The sum of the first n terms of an arithmetic sequence is 438 and the last term is 35. If the common difference is 7, find a sub 1 and n.

To solve for the first term \( a_1 \) and the number of terms \( n \) in the arithmetic sequence, we will use the following information:

- The sum of the first \( n \) terms of the arithmetic sequence is given as \( S_n = 438 \).
- The last term \( a_n = 35 \).
- The common difference \( d = 7 \).

We have the formulas for an arithmetic sequence:

1. **Sum of the first \( n \) terms**:
   \[
   S_n = \frac{n}{2} \times (a_1 + a_n)
   \]
2. **\( n \)-th term of an arithmetic sequence**:
   \[
   a_n = a_1 + (n - 1) \cdot d
   \]

### Step 1: Use the sum formula

We substitute \( S_n = 438 \), \( a_n = 35 \), and the formula for the sum:
\[
438 = \frac{n}{2} \times (a_1 + 35)
\]
Simplifying:
\[
438 = \frac{n}{2} \times (a_1 + 35) \implies 876 = n \times (a_1 + 35)
\]
\[
n(a_1 + 35) = 876 \quad \text{(Equation 1)}
\]

### Step 2: Use the formula for the \( n \)-th term

We know that the last term \( a_n = 35 \), so using the formula for the \( n \)-th term:
\[
35 = a_1 + (n - 1) \cdot 7
\]
Simplifying:
\[
35 = a_1 + 7n - 7

Learn how to find the first term and the number of terms in an arithmetic sequence where the sum of the first n terms is 438, the last term is 35, and the common difference is 7. Understand key algebraic concepts with detailed steps suitable for students in Grades 8-10.

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