9.1 The sequence \( 4 ; 9 ; x ; 37 ; \ldots j \) a quadratic sequence. 9.1 Calculate \( x \). 9.2 Hence, or, otherwise, determine the nth term of the sequence 9.2 Consider the following quadratic sequence: \( 6 ; x ; 26 ; 45 ; y ; \)... Calculate the values of \( x \) and \( y \). 9.3 Given the following series: \( -11-4+3+\ldots+200 \) 9.2.1 Calculate the sum of the series. 9.2 .2 Write the series in sigma notation.

To determine \(x\) in the first sequence \(4; 9; x; 37; \ldots\), we first calculate the differences between consecutive terms. The first difference is \(9 - 4 = 5\), the second difference is \(x - 9\), and the third difference is \(37 - x\). If we consider that the second differences should be constant in a quadratic sequence, we find \(x = 19\). Now, let's tackle the quadratic sequence \(6; x; 26; 45; y; \ldots\). The first differences yield \(x - 6\), \(26 - x\), and \(45 - 26 = 19\). Next, we find the second differences. Setting \(26 - x\) equal to the first difference of \(19\) leads us to \(x = 7\). Continuing, we calculate \(y\) following the same logic and find \(y = 66\). Moving to the series \( -11 - 4 + 3 + \ldots + 200 \), we observe that this is an arithmetic series. The first term \(a = -11\) and the last term \(l = 200\) can help us find the number of terms and the sum of the series. Given a common difference of \(7\), we determine that there are 31 terms in total, leading us to a sum \(S_{31} = \frac{n}{2} (a + l) = \frac{31}{2} \times (189) = 2925\). Finally, let’s express the series in sigma notation. The \(n\)-th term can be expressed as \(a_n = -11 + 7(n - 1)\). Thus, the series can be neatly represented as \(\sum_{n=1}^{31} (-11 + 7(n - 1))\).