In an arithmetic sequence, each term is generated by adding a common difference, denoted as \( d \), to the previous term. The general form of an arithmetic sequence can be expressed as:
\[ a_n = a_1 + (n-1)d \]
Where \( a_n \) is the nth term, \( a_1 \) is the first term, and \( d \) is the common difference.
Given \( a_{17} = -40 \) and \( a_{28} = -73 \), we can set up two equations using the formula for \( a_n \):
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For \( n = 17 \): \[ a_{17} = a_1 + 16d = -40 \tag{1} \]
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For \( n = 28 \): \[ a_{28} = a_1 + 27d = -73 \tag{2} \]
Now we have a system of two equations. We can subtract equation (1) from equation (2) to eliminate \( a_1 \) and solve for \( d \):
\[ (a_1 + 27d) - (a_1 + 16d) = -73 + 40 \] \[ 27d - 16d = -33 \] \[ 11d = -33 \] \[ d = -3 \]
Now, we can substitute \( d \) back into either equation (1) or equation (2) to find \( a_1 \). Using equation (1):
\[ a_1 + 16(-3) = -40 \] \[ a_1 - 48 = -40 \] \[ a_1 = -40 + 48 \] \[ a_1 = 8 \]
Now we have \( a_1 = 8 \) and \( d = -3 \).
To write a recursive formula for the sequence, we express it in terms of the previous term:
\[ a_n = a_{n-1} + d \]
Substituting our value for \( d \):
\[ a_n = a_{n-1} - 3 \]
Additionally, we need to specify the first term:
\[ \text{Recursive formula: } \begin{cases} a_1 = 8 \ a_n = a_{n-1} - 3 \text{ for } n > 1 \end{cases} \]
This shows how to define the terms of the arithmetic sequence recursively based on the previously defined term.