QUESTION 2 2.1 The following sequence is a combination of arithmetic and a geometric sequence: \[ 2 ; 3 ; 4 ; 6 ; 6 ; 12 ; \ldots \] 2.1.1 Write down the next two terms of the sequence. 2.1.2 Calculate \( \mathrm{T}_{30}-\mathrm{T}_{31} \). 2.1.3 Write the geometric sequence in sigma notation. 2.2 The middle term of an arithmetic sequence of 31 terms is 50. Determine the sum of the series.

2.1.1 We observe that the given sequence is formed by interlacing an arithmetic sequence (for the odd‐numbered terms) and a geometric sequence (for the even‐numbered terms). For the odd-numbered terms: \[ T_1 = 2,\quad T_3 = 4,\quad T_5 = 6,\ldots \] The pattern is arithmetic with common difference \(2\). In general, for odd-indexed term number (where the \(k\)th odd term is \(T_{2k-1}\)): \[ T_{2k-1} = 2 + (k-1)\cdot 2 = 2k. \] Thus the next odd term, which is \(T_7\) (with \(k=4\)), is: \[ T_7 = 2 \times 4 = 8. \] For the even-numbered terms: \[ T_2 = 3,\quad T_4 = 6,\quad T_6 = 12,\ldots \] The pattern is geometric with common ratio \(2\). In general, for the even-indexed term (where the \(k\)th even term is \(T_{2k}\)): \[ T_{2k} = 3 \cdot 2^{k-1}. \] Thus the next even term, which is \(T_8\) (with \(k=4\)), is: \[ T_8 = 3 \cdot 2^{4-1} = 3 \cdot 2^{3} = 3 \cdot 8 = 24. \] The next two terms of the sequence are: \[ \boxed{8\quad \text{and}\quad 24.} \] 2.1.2 We need to calculate \(T_{30} - T_{31}\). - For \(T_{30}\): Since \(30\) is even, let \(2k = 30\) so that \(k = 15\). Then \[ T_{30} = 3 \cdot 2^{15-1} = 3 \cdot 2^{14}. \] - For \(T_{31}\): Since \(31\) is odd, express it as \(T_{2k-1}\) where \(2k-1 = 31\) so that \(k = 16\). Then \[ T_{31} = 2k = 2 \cdot 16 = 32. \] Thus, \[ T_{30} - T_{31} = 3 \cdot 2^{14} - 32. \] We can leave the answer in this form or simplify further. Calculating \(2^{14}\): \[ 2^{14} = 16384, \] so \[ T_{30} - T_{31} = 3 \cdot 16384 - 32 = 49152 - 32 = 49120. \] Therefore, \[ \boxed{49120}. \] 2.1.3 The geometric sequence is represented by the even-numbered terms: \[ 3,\, 6,\, 12,\, 24,\, \ldots \] In general, the \(k\)th term is given by: \[ T_{2k} = 3 \cdot 2^{k-1}. \] Thus, the sum of the first \(n\) terms of this geometric sequence can be written in sigma notation as: \[ \boxed{\sum_{k=1}^{n} 3 \cdot 2^{\,k-1}}. \] 2.2 Given that the arithmetic sequence has 31 terms and its middle term is 50, note that when there are an odd number of terms in an arithmetic sequence, the middle term is also the average of all terms. Since the arithmetic sequence has 31 terms, the sum \(S\) of the sequence is given by: \[ S = (\text{number of terms}) \times (\text{average term}) = 31 \times 50 = 1550. \] Thus, the sum of the series is: \[ \boxed{1550}. \]