Given the quadratic pattern \( 244,193,148,109 \) (1) Write down the neat term of puthen (2) Determine zormula for the inth term of the pattan? (3) Which term of the pattern will heve a value of soz? (4) Between which Two consecutive terms of the quadratic pattern will have zirst ditterence be. 453 (5) Show that all the terms quadratic pattern are positive

To analyze the quadratic pattern given by the sequence \( 244, 193, 148, 109 \), we will follow the steps outlined in your questions. ### (1) Write down the neat term of the pattern The sequence appears to be quadratic, so we can express the \( n \)-th term in the form: \[ a_n = An^2 + Bn + C \] ### (2) Determine the formula for the \( n \)-th term of the pattern To find the coefficients \( A \), \( B \), and \( C \), we can set up a system of equations using the known terms: 1. For \( n = 1 \): \[ A(1^2) + B(1) + C = 244 \quad \Rightarrow \quad A + B + C = 244 \quad \text{(1)} \] 2. For \( n = 2 \): \[ A(2^2) + B(2) + C = 193 \quad \Rightarrow \quad 4A + 2B + C = 193 \quad \text{(2)} \] 3. For \( n = 3 \): \[ A(3^2) + B(3) + C = 148 \quad \Rightarrow \quad 9A + 3B + C = 148 \quad \text{(3)} \] Now we can solve this system of equations. Subtract equation (1) from (2): \[ (4A + 2B + C) - (A + B + C) = 193 - 244 \] \[ 3A + B = -51 \quad \text{(4)} \] Subtract equation (2) from (3): \[ (9A + 3B + C) - (4A + 2B + C) = 148 - 193 \] \[ 5A + B = -45 \quad \text{(5)} \] Now, subtract equation (4) from (5): \[ (5A + B) - (3A + B) = -45 + 51 \] \[ 2A = 6 \quad \Rightarrow \quad A = 3 \] Substituting \( A = 3 \) back into equation (4): \[ 3(3) + B = -51 \] \[ 9 + B = -51 \quad \Rightarrow \quad B = -60 \] Now substitute \( A \) and \( B \) back into equation (1): \[ 3 - 60 + C = 244 \] \[ C = 244 + 60 - 3 = 301 \] Thus, the formula for the \( n \)-th term is: \[ a_n = 3n^2 - 60n + 301 \] ### (3) Which term of the pattern will have a value of 50? To find which term equals 50, we set the equation: \[ 3n^2 - 60n + 301 = 50 \] \[ 3n^2 - 60n + 251 = 0 \] Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ n = \frac{60 \pm \sqrt{(-60)^2 - 4 \cdot 3 \cdot 251}}{2 \cdot 3} \] \[ n = \frac{60 \pm \sqrt{3600 - 3012}}{6} \] \[ n = \frac{60 \pm \sqrt{588}}{6} \] \[ n = \frac{60 \pm 24.2}{6} \] Calculating the two possible values: \[ n_1 = \frac{84.2}{6} \approx 14.03 \quad \text{(not an integer)} \] \[ n_2 = \frac{35.8}{6} \approx 5.97 \quad \text{(not an integer)} \] Since neither value is an integer, there is no term in the sequence that equals 50. ### (4) Between which two consecutive terms of the quadratic pattern will the first difference be 453? The first difference is given by: \[ a_{n+1} - a_n = (3(n+1)^2 - 60(n+1) + 301) - (3n^2 - 60n + 301) \] \[ = 3((n^2 + 2n + 1) - n^2) - 60 \] \[ = 3(2n + 1) - 60 = 6n + 3 - 60 = 6n - 57 \] We want to find \( n \) such that: \[ 6n - 57 = 453 \] \[ 6n = 510 \quad \Rightarrow \quad n = 85 \] Thus, the first difference of 453 occurs between the 85th and 86th terms. ### (5) Show that all the terms of the quadratic pattern are positive To show that all terms are positive, we need to find the values of \( n \) for which \( a_n > 0 \): \[ 3n^2 - 60n + 301 > 0 \] Using the quadratic formula to find the roots: \[ n = \frac{60 \pm \sqrt{(-60)^2 - 4 \cdot 3 \cdot 301}}{2 \cdot 3} \] \[ = \frac{60 \pm \sqrt{3600 - 36012}}{6} \] \[ = \frac{60 \pm \sqrt{-324}}{6} \] Since the discriminant is negative, the quadratic has no real roots, meaning it does not cross the x-axis. Since the leading coefficient (3) is positive, the quadratic opens upwards and is always positive. Thus, all terms of the quadratic pattern are positive. ### Summary of Answers: 1. Neat term: \( a_n = 3n^2 - 60n + 301 \) 2. Formula for the \( n \)-th term: \( a_n = 3n^2 - 60n + 301 \) 3. No term equals 50. 4. The first difference of 453 occurs between the 85th and 86th terms. 5. All terms are positive.