1. **State the problem:** We are given a geometric progression (GP) where the sum to infinity is 16, and the sum of the first four terms is 15. We need to find the first four terms. 2. **Recall formulas:** - Sum to infinity of a GP: $$S_\infty = \frac{a}{1-r}$$ where $a$ is the first term and $r$ is the common ratio with $|r|<1$. - Sum of first $n$ terms: $$S_n = a \frac{1-r^n}{1-r}$$ 3. **Use given information:** - From sum to infinity: $$16 = \frac{a}{1-r} \implies a = 16(1-r)$$ - From sum of first 4 terms: $$15 = a \frac{1-r^4}{1-r}$$ 4. **Substitute $a$ from the first equation into the second:** $$15 = 16(1-r) \cdot \frac{1-r^4}{1-r}$$ 5. **Simplify:** $$15 = 16(1-r^4)$$ 6. **Solve for $r^4$:** $$1-r^4 = \frac{15}{16} \implies r^4 = 1 - \frac{15}{16} = \frac{1}{16}$$ 7. **Find $r$:** Since $|r|<1$, $$r = \frac{1}{2}$$ 8. **Find $a$:** $$a = 16(1 - \frac{1}{2}) = 16 \times \frac{1}{2} = 8$$ 9. **Find first four terms:** $$a = 8$$ $$ar = 8 \times \frac{1}{2} = 4$$ $$ar^2 = 8 \times \left(\frac{1}{2}\right)^2 = 8 \times \frac{1}{4} = 2$$ $$ar^3 = 8 \times \left(\frac{1}{2}\right)^3 = 8 \times \frac{1}{8} = 1$$ **Final answer:** The first four terms are $8, 4, 2, 1$.