In an arithmetic progression (AP), the \( n \)-th term can be expressed as:
\[ a_n = a + (n - 1)d \]
where \( a \) is the first term and \( d \) is the common difference.
Given:
- The 4th term \( a_4 = 13 \)
- The 10th term \( a_{10} = 31 \)
We can write the equations for these terms using the formula:
-
For the 4th term: \[ a + 3d = 13 \quad \text{(1)} \]
-
For the 10th term: \[ a + 9d = 31 \quad \text{(2)} \]
Now we can solve these two equations. We'll subtract equation (1) from equation (2):
\[ (a + 9d) - (a + 3d) = 31 - 13 \]
This simplifies to:
\[ 6d = 18 \]
Dividing both sides by 6 gives:
\[ d = 3 \]
Now substituting \( d = 3 \) back into equation (1) to find \( a \):
\[ a + 3(3) = 13 \]
This simplifies to:
\[ a + 9 = 13 \]
So:
\[ a = 4 \]
Now that we have \( a = 4 \) and \( d = 3 \), we can find the 21st term using the formula:
\[ a_{21} = a + (21 - 1)d \] \[ a_{21} = 4 + 20(3) \] \[ a_{21} = 4 + 60 \] \[ a_{21} = 64 \]
Thus, the 21st term of the arithmetic progression is:
\[ \boxed{64} \]