ 5. Amount cheated: $$205 \times 0.0778 \approx 15.94$$ Rounded to nearest whole number: 16. --- ### Final answers: 1. Positive common ratio $r \approx 0.794$. 2. A. Average percentage error $\approx -9\%$. 2. B. Amount cheated $\approx 16$.