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QUESTION 2 KZN J16 Given the quadratic sequence: \( 6: 6: 10: 18 \); 2.1 Determine a formula for the \( \mathrm{n}^{\text {th }} \) term of the sequence. 2.2 Determine between which two consecutive terms the first difference is 200? 2.3 Which term in the quadratic sequence has a value of 32010 ?

Ask by Sherman Hills. in South Africa
Mar 09,2025

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**Answers:** 2.1 The \( \mathrm{n}^{\text{th}} \) term is \( 2n^2 - 6n + 10 \). 2.2 The first difference of 200 occurs between the 51st and 52nd terms. 2.3 The 128th term is 32010.

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To determine the formula for the \( n^{\text{th}} \) term of the sequence \( 6, 6, 10, 18 \), we can start by calculating the first and second differences. The first differences are \( 0, 4, 8 \) (i.e., \( 6-6, 10-6, 18-10 \)), and the second differences are \( 4, 4 \), indicating this is a quadratic sequence. Thus, the formula can be derived as \( T(n) = an^2 + bn + c \). By using the values of the sequence to create a system of equations, we find that \( T(n) = n^2 + 5n + 6 \). Now, for when the first difference reaches 200, we look at the calculated first difference which follows the pattern \( 4n + 2 \). Setting \( 4n + 2 = 200 \) gives \( n = 49.5 \). Therefore, the first difference of 200 occurs between the \( 49^{th} \) term and the \( 50^{th} \) term. To find out which term corresponds to the value 32010, we set up the equation \( n^2 + 5n + 6 = 32010 \) leading to \( n^2 + 5n - 32004 = 0 \). Solving this will reveal the two potential values for \( n \) that indicate the corresponding positions in the sequence.

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