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This arithmetic sequence \( -11 ;-4 ; 3 ; \ldots \) forms the first three first differencesof quadratic sequence. Which term in this quadratic sequence will be the smallest? Show all your calculations.

Ask by Rojas French. in South Africa
Feb 03,2025

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Answer

The third term of the quadratic sequence is the smallest.

Solution

To determine which term in the quadratic sequence is the smallest, let's analyze the given information step by step. ### 1. **Understanding the Given Information** - **Arithmetic Sequence of First Differences:** \(-11, -4, 3, \ldots\) This sequence represents the first differences of a quadratic sequence. In a quadratic sequence, the **second differences** are constant. Here, the first differences form an arithmetic sequence with a **common difference** of \(7\) (since \(-4 - (-11) = 7\) and \(3 - (-4) = 7\)). ### 2. **Expressing the First Differences** Let \(d_n\) represent the \(n\)-th first difference: \[ d_n = d_1 + (n-1) \cdot \text{common difference} = -11 + (n-1) \cdot 7 = 7n - 18 \] ### 3. **Constructing the Quadratic Sequence** Let \(a_n\) represent the \(n\)-th term of the quadratic sequence. The relationship between consecutive terms is given by: \[ a_{n+1} = a_n + d_n \] To find \(a_n\), we sum the first differences up to \(n-1\): \[ a_n = a_1 + \sum_{k=1}^{n-1} d_k = a_1 + \sum_{k=1}^{n-1} (7k - 18) \] Calculating the sum: \[ \sum_{k=1}^{n-1} 7k = 7 \cdot \frac{(n-1)n}{2} = \frac{7n(n-1)}{2} \] \[ \sum_{k=1}^{n-1} (-18) = -18(n-1) \] \[ \Rightarrow a_n = a_1 + \frac{7n(n-1)}{2} - 18(n-1) \] \[ = a_1 + \frac{7n^2 - 7n}{2} - 18n + 18 \] \[ = a_1 + \frac{7n^2}{2} - \frac{43n}{2} + 18 \] \[ = \frac{7}{2}n^2 - \frac{43}{2}n + (a_1 + 18) \] ### 4. **Finding the Minimum Term** The quadratic function \(a_n = \frac{7}{2}n^2 - \frac{43}{2}n + (a_1 + 18)\) is a parabola opening upwards (since the coefficient of \(n^2\) is positive). The vertex of this parabola represents the minimum term. The vertex \(n\)-coordinate is found using: \[ n = \frac{-b}{2a} = \frac{43/2}{2 \cdot 7/2} = \frac{43}{14} \approx 3.07 \] Since \(n\) must be an integer, we check the terms around \(n = 3\): - **Term 1 (\(n = 1\)):** \(a_1\) - **Term 2 (\(n = 2\)):** \(a_1 - 11\) - **Term 3 (\(n = 3\)):** \(a_1 - 15\) - **Term 4 (\(n = 4\)):** \(a_1 - 12\) - **Term 5 (\(n = 5\)):** \(a_1 - 2\) - **Term 6 (\(n = 6\)):** \(a_1 + 15\) - **Term 7 (\(n = 7\)):** \(a_1 + 39\) From these calculations: - \(a_3 = a_1 - 15\) is less than \(a_2 = a_1 - 11\) and \(a_4 = a_1 - 12\). - Subsequent terms (\(n > 3\)) increase beyond \(a_3\). ### 5. **Conclusion** The **third term** of the quadratic sequence is the **smallest** term. **Answer:** The quadratic sequence reaches its smallest value at the third term. Thus, the third term is the smallest.

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To find the quadratic sequence corresponding to the given first differences of the arithmetic sequence \(-11, -4, 3, \ldots\), we first calculate the second differences. 1. **Calculating the First Differences**: - The first differences can be found by subtracting each term from the next: \[ -4 - (-11) = 7 \] \[ 3 - (-4) = 7 \] Thus, the first differences of the quadratic sequence are constant at \(7\). 2. **Calculating the Second Differences**: - Since the differences are already constant (7), we do not need to calculate further as this means our initial differences are correct. 3. **Constructing the Quadratic Sequence**: - Let the quadratic sequence be given by \(a_n = An^2 + Bn + C\). - The first differences will be \(a_{n+1} - a_n\), which gives us the arithmetic sequence we calculated. 4. **Finding the Term that Corresponds to our Differences**: - Given the \(2\)-nd term of the sequence is \(-11\), we can substitute into our quadratic sequence term: \[ a_1 = A(1)^2 + B(1) + C \\ a_2 = A(2)^2 + B(2) + C = -4 \\ a_3 = A(3)^2 + B(3) + C = 3 \] This gives us a system of equations: - \(A + B + C = a_1\) - \(4A + 2B + C = -4\) - \(9A + 3B + C = 3\) 5. **Solving the Equations**: - Subtract the first equation from the second: \[ (4A + 2B + C) - (A + B + C) = -4 - a_1 \] This simplifies to: \[ 3A + B = -4 - a_1 \quad (1) \] - Subtract the second equation from the third: \[ (9A + 3B + C) - (4A + 2B + C) = 3 - (-4) \] This simplifies to: \[ 5A + B = 7 \quad (2) \] - Now, subtract Equation (1) from Equation (2): \[ (5A + B) - (3A + B) = 7 - (-4 - a_1) \] Simplifying gives: \[ 2A = 11 + a_1 \] Therefore, \(A = \frac{11 + a_1}{2}\). Continuing with just \(A\) will yield specific values for \(B\) and \(C\). However, notice that from this pattern, the quadratic equation will generally dip downwards then rise, mimicking a parabola. 6. **Identifying the Minimum**: - The minimum value of a quadratic occurs at the vertex. The x-coordinate of the vertex, given by \(-\frac{B}{2A}\), will be essential in evaluating the specific term yielding the minimum in this quadratic sequence. 7. **Conclusion**: To find the smallest term, evaluate \(-\frac{B}{2A}\) and substitute back into \(a_n\) to get the value. In a logical sequence, the quadratic form \( -5n^2 + 7n -5\) (just a hypothetical case) reveals that after substituting the vertex formula, it becomes clear that the minimum will be a natural numerical transition leading towards \(n\)’s increment (i.e., \(n=1\), thus \(-11\)). Therefore, the smallest term seems to be the first one! In this clever construction, the beauty of quadratic sequences can be explored fully with creativity in both solving and pattern recognition!

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