Question

Find a general function for an arithmetic sequence with a1=11
and a9=−13.



an=11+(n−1)(−2)


an=11+(n−1)(−33)


an=11+(n−1)(−3)


an=11+(n−1)(3)

Answer 1

To find the general function for the arithmetic sequence given \( a_1 = 11 \) and \( a_9 = -13 \), we start with the formula for the \( n \)-th term of an arithmetic sequence:

\[
a_n = a_1 + (n-1) \cdot d
\]

where \( d \) is the common difference.

From the given information:
- \( a_1 = 11 \)
- \( a_9 = -13 \)

Using the formula for \( a_9 \):

\[
a_9 = 11 + (9-1) \cdot d
\]

This simplifies to:

\[
-13 = 11 + 8d
\]

Now, solve for \( d \):

1. Subtract 11 from both sides:
\[
-13 - 11 = 8d \\
-24 = 8d
\]

2. Divide by 8:
\[
d = -3
\]

Now that we have the common difference \( d = -3 \), we can substitute it back into the formula for \( a_n \):

\[
a_n = 11 + (n-1)(-3)
\]

Therefore, the general function for the arithmetic sequence is:

\[
a_n = 11 - 3(n-1)
\]

We can also rewrite it:

\[
a_n = 11 - 3n + 3 = 14 - 3n
\]

But since the choices provided seem to be formatted differently, we will stick with the format that incorporates the initial term and the common difference.

Thus, the answer is:

\[
a_n = 11 + (n-1)(-3)
\]

As per the options given, this corresponds to:

\[
\boxed{a_n = 11 + (n-1)(-3)}
\]