Question

If the 7th term of an arithmetic sequence is 42 and the 15th term is 90, what is the first term?

Answer 1

To solve the problem, we need to first understand the formula for the \( n \)-th term of an arithmetic sequence, which is given by:

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

where \( a \) is the first term, \( d \) is the common difference, and \( n \) is the position of the term in the sequence.

Given the information from the problem:
- The 7th term (\( a_7 \)) is 42.
- The 15th term (\( a_{15} \)) is 90.

We can set up equations for these terms:

1. For the 7th term:
\[
a_7 = a + (7 - 1)d = a + 6d = 42 \tag{1}
\]

2. For the 15th term:
\[
a_{15} = a + (15 - 1)d = a + 14d = 90 \tag{2}
\]

Now we have a system of two equations:

\[
\begin{align*}
a + 6d &= 42 \quad \text{(1)} \\
a + 14d &= 90 \quad \text{(2)}
\end{align*}
\]

Next, we will solve this system of equations. We can subtract equation (1) from equation (2):

\[
(a + 14d) - (a + 6d) = 90 - 42
\]

This simplifies to:

\[
14d - 6d = 48
\]

\[
8d = 48
\]

Dividing both sides by 8 gives:

\[
d = 6
\]

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

\[
a + 6d = 42
\]

Substituting \( d = 6 \):

\[
a + 6(6) = 42
\]

\[
a + 36 = 42
\]

Subtracting 36 from both sides gives:

\[
a = 42 - 36 = 6
\]

Thus, the first term \( a \) of the arithmetic sequence is:

\[
\boxed{6}
\]

We can double-check our solution by calculating the 7th and 15th terms using \( a = 6 \) and \( d = 6 \):

- 7th term:
\[
a_7 = 6 + 6(6) = 6 + 36 = 42
\]

- 15th term:
\[
a_{15} = 6 + 14(6) = 6 + 84 = 90
\]

Both terms are correct, confirming that the first term is indeed \( \boxed{6} \).