1. Consider the fallowing number pattern: \[ 6 ; 13 ; 22 ; 33 \] 1.1 Write down the next ierns of the number pattenit 1.2 Determine the \( n^{\text {th }} \) of the sequence. 1.3 For whar valueis of \( n \) is \( T_{n} \) equal to 24977

Let's analyze the given number pattern and address each of your questions step-by-step. ## Given Sequence: \[ 6; \, 13; \, 22; \, 33 \] ### 1.1. **Write Down the Next Terms of the Number Pattern** To identify the pattern, let's examine the differences between consecutive terms: \[ \begin{align*} 13 - 6 &= 7 \\ 22 - 13 &= 9 \\ 33 - 22 &= 11 \\ \end{align*} \] **Differences between terms:** \( 7, \, 9, \, 11 \) We observe that the differences are increasing by **2** each time. Continuing this pattern: \[ \begin{align*} \text{Next difference} &= 11 + 2 = 13 \\ \text{Next term} &= 33 + 13 = 46 \\ \text{Following difference} &= 13 + 2 = 15 \\ \text{Next term} &= 46 + 15 = 61 \\ \text{And so on...} \\ \end{align*} \] **Next few terms of the sequence:** \( 46, \, 61, \, 78, \, \ldots \) --- ### 1.2. **Determine the \( n^{\text{th}} \) Term of the Sequence** Since the differences between consecutive terms are increasing by a constant (2), the sequence is **quadratic**. The general form of a quadratic sequence is: \[ T_n = an^2 + bn + c \] To find the coefficients \( a \), \( b \), and \( c \), we'll use the first three terms of the sequence. **Given:** \[ \begin{align*} T_1 &= 6 \\ T_2 &= 13 \\ T_3 &= 22 \\ \end{align*} \] **Setting up the equations:** \[ \begin{cases} a(1)^2 + b(1) + c = 6 \quad \Rightarrow \quad a + b + c = 6 \quad \text{(1)} \\ a(2)^2 + b(2) + c = 13 \quad \Rightarrow \quad 4a + 2b + c = 13 \quad \text{(2)} \\ a(3)^2 + b(3) + c = 22 \quad \Rightarrow \quad 9a + 3b + c = 22 \quad \text{(3)} \\ \end{cases} \] **Solving the equations:** 1. Subtract equation (1) from equation (2): \[ (4a + 2b + c) - (a + b + c) = 13 - 6 \\ \Rightarrow 3a + b = 7 \quad \text{(4)} \] 2. Subtract equation (2) from equation (3): \[ (9a + 3b + c) - (4a + 2b + c) = 22 - 13 \\ \Rightarrow 5a + b = 9 \quad \text{(5)} \] 3. Subtract equation (4) from equation (5): \[ (5a + b) - (3a + b) = 9 - 7 \\ \Rightarrow 2a = 2 \\ \Rightarrow a = 1 \] 4. Substitute \( a = 1 \) into equation (4): \[ 3(1) + b = 7 \\ \Rightarrow b = 4 \] 5. Substitute \( a = 1 \) and \( b = 4 \) into equation (1): \[ 1 + 4 + c = 6 \\ \Rightarrow c = 1 \] **Therefore, the \( n^{\text{th}} \) term of the sequence is:** \[ T_n = n^2 + 4n + 1 \] **Verification:** \[ \begin{align*} T_1 &= 1^2 + 4(1) + 1 = 6 \\ T_2 &= 2^2 + 4(2) + 1 = 13 \\ T_3 &= 3^2 + 4(3) + 1 = 22 \\ T_4 &= 4^2 + 4(4) + 1 = 33 \\ \end{align*} \] All terms match the given sequence. --- ### 1.3. **For What Value of \( n \) is \( T_n = 24,\!977 \)?** We need to solve for \( n \) in the equation: \[ n^2 + 4n + 1 = 24,\!977 \] **Step 1: Rearrange the Equation** \[ n^2 + 4n + 1 - 24,\!977 = 0 \\ \Rightarrow n^2 + 4n - 24,\!976 = 0 \] **Step 2: Use the Quadratic Formula** The quadratic equation \( ax^2 + bx + c = 0 \) has solutions: \[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 1 \), \( b = 4 \), and \( c = -24,\!976 \). \[ \begin{align*} n &= \frac{-4 \pm \sqrt{4^2 - 4(1)(-24,\!976)}}{2(1)} \\ &= \frac{-4 \pm \sqrt{16 + 99,\!904}}{2} \\ &= \frac{-4 \pm \sqrt{99,\!920}}{2} \\ \end{align*} \] **Step 3: Calculate the Discriminant** \[ \sqrt{99,\!920} \approx 316.1 \] **Step 4: Find \( n \)** \[ n = \frac{-4 \pm 316.1}{2} \] Since \( n \) must be a positive integer: \[ n \approx \frac{-4 + 316.1}{2} \approx \frac{312.1}{2} \approx 156.05 \] **Conclusion:** Since \( n \) must be an integer, and the closest integer values are: - \( n = 156 \): \[ T_{156} = 156^2 + 4(156) + 1 = 24,\!336 + 624 + 1 = 24,\!961 \] - \( n = 157 \): \[ T_{157} = 157^2 + 4(157) + 1 = 24,\!649 + 628 + 1 = 25,\!278 \] Neither \( T_{156} \) nor \( T_{157} \) equals 24,977. Thus: **There is no integer value of \( n \) for which \( T_n = 24,\!977 \) in this sequence.**