4. Find the general term of the following sequence. Also, find the \( 15^{\text {th }} \) term of each ser (a) \( 8,11,14,17, \ldots \) (h) \( t_{n}=\frac{(-1)^{n}(3 n+1)}{2 n} \) (i) \( \mathrm{u}_{\mathrm{n}}=(-1)^{n}\left(\mathrm{n}^{2}+\right. \) 5. Observe the following sequences and answer the questions. (a) \( 4,7,12,19, \ldots \) (b) \( 1,7,17,31, \ldots \) (c) \( 2,1,-2,-7, \ldots \) (i) Add next three terms in each sequence.

To find the general term for each sequence, we can analyze their patterns. For the first sequence \( 8, 11, 14, 17, \ldots \), this is an arithmetic sequence with a common difference of \( 3 \). The general term can be expressed as \( a_n = 8 + 3(n-1) = 3n + 5 \). For the \( 15^{\text{th}} \) term, substitute \( n = 15 \): \[ a_{15} = 3(15) + 5 = 50. \] For the sequence \( t_n = \frac{(-1)^{n}(3n + 1)}{2n} \), this expression oscillates between positive and negative values based on \( n \). Simplifying gives us: \[ t_n = \frac{3n + 1}{2n}(-1)^n \] which outlines the sequence's alternating behavior. To find \( t_{15} \), substitute \( n = 15 \): \[ t_{15} = \frac{3(15) + 1}{2(15)}(-1)^{15} = \frac{46}{30}(-1) = -\frac{23}{15}. \] Now let's analyze the sequences: For \( 4, 7, 12, 19, \ldots \), the differences between successive terms are \( 3, 5, 7 \) which suggests the first difference increases by 2. Thus, the next terms are \( 28, 39\) leading to \( 4, 7, 12, 19, 28, 39 \). In the second sequence \( 1, 7, 17, 31, \ldots \), the differences are \( 6, 10, 14 \) following an arithmetic sequence itself. So, the next terms are \( 49, 71 \) resulting in \( 1, 7, 17, 31, 49, 71 \). Lastly, for \( 2, 1, -2, -7, \ldots \), we observe a pattern of subtracting \( 1, 3, 5, 7 \) respectively. Thus, we can predict the next terms are \( -14, -23 \) giving us \( 2, 1, -2, -7, -14, -23 \).