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,
- \(d\) is the common difference,
- \(n\) is the term number.
Given:
- The 7th term \(a_7 = 34\)
- The 20th term \(a_{20} = 112\)
Using the formula for the \(n\)th term, we can write the following equations:
-
For the 7th term: \[ a + 6d = 34 \quad \text{(1)} \]
-
For the 20th term: \[ a + 19d = 112 \quad \text{(2)} \]
Now, we can solve these two equations simultaneously.
Subtract equation (1) from equation (2):
\[ (a + 19d) - (a + 6d) = 112 - 34 \]
This simplifies to:
\[ 13d = 78 \]
Now, solve for \(d\):
\[ d = \frac{78}{13} = 6 \]
Now that we have \(d\), we can substitute it back into equation (1) to find \(a\):
\[ a + 6(6) = 34 \] \[ a + 36 = 34 \] \[ a = 34 - 36 = -2 \]
Now we have both the first term \(a\) and the common difference \(d\):
- First term \(a = -2\)
- Common difference \(d = 6\)
The general formula for the \(n\)th term of this AP can now be written as:
\[ a_n = -2 + (n-1) \cdot 6 \]
To summarize, the AP is:
\[ -2, 4, 10, 16, 22, 28, 34, 40, 46, 52, 58, 64, 70, 76, 82, 88, 94, 100, 106, 112, \ldots \]