Arithmetic Rule 5F5Ed7

1. **State the problem:** We are given two sequences: \(I = 0, 3, 6, x, 8, 12\) and \(r = 5, 8, 12, 17, 24, y\). We need to find the arithmetic rule for each sequence and then find the values of \(x\) and \(y\). 2. **Find the arithmetic rule for sequence \(I\):** - Check the differences between consecutive terms: \[3 - 0 = 3, \quad 6 - 3 = 3\] - The common difference is \(3\), so \(I\) is an arithmetic sequence with rule: \[I_n = I_1 + (n-1) \times 3 = 0 + (n-1) \times 3 = 3(n-1)\] 3. **Find \(x\) in sequence \(I\):** - The terms given are \(I_1=0, I_2=3, I_3=6, I_4=x, I_5=8, I_6=12\). - Using the rule, \(I_4 = 3(4-1) = 3 \times 3 = 9\). - But the given \(I_5=8\) does not fit the arithmetic sequence rule (should be \(3(5-1)=12\)). - Since \(I_5=8\) is given, it contradicts the arithmetic sequence rule. - Possibly a typo or \(x\) is to be found assuming the sequence is arithmetic ignoring the given \(8\). - If we assume the sequence is arithmetic with difference 3, then \(x = 9\). 4. **Find the arithmetic rule for sequence \(r\):** - Differences between terms: \[8 - 5 = 3, \quad 12 - 8 = 4, \quad 17 - 12 = 5, \quad 24 - 17 = 7\] - Differences are not constant, so \(r\) is not arithmetic. - Check if \(r\) follows another pattern or if the problem means to find a rule relating \(I\) and \(r\). 5. **Check if \(r\) depends on \(I\):** - Given pairs \((I, r)\): \[(0,5), (3,8), (6,12), (x,17), (8,24), (12,y)\] - Try to find a linear relation \(r = aI + b\). - Use points \((0,5)\) and \((3,8)\): \[8 = 3a + b\] \[5 = 0a + b \Rightarrow b = 5\] - Substitute \(b=5\) into first: \[8 = 3a + 5 \Rightarrow 3a = 3 \Rightarrow a = 1\] - So \(r = I + 5\). 6. **Check if this fits other points:** - For \(I=6\), \(r = 6 + 5 = 11\), but given \(r=12\) (close but not exact). - For \(I=8\), \(r=8+5=13\), but given \(r=24\) (does not fit). - So linear relation does not fit all points. 7. **Try quadratic relation \(r = aI^2 + bI + c\):** - Use points \((0,5), (3,8), (6,12)\): \[r(0) = c = 5\] \[r(3) = 9a + 3b + 5 = 8\] \[r(6) = 36a + 6b + 5 = 12\] - From second: \[9a + 3b = 3\] - From third: \[36a + 6b = 7\] - Multiply second by 2: \[18a + 6b = 6\] - Subtract from third: \[(36a + 6b) - (18a + 6b) = 7 - 6 \Rightarrow 18a = 1 \Rightarrow a = \frac{1}{18}\] - Substitute \(a\) into second: \[9 \times \frac{1}{18} + 3b = 3 \Rightarrow \frac{1}{2} + 3b = 3 \Rightarrow 3b = \frac{5}{2} \Rightarrow b = \frac{5}{6}\] 8. **Quadratic rule:** \[r = \frac{1}{18} I^2 + \frac{5}{6} I + 5\] 9. **Find \(y\) when \(I=12\):** \[r = \frac{1}{18} \times 144 + \frac{5}{6} \times 12 + 5 = 8 + 10 + 5 = 23\] 10. **Summary:** - Arithmetic rule for \(I\) is \(I_n = 3(n-1)\). - \(x = I_4 = 9\). - Quadratic rule for \(r\) is \(r = \frac{1}{18} I^2 + \frac{5}{6} I + 5\). - \(y = r(12) = 23\).