In an arithmetic series, where S40 = 1900 and u40 = 106, find u1 and the common difference

In an arithmetic series, the sum of the first \(n\) terms is given by the formula:

\[
S_n = \frac{n}{2} \times (u_1 + u_n)
\]

And the general term of the series is given by:

\[
u_n = u_1 + (n - 1) \times d
\]

Where:
- \(u_1\) is the first term,
- \(u_n\) is the \(n\)-th term,
- \(d\) is the common difference,
- \(S_n\) is the sum of the first \(n\) terms.

### Step 1: Using the sum formula
We are given that:
- \(S_{40} = 1900\),
- \(u_{40} = 106\).

Substitute these values into the sum formula:

\[
1900 = \frac{40}{2} \times (u_1 + 106)
\]

Simplifying:

\[
1900 = 20 \times (u_1 + 106)
\]

\[
u_1 + 106 = \frac{1900}{20} = 95
\]

\[
u_1 = 95 - 106 = -11
\]

### Step 2: Using the general term formula
We now use the general term formula for \(u_{40} = 106\):

\[
106 = u_1 + 39d
\]

Substitute \(u_1 = -11\) into this equation:

\[
106 = -11 + 39d
\]

\[
39d = 106 + 11 = 117
\]

\[
d = \frac{117}{39} = 3
\]

### Final Answer:
- First term, \(u_1 = -11\),
- Common difference, \(d = 3\).

Would you like more details or have any questions?

Here are 5 related questions to explore:
1. How would the series change if \(d\) were negative?
2. Can you calculate the sum \(S_{50}\) if \(u_1 = -11\) and \(d = 3\)?
3. How does the sum of an arithmetic series differ from a geometric series?
4. What is the formula for the sum of an arithmetic series when the first and last terms are known?
5. How would you solve for \(d\) if only \(S_{n}\) and \(u_1\) are given?

**Tip:** In an arithmetic sequence, the common difference \(d\) can be found by subtracting any two consecutive terms.

In an arithmetic series, where S40 = 1900 and u40 = 106, find u1 and the common difference.

Learn how to find the first term (u1) and the common difference (d) of an arithmetic series given S40 = 1900 and u40 = 106. This problem involves key mathematical concepts such as arithmetic series, sequence formulas, and algebra. Suitable for students in Grades 9-11.

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