1. **State the problem:** We have an arithmetic progression (AP) where the first term $a_1 = 2$. The sum of the first 8 terms $S_8 = 156$. We need to find the common difference $d$ and the number of terms $n$ when the sum $S_n = 416$. 2. **Recall the formula for the sum of the first $n$ terms of an AP:** $$S_n = \frac{n}{2} [2a_1 + (n-1)d]$$ 3. **Use the given information for the first 8 terms:** $$156 = \frac{8}{2} [2 \times 2 + (8-1)d]$$ Simplify: $$156 = 4 [4 + 7d]$$ $$156 = 16 + 28d$$ 4. **Solve for $d$:** $$28d = 156 - 16 = 140$$ $$d = \frac{140}{28} = 5$$ 5. **Now find $n$ such that $S_n = 416$:** Use the sum formula: $$416 = \frac{n}{2} [2 \times 2 + (n-1) \times 5]$$ Simplify inside the bracket: $$416 = \frac{n}{2} [4 + 5(n-1)] = \frac{n}{2} [4 + 5n - 5] = \frac{n}{2} (5n - 1)$$ Multiply both sides by 2: $$832 = n(5n - 1)$$ Rewrite: $$5n^2 - n - 832 = 0$$ 6. **Solve the quadratic equation:** Use the quadratic formula: $$n = \frac{1 \pm \sqrt{(-1)^2 - 4 \times 5 \times (-832)}}{2 \times 5} = \frac{1 \pm \sqrt{1 + 16640}}{10} = \frac{1 \pm \sqrt{16641}}{10}$$ Calculate the square root: $$\sqrt{16641} = 129$$ So: $$n = \frac{1 \pm 129}{10}$$ Two possible values: $$n = \frac{1 + 129}{10} = \frac{130}{10} = 13$$ $$n = \frac{1 - 129}{10} = \frac{-128}{10} = -12.8$$ (not valid since $n$ must be positive) 7. **Final answers:** - Common difference $d = 5$ - Number of terms $n = 13$ when the sum is 416