In this problem, we are tasked with finding the 18th term of a sequence given the 4th term \( a_4 = -2 \) and the 6th term \( a_6 = 6 \). To approach this, we start by using the general formula for an arithmetic sequence, which is expressed as:

\[ a_n = a_1 + d(n - 1) \]

Here, \( a_1 \) is the first term and \( d \) is the common difference. Since we only have the values for \( a_4 \) and \( a_6 \), we can set up two equations based on the general formula:

\[ a_4 = a_1 + 3d = -2 \quad (1) \]

\[ a_6 = a_1 + 5d = 6 \quad (2) \]

Now, we have a system of equations with two unknowns, \( a_1 \) and \( d \). To solve this system, we can use the elimination method. By subtracting equation (1) from equation (2), we eliminate \( a_1 \):

\[ (a_1 + 5d) - (a_1 + 3d) = 6 - (-2) \]

\[ 2d = 8 \]

\[ d = 4 \]

Now that we have the common difference \( d = 4 \), we can substitute this value back into either equation to find \( a_1 \). Using equation (1):

\[ a_1 + 3(4) = -2 \]

\[ a_1 + 12 = -2 \]

\[ a_1 = -14 \]

With both \( a_1 \) and \( d \) known, we can now write the general formula for the sequence:

\[ a_n = -14 + 4(n - 1) \]

To find the 18th term \( a_{18} \), we substitute \( n = 18 \) into the general formula:

\[ a_{18} = -14 + 4(18 - 1) \]

\[ a_{18} = -14 + 4 \times 17 \]

\[ a_{18} = -14 + 68 \]

\[ a_{18} = 54 \]

Thus, the 18th term of the sequence is \( 54 \). This problem illustrates how to derive a general formula from known terms in an arithmetic sequence and subsequently use that formula to find other terms in the sequence.