Let's denote the first term of the arithmetic series by [tex]\( a \)[/tex] and the common difference by [tex]\( d \)[/tex].
We know from the problem statement that:
- The 5th term [tex]\( a_5 \)[/tex] is 17.
- The 10th term [tex]\( a_{10} \)[/tex] is 42.
The [tex]\( n \)[/tex]-th term of an arithmetic series can be expressed using the formula:
[tex]\[ a_n = a + (n-1)d \][/tex]
For [tex]\( a_5 \)[/tex]:
[tex]\[ a + 4d = 17 \quad \text{(1)} \][/tex]
For [tex]\( a_{10} \)[/tex]:
[tex]\[ a + 9d = 42 \quad \text{(2)} \][/tex]
We need to solve these two equations to find [tex]\( a \)[/tex] and [tex]\( d \)[/tex].
First, let's eliminate [tex]\( a \)[/tex] by subtracting equation (1) from equation (2):
[tex]\[ (a + 9d) - (a + 4d) = 42 - 17 \][/tex]
[tex]\[ a + 9d - a - 4d = 25 \][/tex]
[tex]\[ 5d = 25 \][/tex]
[tex]\[ d = 5 \][/tex]
Now that we have [tex]\( d \)[/tex], we can substitute it back into equation (1) to find [tex]\( a \)[/tex]:
[tex]\[ a + 4 \cdot 5 = 17 \][/tex]
[tex]\[ a + 20 = 17 \][/tex]
[tex]\[ a = 17 - 20 \][/tex]
[tex]\[ a = -3 \][/tex]
Now that we have both [tex]\( a \)[/tex] and [tex]\( d \)[/tex], we can determine the 20th term [tex]\( a_{20} \)[/tex]:
[tex]\[ a_{20} = a + 19d \][/tex]
[tex]\[ a_{20} = -3 + 19 \cdot 5 \][/tex]
[tex]\[ a_{20} = -3 + 95 \][/tex]
[tex]\[ a_{20} = 92 \][/tex]
Hence, the 20th term of the arithmetic series is:
[tex]\[ \boxed{92} \][/tex]
