Determine the constant difference or the constant ratio of the following sequences and state whether the sequence is numeric or geometric.

    *   Pattern 1: 6 sweets
    *   Pattern 2: 10 sweets
    *   Pattern 3: 14 sweets

    The constant difference is $10-6=4$.  
    The rule for the $n$-th term is $T_n = a_1 + (n-1)d = 6 + (n-1)4 = 6 + 4n - 4 = 4n + 2$.

    *   For Pattern 10: $T_{10} = 4(10) + 2 = 40 + 2 = 42$
    *   For Pattern 48: $T_{48} = 4(48) + 2 = 192 + 2 = 194$

    *   Pattern 1: 11 toothpicks (3 bottom, 2 top, 4 vertical, 2 slanted)
    *   Pattern 2: 20 toothpicks (6 bottom, 4 top, 6 vertical, 4 slanted)
    *   Pattern 3: 29 toothpicks (9 bottom, 6 top, 8 vertical, 6 slanted)

    The constant difference is $20-11=9$.  
    The rule for the $n$-th term is $T_n = a_1 + (n-1)d = 11 + (n-1)9 = 11 + 9n - 9 = 9n + 2$.

    *   For Pattern 10: $T_{10} = 9(10) + 2 = 90 + 2 = 92$
    *   For Pattern 48: $T_{48} = 9(48) + 2 = 432 + 2 = 434$

$$\begin{array}{|c|c|c|c|c|c|}
\hline  
\text{Pattern number} & 1 & 2 & 3 & 10 & 48 \\
\hline  
\text{Number of sweets} & \textbf{6} & \textbf{10} & \textbf{14} & \textbf{42} & \textbf{194} \\
\hline  
\text{Number of toothpicks} & \textbf{11} & \textbf{20} & \textbf{29} & \textbf{92} & \textbf{434} \\
\hline  
\end{array}$$

    The difference between consecutive terms is $12-9=3$, $15-12=3$. The rule is to **add 3 to the previous term**.

    This is an arithmetic sequence with first term $a_1=9$ and common difference $d=3$.  
    The formula for the $n$-th term is $T_n = a_1 + (n-1)d$.
    $$
    T_n = 9 + (n-1)3
    $$
    $$
    T_n = 9 + 3n - 3
    $$
    $$
    T_n = \boxed{\textbf{3n + 6}}
    $$

    The difference between consecutive terms is $-7 - (-5) = -2$, $-9 - (-7) = -2$. The rule is to **subtract 2 from the previous term**.

    This is an arithmetic sequence with first term $a_1=-5$ and common difference $d=-2$.  
    The formula for the $n$-th term is $T_n = a_1 + (n-1)d$.
    $$
    T_n = -5 + (n-1)(-2)
    $$
    $$
    T_n = -5 - 2n + 2
    $$
    $$
    T_n = \boxed{\textbf{-2n - 3}}
    $$