Question

an arithmetic sequence has a second term of 9 and a fourth term of 21. what is the 40th term in the sequence

Answer 1

In an arithmetic sequence, each term can be expressed as:

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

where \( a_n \) is the nth term, \( a_1 \) is the first term, \( d \) is the common difference, and \( n \) is the term number.

Given:
- The second term \( a_2 \) is 9.
- The fourth term \( a_4 \) is 21.

We can set up the equations for the second and fourth terms using the formula:

1. For \( a_2 \):
\[
a_1 + d = 9 \quad (1)
\]

2. For \( a_4 \):
\[
a_1 + 3d = 21 \quad (2)
\]

Now, we can solve these equations simultaneously. First, we'll eliminate \( a_1 \) by rearranging equation (1):

From (1):
\[
a_1 = 9 - d
\]

Now substitute \( a_1 \) into equation (2):

\[
(9 - d) + 3d = 21
\]

Combine like terms:

\[
9 + 2d = 21
\]

Now, isolate \( d \):

\[
2d = 21 - 9
\]
\[
2d = 12
\]
\[
d = 6
\]

Now that we have \( d \), we can substitute it back into equation (1) to find \( a_1 \):

\[
a_1 + 6 = 9
\]
\[
a_1 = 9 - 6 = 3
\]

Now we have \( a_1 = 3 \) and \( d = 6 \).

To find the 40th term \( a_{40} \), we use the formula:

\[
a_{40} = a_1 + (40-1)d
\]
\[
a_{40} = 3 + 39 \cdot 6
\]
\[
a_{40} = 3 + 234
\]
\[
a_{40} = 237
\]

Thus, the 40th term in the sequence is:

\[
\boxed{237}
\]