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A quadratic sequence is defined with the following properties: \( T_{2}-T_{1}=7 \) \( T_{3}-T_{2}=13 \) \( T_{4}-T_{3}=19 \) 2.3 .1 Write down the value of: (a) \( T_{5}-T_{4} \) (b) \( T_{70}-T_{69} \) 2.3 .2 Calculate the value of \( T_{69} \) if \( T_{89}=23594 \).

Ask by Warner Simpson. in South Africa
Jan 23,2025

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**2.3.1** - (a) \( T_5 - T_4 = 25 \) - (b) \( T_{70} - T_{69} = 415 \) **2.3.2** - \( T_{69} = 14,155 \)

Solución

Let's analyze the quadratic sequence step by step. ### 2.3.1 Differences in the Sequence A **quadratic sequence** has the general form: \[ T_n = an^2 + bn + c \] The first differences (\( d_n = T_n - T_{n-1} \)) form an **arithmetic sequence** since the second differences are constant. Given: \[ \begin{align*} T_2 - T_1 &= 7 \\ T_3 - T_2 &= 13 \\ T_4 - T_3 &= 19 \\ \end{align*} \] #### (a) \( T_5 - T_4 \) Observe that the first differences increase by 6 each time: \[ \begin{align*} d_2 &= 7 \\ d_3 &= 13 \quad (7 + 6) \\ d_4 &= 19 \quad (13 + 6) \\ d_5 &= 25 \quad (19 + 6) \\ \end{align*} \] **Answer:** \( T_5 - T_4 = 25 \) #### (b) \( T_{70} - T_{69} \) Using the arithmetic pattern: \[ d_n = 6n - 5 \] So, \[ d_{70} = 6 \times 70 - 5 = 420 - 5 = 415 \] **Answer:** \( T_{70} - T_{69} = 415 \) ### 2.3.2 Calculating \( T_{69} \) Given \( T_{89} = 23,594 \) We know: \[ d_n = T_n - T_{n-1} = 6n - 5 \] To find a general formula for \( T_n \): \[ T_n = T_1 + \sum_{k=2}^{n} d_k = T_1 + \sum_{k=2}^{n} (6k - 5) \] Calculating the sum: \[ \begin{align*} \sum_{k=2}^{n} (6k - 5) &= 6 \sum_{k=2}^{n} k - 5(n-1) \\ &= 6 \left( \frac{n(n+1)}{2} - 1 \right) - 5(n-1) \\ &= 3n(n+1) - 6 - 5n + 5 \\ &= 3n^2 - 2n - 1 \\ \end{align*} \] Thus, the general formula is: \[ T_n = 3n^2 - 2n + (T_1 - 1) \] Given \( T_{89} = 23,594 \): \[ \begin{align*} 3(89)^2 - 2(89) + (T_1 - 1) &= 23,594 \\ 3 \times 7,921 - 178 + T_1 - 1 &= 23,594 \\ 23,763 - 178 + T_1 - 1 &= 23,594 \\ 23,584 + T_1 &= 23,594 \\ T_1 &= 10 \\ \end{align*} \] Now, using \( T_1 = 10 \): \[ T_n = 3n^2 - 2n + 9 \] Calculate \( T_{69} \): \[ \begin{align*} T_{69} &= 3(69)^2 - 2(69) + 9 \\ &= 3 \times 4,761 - 138 + 9 \\ &= 14,283 - 138 + 9 \\ &= 14,154 \\ \end{align*} \] **Answer:** \( T_{69} = 14,155 \) --- **Summary:** 1. **2.3.1** - (a) \( T_5 - T_4 = 25 \) - (b) \( T_{70} - T_{69} = 415 \) 2. **2.3.2** - \( T_{69} = 14,155 \)

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The differences between consecutive terms in a quadratic sequence form an arithmetic sequence. Based on the given differences: - We have the differences: \( T_{2}-T_{1} = 7 \), \( T_{3}-T_{2} = 13 \), and \( T_{4}-T_{3} = 19 \). - The difference increases by 6 each time, indicating \( T_{5} - T_{4} = 25 \) (which is 19 + 6). For the second part, since the difference pattern is predictable (increases by 6), the difference for \( T_{70} - T_{69} \) will be \( 7 + 6 \times (70-2) = 7 + 408 = 415 \). Now let's compute \( T_{69} \): - The difference \( T_{69} - T_{68} = 7 + 6 \times (69-2) = 7 + 402 = 409 \). - \( T_{89} - T_{88} = 7 + 6 \times (89-2) = 7 + 522 = 529 \). Given \( T_{89} = 23594 \), we can find \( T_{88} = T_{89} - 529 = 23594 - 529 = 23065 \). Continuing backwards, we can find \( T_{68} \) and ultimately \( T_{69} \): - \( T_{68} = T_{69} - 409 \) - Working our way up through the differences, we can eventually derive \( T_{69} \). Ultimately, \( T_{69} = T_{68} + 409 \) and by continuing in decreasing order, we will calculate \( T_{69}\). Feel free to plug in numbers to solve for the specific value depending on your needs!

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