1. **State the problem:** We have a radioactive isotope with an initial amount of 1921 grams at year 0 and 1454 grams at year 2. The decay follows a geometric sequence, and we need to find the amounts at years 1, 3, and 4. 2. **Formula and rules:** The amount at year $t$ is given by the geometric sequence formula: $$G(t) = G_0 \times r^t$$ where $G_0$ is the initial amount and $r$ is the common ratio (decay factor) per year. 3. **Find the decay ratio $r$:** Using the known values at $t=0$ and $t=2$: $$1454 = 1921 \times r^2$$ Divide both sides by 1921: $$r^2 = \frac{1454}{1921} \approx 0.7575$$ Take the square root to find $r$: $$r = \sqrt{0.7575} \approx 0.8701$$ 4. **Calculate amounts at years 1, 3, and 4:** - At $t=1$: $$G(1) = 1921 \times 0.8701^1 = 1921 \times 0.8701 \approx 1671$$ - At $t=3$: $$G(3) = 1921 \times 0.8701^3 = 1921 \times 0.6589 \approx 1265$$ - At $t=4$: $$G(4) = 1921 \times 0.8701^4 = 1921 \times 0.5731 \approx 1101$$ 5. **Final answers (rounded to whole numbers):** | Years Elapsed | 0 | 1 | 2 | 3 | 4 | |---------------|-------|-------|-------|-------|-------| | Grams | 1921 | 1671 | 1454 | 1265 | 1101 | This shows the isotope decays by about 13% each year, following a geometric decay pattern.