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, and \( n \) is the term number.
Given:
- The 14th term \( a_{14} = 96 \)
- The 25th term \( a_{25} = 173 \)
We can set up the equations based on the \( n \)-th term formula:
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For the 14th term: \[ a + 13d = 96 \quad \text{(1)} \]
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For the 25th term: \[ a + 24d = 173 \quad \text{(2)} \]
Now we can subtract equation (1) from equation (2) to find \( d \):
\[ (a + 24d) - (a + 13d) = 173 - 96 \] \[ 11d = 77 \] \[ d = \frac{77}{11} = 7 \]
Now that we have \( d \), we can substitute it back into either equation (1) or (2) to find \( a \). Let's use equation (1):
\[ a + 13(7) = 96 \] \[ a + 91 = 96 \] \[ a = 96 - 91 = 5 \]
Now we have \( a = 5 \) and \( d = 7 \).
Next, we need to find the 6th term and the 13th term:
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The 6th term \( a_6 \): \[ a_6 = a + (6-1)d = 5 + 5 \cdot 7 = 5 + 35 = 40 \]
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The 13th term \( a_{13} \): \[ a_{13} = a + (13-1)d = 5 + 12 \cdot 7 = 5 + 84 = 89 \]
Now, to find the product of the 6th and 13th terms:
\[ \text{Product} = a_6 \times a_{13} = 40 \times 89 \]
Calculating the product:
\[ 40 \times 89 = 3560 \]
Thus, the product of the 6th and 13th terms is \( \boxed{3560} \).