1. **State the problem:** We are given the sum of the first 10 terms ($S_{10}$) of an arithmetic progression (AP) as 460 and the sum of the next 10 terms ($S_{20} - S_{10}$) as 1260. We need to find the first term $a$ and the common difference $d$. 2. **Recall the formula for the sum of the first $n$ terms of an AP:** $$S_n = \frac{n}{2} [2a + (n-1)d]$$ 3. **Write the equations from the given sums:** - For the first 10 terms: $$460 = \frac{10}{2} [2a + (10-1)d] = 5(2a + 9d)$$ This simplifies to: $$5(2a + 9d) = 460 \implies 2a + 9d = 92 \quad (1)$$ - For the first 20 terms: $$S_{20} = S_{10} + \text{sum of next 10 terms} = 460 + 1260 = 1720$$ Using the sum formula: $$1720 = \frac{20}{2} [2a + (20-1)d] = 10(2a + 19d)$$ This simplifies to: $$10(2a + 19d) = 1720 \implies 2a + 19d = 172 \quad (2)$$ 4. **Solve the system of equations:** From (1): $$2a = 92 - 9d$$ Substitute into (2): $$92 - 9d + 19d = 172$$ $$92 + 10d = 172$$ $$10d = 80 \implies d = 8$$ 5. **Find $a$ using $d=8$ in (1):** $$2a + 9(8) = 92$$ $$2a + 72 = 92$$ $$2a = 20 \implies a = 10$$ **Final answer:** $$a = 10, \quad d = 8$$