Question

In an arithmetic sequence, a17 = -40 and
a28 = -73. Explain how to use this information to write a recursive formula for this sequence.

Answer 1

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 \( a_n \):

1. For \( n = 17 \):
\[
a_{17} = a_1 + 16d = -40 \tag{1}
\]

2. 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.