Jumlah Deret Aritmatika 93064D

1. **Problem 1: Find the sum of the first 20 terms of the sequence 21 + 16 + 11 + ...** 2. This is an arithmetic sequence where the first term $a_1 = 21$ and the common difference $d = 16 - 21 = -5$. 3. The formula for the sum of the first $n$ terms of an arithmetic sequence is: $$S_n = \frac{n}{2} (2a_1 + (n-1)d)$$ 4. Substitute $n=20$, $a_1=21$, and $d=-5$: $$S_{20} = \frac{20}{2} (2 \times 21 + (20-1)(-5))$$ 5. Simplify inside the parentheses: $$2 \times 21 = 42$$ $$(20-1) = 19$$ $$19 \times (-5) = -95$$ 6. So: $$S_{20} = 10 (42 - 95) = 10 \times (-53) = -530$$ --- 1. **Problem 2: Find the sum of the sequence 3 + 1 + (-1) + ... + (-61)** 2. This is an arithmetic sequence with $a_1 = 3$ and $d = 1 - 3 = -2$. 3. We need to find the number of terms $n$ such that the last term $a_n = -61$. 4. The $n$th term formula is: $$a_n = a_1 + (n-1)d$$ 5. Substitute $a_n = -61$, $a_1=3$, $d=-2$: $$-61 = 3 + (n-1)(-2)$$ 6. Simplify: $$-61 - 3 = (n-1)(-2)$$ $$-64 = -2(n-1)$$ 7. Divide both sides by $-2$: $$\cancel{-2}(n-1) = \frac{-64}{\cancel{-2}}$$ $$n-1 = 32$$ $$n = 33$$ 8. Now find the sum of 33 terms: $$S_{33} = \frac{33}{2} (2 \times 3 + (33-1)(-2))$$ 9. Simplify: $$2 \times 3 = 6$$ $$(33-1) = 32$$ $$32 \times (-2) = -64$$ 10. So: $$S_{33} = \frac{33}{2} (6 - 64) = \frac{33}{2} (-58) = 33 \times (-29) = -957$$ --- 1. **Problem 3: Given the 7th term is 10 times the 13th term and the 13th term is -2, find the sum of the first 25 terms of the arithmetic sequence.** 2. Let the first term be $a_1$ and common difference $d$. 3. The $n$th term formula: $$a_n = a_1 + (n-1)d$$ 4. Given: $$a_7 = 10 \times a_{13}$$ and $$a_{13} = -2$$ 5. Substitute $a_7$ and $a_{13}$: $$a_1 + 6d = 10 (a_1 + 12d)$$ 6. Expand right side: $$a_1 + 6d = 10a_1 + 120d$$ 7. Rearrange: $$a_1 + 6d - 10a_1 - 120d = 0$$ $$-9a_1 - 114d = 0$$ $$9a_1 + 114d = 0$$ 8. Simplify by dividing by 3: $$3a_1 + 38d = 0$$ $$3a_1 = -38d$$ $$a_1 = -\frac{38}{3} d$$ 9. Use $a_{13} = -2$: $$a_1 + 12d = -2$$ Substitute $a_1$: $$-\frac{38}{3} d + 12d = -2$$ 10. Combine terms: $$-\frac{38}{3} d + \frac{36}{3} d = -2$$ $$-\frac{2}{3} d = -2$$ 11. Multiply both sides by $-\frac{3}{2}$: $$d = -2 \times -\frac{3}{2} = 3$$ 12. Find $a_1$: $$a_1 = -\frac{38}{3} \times 3 = -38$$ 13. Find sum of first 25 terms: $$S_{25} = \frac{25}{2} (2a_1 + (25-1)d)$$ $$= \frac{25}{2} (2 \times -38 + 24 \times 3)$$ $$= \frac{25}{2} (-76 + 72) = \frac{25}{2} (-4) = 25 \times (-2) = -50$$ --- 1. **Problem 4: Pak Karto's profit increases by the same amount each month. The total profit after 3 months is 480000 and after 12 months is 2568000. Find the total profit after 3 years (36 months).** 2. This is an arithmetic sequence where the sum after $n$ months is: $$S_n = \frac{n}{2} (2a_1 + (n-1)d)$$ 3. Given: $$S_3 = 480000$$ $$S_{12} = 2568000$$ 4. Write equations: $$S_3 = \frac{3}{2} (2a_1 + 2d) = 480000$$ $$S_{12} = \frac{12}{2} (2a_1 + 11d) = 2568000$$ 5. Simplify: $$\frac{3}{2} (2a_1 + 2d) = 480000 \Rightarrow 3(a_1 + d) = 480000 \Rightarrow a_1 + d = 160000$$ $$6 (2a_1 + 11d) = 2568000 \Rightarrow 2a_1 + 11d = 428000$$ 6. From $a_1 + d = 160000$, express $a_1$: $$a_1 = 160000 - d$$ 7. Substitute into second equation: $$2(160000 - d) + 11d = 428000$$ $$320000 - 2d + 11d = 428000$$ $$320000 + 9d = 428000$$ $$9d = 108000$$ $$d = 12000$$ 8. Find $a_1$: $$a_1 = 160000 - 12000 = 148000$$ 9. Find sum after 36 months: $$S_{36} = \frac{36}{2} (2 \times 148000 + 35 \times 12000)$$ $$= 18 (296000 + 420000) = 18 \times 716000 = 12888000$$ **Final answers:** - Sum of first 20 terms (Problem 1): $-530$ - Sum of sequence to $-61$ (Problem 2): $-957$ - Sum of first 25 terms (Problem 3): $-50$ - Total profit after 3 years (Problem 4): $12888000$