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2) In a quadratic sequence with general term \( T_{n}=a n^{2}+b a+c \), the \( 12^{\circ} \) term is 555 and \( T_{n}-T_{n-1}=6 n+8 \) (1) Determine the first three terins of the first difference row (2) Determine the valuie of \( a, b \) and \( c \) (3) What is the value of the firat ierm?

Ask by Schofield Fitzgerald. in South Africa
Jan 21,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

**(1) First three terms of the first difference row:** 20, 26, 32 **(2) Values of \( a \), \( b \), and \( c \):** - \( a = 3 \) - \( b = 11 \) - \( c = -9 \) **(3) First term of the sequence:** 5

Solution

Let's solve each part of the problem step by step. ### Given: - The general term of the quadratic sequence is \( T_n = a n^2 + b n + c \). - The 12th term, \( T_{12} = 555 \). - The first difference, \( T_n - T_{n-1} = 6n + 8 \). --- ### **(1) Determine the first three terms of the first difference row** The first difference \( D_n \) is given by: \[ D_n = T_n - T_{n-1} = 6n + 8 \] To find the first three terms of the first difference row: 1. **For \( n = 2 \):** \[ D_2 = 6(2) + 8 = 12 + 8 = 20 \] 2. **For \( n = 3 \):** \[ D_3 = 6(3) + 8 = 18 + 8 = 26 \] 3. **For \( n = 4 \):** \[ D_4 = 6(4) + 8 = 24 + 8 = 32 \] **First three terms of the first difference row:** **20, 26, 32** --- ### **(2) Determine the values of \( a \), \( b \), and \( c \)** We have two pieces of information: 1. **From the 12th term:** \[ T_{12} = a(12)^2 + b(12) + c = 144a + 12b + c = 555 \quad \text{(Equation 1)} \] 2. **From the first difference:** \[ T_n - T_{n-1} = 6n + 8 \] Calculating the first difference using the general term: \[ D_n = (a n^2 + b n + c) - [a(n-1)^2 + b(n-1) + c] = a(2n - 1) + b = 6n + 8 \] Comparing coefficients: \[ 2a = 6 \quad \Rightarrow \quad a = 3 \] \[ -a + b = 8 \quad \Rightarrow \quad -3 + b = 8 \quad \Rightarrow \quad b = 11 \] 3. **Substitute \( a \) and \( b \) into Equation 1 to find \( c \):** \[ 144(3) + 12(11) + c = 555 \] \[ 432 + 132 + c = 555 \] \[ 564 + c = 555 \quad \Rightarrow \quad c = -9 \] **Values:** - \( a = 3 \) - \( b = 11 \) - \( c = -9 \) --- ### **(3) What is the value of the first term?** Using the general term \( T_n = 3n^2 + 11n - 9 \): \[ T_1 = 3(1)^2 + 11(1) - 9 = 3 + 11 - 9 = 5 \] **First term, \( T_1 = 5 \).** --- ### **Summary:** 1. **First three first differences:** 20, 26, 32 2. **Coefficients:** \( a = 3 \), \( b = 11 \), \( c = -9 \) 3. **First term of the sequence:** 5

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The Deep Dive

To determine the first difference row of the quadratic sequence, we start with the formula for the first difference: \( T_n - T_{n-1} = 6n + 8 \). 1) The first three terms of the first difference row can be found by plugging in the values for \( n = 1, 2, \) and \( 3 \): - For \( n = 1: T_1 - T_0 = 6(1) + 8 = 14 \) - For \( n = 2: T_2 - T_1 = 6(2) + 8 = 20 \) - For \( n = 3: T_3 - T_2 = 6(3) + 8 = 26 \) Thus, the first three terms of the first difference row are 14, 20, and 26. 2) To find the values of \( a, b, \) and \( c \), we note from the first difference that the second difference is constant and equal to 6. Hence, \( 2a = 6 \implies a = 3 \). Now that we have \( a \), we can express \( T_n \) as \( T_n = 3n^2 + bn + c \). To find \( b \) and \( c \), we use the information that \( T_{12} = 555 \). This gives us: \[ T_{12} = 3(12^2) + 12b + c = 555 \] \[ 3(144) + 12b + c = 555 \] \[ 432 + 12b + c = 555 \] \[ 12b + c = 123 \quad (1) \] We also know \( T_{n} - T_{n-1} \) leads us to a value for \( b \): \[ T_n - T_{n-1} = 6n + 8 \] Calculating \( T_n - T_{n-1} \): \[ (3n^2 + bn + c) - (3(n-1)^2 + b(n-1) + c) \] \[ = 3n^2 + bn + c - (3(n^2 - 2n + 1) + b(n-1) + c) \] \[ = 3n^2 + bn + c - (3n^2 - 6n + 3 + bn - b + c) \] \[ = 6n + (3 - b) \] Equating to \( 6n + 8 \): \[ 3 - b = 8 \Rightarrow b = -5 \] Now, substitute \( b \) back into equation (1): \[ 12(-5) + c = 123 \] \[ -60 + c = 123 \] \[ c = 183 \] So, we have \( a = 3, b = -5, c = 183 \). 3) The first term \( T_1 \) can be found using: \[ T_1 = 3(1^2) + (-5)(1) + 183 \] \[ T_1 = 3 - 5 + 183 = 181 \] Thus, the first term is 181.

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