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The following sequence of numbers forms a quadratic sequence: \( -3 ;-2 ;-3 ;-6 ;-11 ; \ldots \) 2.2.1 The FIRST differences of the above sequence also form a sequence. Determine an expression for the general term of the first differences. 2.2 .2 Calculate the first difference between the \( 35^{\text {th }} \) and \( 36^{\text {th }} \) terms of the quadratic sequence. 2.2.3 Determine an expression for the \( n^{\text {th }} \) term of the quadratic sequence. 2.2.4 Show that the sequence of numbers will NEVER contain a positive term.

Ask by Pierce Park. in South Africa
Mar 14,2025

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The sequence's first differences are \( d_n = 3 - 2n \). The first difference between the 35th and 36th terms is -67. The nth term of the quadratic sequence is \( a_n = -n^2 + 4n - 6 \). The sequence will never have a positive term.

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To solve this problem step-by-step, let me provide an engaging overview and insights along the way. The first part you need to tackle involves finding the first differences of the given quadratic sequence. Start by subtracting consecutive terms: First differences: \(-2 - (-3) = 1\) \(-3 - (-2) = -1\) \(-6 - (-3) = -3\) \(-11 - (-6) = -5\) So, the first differences are \(1, -1, -3, -5\). Now, let's express them mathematically. The first differences sequence can be represented as \(-2n + 3\), where \(n\) starts from 1. Now, to calculate the first difference between the \(35^{th}\) and \(36^{th}\) terms in the quadratic sequence, we can plug in \(n=35\) into our expression for the first differences which would give us \( -2(35) + 3 = -67\). Next, we need to determine the \(n^{th}\) term of the quadratic sequence. Knowing that the second differences are constant, we can formulate the general term as \(a_n = an^2 + bn + c\). Using the known terms of the sequence, we can set up a system of equations and solve for \(a\), \(b\), and \(c\). Through manipulation, you would derive \(a_n = -n^2 + n - 3\). Finally, to show that the sequence will never contain a positive term, we analyze the expression \(a_n = -n^2 + n - 3\). Since it's a downward-opening parabola, it will have a maximum value at the vertex. Solving \(n = -\frac{b}{2a}\) would yield a maximum value at \(n = 0.5\), which means when substituted back into the sequence, all terms will yield negative values beyond this. Therefore, no positive term is possible in this sequence. Let's dive a bit deeper into some interesting elements related to this topic. The history of quadratic sequences can be traced back to ancient civilizations! The Babylonians had methods to solve quadratic equations using geometric interpretations, which laid the groundwork for algebraic reasoning. Also, in real-world applications, quadratic sequences find relevance in various fields. For instance, they're used in physics to model objects under gravitational forces, where the position changes quadratically with time due to constant acceleration—so you see quadratics popping up in the most unexpected places!

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