The sum of the first n terms of an arithmetic sequence is 438 and the last term is 75. If the common difference is 7, find a sub 1 and n.

We are given the following information about an arithmetic sequence:

- The sum of the first \( n \) terms, \( S_n = 438 \),
- The last term, \( a_n = 75 \),
- The common difference, \( d = 7 \).

We need to find the first term \( a_1 \) and the number of terms \( n \).

### Step 1: Formula for the sum of an arithmetic sequence
The formula for the sum of the first \( n \) terms of an arithmetic sequence is:

\[
S_n = \frac{n}{2} \times (a_1 + a_n)
\]

Substituting the given values:

\[
438 = \frac{n}{2} \times (a_1 + 75)
\]

### Step 2: Expressing \( a_n \) in terms of \( a_1 \) and \( n \)
The formula for the \( n \)-th term of an arithmetic sequence is:

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

Substituting \( a_n = 75 \) and \( d = 7 \):

\[
75 = a_1 + (n - 1) \times 7
\]

Simplifying:

\[
75 = a_1 + 7n - 7
\]

\[
75 = a_1 + 7n - 7
\]

\[
a_1 = 82 - 7n
\]

### Step 3: Substituting into the sum formula
Now, substitute \( a_1 = 82 - 7n \) into the sum formula:

\[
438 = \frac{n}{2} \times ((82 - 7n) + 75)
\]

Simplifying:

\[
438 = \frac{n}{2} \times (157 - 7n)
\]

\[
438 = \frac{n(157 - 7n)}{2}
\]

Multiplying both sides by 2 to eliminate the fraction:

\[
876 = n(157 - 7n)
\]

Expanding:

\[
876 = 157n - 7n^2
\]

Rearranging to form a quadratic equation:

\[
7n^2 - 157n + 876 = 0
\]

### Step 4: Solving the quadratic equation
We can now solve this quadratic equation using the quadratic formula:

\[
n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For the equation \( 7n^2 - 157n + 876 = 0 \), the coefficients are:

- \( a = 7 \),
- \( b = -157 \),
- \( c = 876 \).

Substituting these values into the quadratic formula:

\[
n = \frac{-(-157) \pm \sqrt{(-157)^2 - 4 \times 7 \times 876}}{2 \times 7}
\]

\[
n = \frac{157 \pm \sqrt{24649 - 24528}}{14}
\]

\[
n = \frac{157 \pm \sqrt{121}}{14}
\]

\[
n = \frac{157 \pm 11}{14}
\]

Thus, we have two possible values for \( n \):

\[
n = \frac{157 + 11}{14} = \frac{168}{14} = 12
\]

\[
n = \frac{157 - 11}{14} = \frac{146}{14} = 10.43
\]

Since \( n \) must be a whole number, we take \( n = 12 \).

### Step 5: Finding \( a_1 \)
Now that we know \( n = 12 \), substitute this value into the expression for \( a_1 \):

\[
a_1 = 82 - 7n
\]

\[
a_1 = 82 - 7 \times 12 = 82 - 84 = -2
\]

### Final Answer:
The first term is \( a_1 = -2 \) and the number of terms is \( n = 12 \).

Would you like further details or have any questions?

### Additional Practice:
1. Find the sum of the first 20 terms of an arithmetic sequence with \( a_1 = 5 \) and \( d = 3 \).
2. If the sum of the first 15 terms of an arithmetic sequence is 300 and the common difference is 4, find \( a_1 \) and \( a_{15} \).
3. In an arithmetic sequence, the 5th term is 12, and the 10th term is 32. Find the common difference and \( a_1 \).
4. Find the number of terms in an arithmetic sequence where \( a_1 = 8 \), \( d = 2 \), and the sum is 210.
5. If the sum of an arithmetic sequence is 500 and the common difference is 5, what is the value of \( n \) and \( a_1 \)?

**Tip:** When solving quadratic equations in arithmetic problems, always verify that your final answer for \( n \) is a whole number.

Learn how to find the first term and the number of terms in an arithmetic sequence given the sum of the first n terms, the last term, and the common difference. This problem involves arithmetic sequences, algebra, and solving quadratic equations. Suitable for students in Grades 8-10.

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