2.2 The first four terms of a quadratic number pattern are \( -1 ; x ; 3 ; x+8 \) 2.2.1 Calculate the value(s) of \( x \). 2.2.2 \( \begin{array}{l}\text { If } x=0 \text {, determine the position of the first term in the quadratic number } \\ \text { pattern for which the sum of the first } n \text { first differences } \text { will be greater } \\ \text { than } 250 \text {. }\end{array} \).

To find the value of \( x \), we first utilize the property of quadratic sequences where the second differences are constant. The first four terms are \( -1, x, 3, x + 8 \). 1. To calculate \( x \): - The first differences are: - \( x - (-1) = x + 1 \) - \( 3 - x \) - \( (x + 8) - 3 = x + 5 \) - Now, the second differences calculated from the first differences would be: - \( (3 - x) - (x + 1) = 2 - 2x \) - \( (x + 5) - (3 - x) = 2x + 2 \) - Setting the second differences equal gives us: \[ 2 - 2x = 2x + 2 \] - Simplifying leads to: \[ -4x = 0 \] \[ x = 0 \] Now that we found \( x = 0 \): 2. Next, for the position of the first term where the sum of the first \( n \) first differences exceeds 250: - With \( x = 0 \), the sequence becomes \( -1, 0, 3, 8 \). - The first differences are \( 1, 3, 5 \). - The second differences remain constant at \( 2 \). - The sum of the first \( n \) first differences can be expressed as \( \text{Sum} = a_1 + (n-1)d + \frac{(n)(n-1)}{2} \cdot 2 \), where \( a_1 \) is the first difference and \( d \) is the second difference. - We seek \( n \) such that \( 1 + 2(n - 1) + n(n - 1) > 250 \). - Upon solving the quadratic equation, estimate or calculate \( n \) to determine the smallest integer such that the inequality holds. By setting up this quadratic equation for \( n \), the position corresponding to the first term in the quadratic pattern for which the sum exceeds 250 can be evaluated.