(2) the quadratic pattern: \( -148 ;-92 ;-32 ; 32 ; \ldots \) d) Find the greatest term of the quadratic number pattern: \( 5199 ; 5396 ; 5591 ; 5784 ; 5975 \) Determine the position and value of the lowest term of the number pattern defined by \( T_{n}=\left(n-\frac{151}{4}\right)^{2}-1000 \). \( \quad \frac{151}{2} \) is the lodest In a linear number pattern, the 7 th term is 16 and the 20 th term is 42 . Determine the \( f \) terms.

Let's break down the problem step by step. ### Part 1: Finding the Greatest Term of the Quadratic Number Pattern The given quadratic number pattern is: \[ 5199, 5396, 5591, 5784, 5975 \] To find the greatest term, we can observe the pattern and check the differences between consecutive terms: 1. \( 5396 - 5199 = 197 \) 2. \( 5591 - 5396 = 195 \) 3. \( 5784 - 5591 = 193 \) 4. \( 5975 - 5784 = 191 \) The differences are decreasing, indicating that the sequence is quadratic. The last term, \( 5975 \), is the greatest term in this pattern. ### Part 2: Finding the Position and Value of the Lowest Term of the Number Pattern The number pattern is defined by: \[ T_n = \left(n - \frac{151}{4}\right)^2 - 1000 \] To find the lowest term, we need to determine the vertex of the quadratic function, which occurs at: \[ n = \frac{151}{4} \] Calculating this value: \[ n = 37.75 \] Since \( n \) must be an integer, we will check the values at \( n = 37 \) and \( n = 38 \). 1. For \( n = 37 \): \[ T_{37} = \left(37 - \frac{151}{4}\right)^2 - 1000 = \left(37 - 37.75\right)^2 - 1000 = (-0.75)^2 - 1000 = 0.5625 - 1000 = -999.4375 \] 2. For \( n = 38 \): \[ T_{38} = \left(38 - \frac{151}{4}\right)^2 - 1000 = \left(38 - 37.75\right)^2 - 1000 = (0.25)^2 - 1000 = 0.0625 - 1000 = -999.9375 \] The lowest term occurs at \( n = 38 \) with a value of \( -999.9375 \). ### Part 3: Finding the Linear Number Pattern In a linear number pattern, the 7th term is 16 and the 20th term is 42. We can express the \( n \)-th term of a linear sequence as: \[ T_n = a + (n-1)d \] where \( a \) is the first term and \( d \) is the common difference. From the information given: 1. \( T_7 = a + 6d = 16 \) 2. \( T_{20} = a + 19d = 42 \) We can set up the following system of equations: 1. \( a + 6d = 16 \) (Equation 1) 2. \( a + 19d = 42 \) (Equation 2) Now, we can solve this system of equations. Subtract Equation 1 from Equation 2: \[ (a + 19d) - (a + 6d) = 42 - 16 \] \[ 13d = 26 \implies d = 2 \] Now substitute \( d = 2 \) back into Equation 1: \[ a + 6(2) = 16 \implies a + 12 = 16 \implies a = 4 \] Thus, the \( n \)-th term of the linear pattern is: \[ T_n = 4 + (n-1) \cdot 2 = 2n + 2 \] ### Summary of Results 1. The greatest term of the quadratic number pattern is \( 5975 \). 2. The lowest term of the number pattern \( T_n = \left(n - \frac{151}{4}\right)^2 - 1000 \) occurs at \( n = 38 \) with a value of \( -999.9375 \). 3. The linear number pattern is defined by \( T_n = 2n + 2 \).