1. **Stating the problem:** Given an arithmetic progression (a_n) with the conditions: $$a_4 - a_2 = 4$$ $$a_1 + a_3 + a_5 + a_6 = 30$$ Calculate the sum of the first 20 terms, $S_{20}$. 2. **Recall the formula for arithmetic progression:** The $n$-th term of an arithmetic progression is: $$a_n = a_1 + (n-1)d$$ where $a_1$ is the first term and $d$ is the common difference. The sum of the first $n$ terms is: $$S_n = \frac{n}{2} (2a_1 + (n-1)d)$$ 3. **Use the given conditions to find $a_1$ and $d$:** From the first condition: $$a_4 - a_2 = 4$$ Substitute the formula for terms: $$[a_1 + 3d] - [a_1 + d] = 4$$ Simplify: $$a_1 + 3d - a_1 - d = 4$$ $$2d = 4$$ $$d = 2$$ 4. **Use the second condition:** $$a_1 + a_3 + a_5 + a_6 = 30$$ Substitute terms: $$a_1 + (a_1 + 2d) + (a_1 + 4d) + (a_1 + 5d) = 30$$ Simplify: $$4a_1 + (2d + 4d + 5d) = 30$$ $$4a_1 + 11d = 30$$ Substitute $d=2$: $$4a_1 + 11 \times 2 = 30$$ $$4a_1 + 22 = 30$$ $$4a_1 = 8$$ $$a_1 = 2$$ 5. **Calculate $S_{20}$:** $$S_{20} = \frac{20}{2} [2a_1 + (20-1)d] = 10 [2 \times 2 + 19 \times 2]$$ $$= 10 [4 + 38] = 10 \times 42 = 420$$ **Final answer:** $$S_{20} = 420$$