To determine the first term [tex]\(a\)[/tex] and the common difference [tex]\(d\)[/tex] of an arithmetic series given that the sum of the first 710 terms is 27.5 and the 10th term is 5, we follow a systematic approach:
### Step 1: Define the Formulas
1. The sum [tex]\(S_n\)[/tex] of the first [tex]\(n\)[/tex] terms of an arithmetic series can be given by:
[tex]\[
S_n = \frac{n}{2} \left(2a + (n-1)d\right)
\][/tex]
2. The [tex]\(n\)[/tex]th term of an arithmetic series can be defined as:
[tex]\[
a_n = a + (n-1)d
\][/tex]
### Step 2: Substitute the Known Values
Given:
- [tex]\(n = 710\)[/tex]
- [tex]\(S_{710} = 27.5\)[/tex]
- The 10th term [tex]\(a_{10} = 5\)[/tex]
### Step 3: Set Up the Equations
1. Using the given sum of the first 710 terms:
[tex]\[
27.5 = \frac{710}{2} (2a + 709d)
\][/tex]
Simplify this equation:
[tex]\[
27.5 = 355 (2a + 709d)
\][/tex]
[tex]\[
2a + 709d = \frac{27.5}{355}
\][/tex]
2. For the 10th term:
[tex]\[
a + 9d = 5
\][/tex]
### Step 4: Solve the System of Equations
We now have two equations:
1. [tex]\(2a + 709d = \frac{27.5}{355}\)[/tex]
2. [tex]\(a + 9d = 5\)[/tex]
We solve these equations simultaneously.
### Step 5: Find the Values of [tex]\(a\)[/tex] and [tex]\(d\)[/tex]
From the equations, solving for [tex]\(a\)[/tex] and [tex]\(d\)[/tex]:
1. From the second equation:
[tex]\[
a = 5 - 9d
\][/tex]
2. Substitute [tex]\(a = 5 - 9d\)[/tex] into the first equation:
[tex]\[
2(5 - 9d) + 709d = \frac{27.5}{355}
\][/tex]
[tex]\[
10 - 18d + 709d = \frac{27.5}{355}
\][/tex]
[tex]\[
691d + 10 = \frac{27.5}{355}
\][/tex]
[tex]\[
691d = \frac{27.5}{355} - 10
\][/tex]
[tex]\[
d = \frac{\frac{27.5}{355} - 10}{691}
\][/tex]
Thus:
[tex]\[
d = -0.0143596746906912
\][/tex]
And substituting [tex]\(d\)[/tex] back into [tex]\(a = 5 - 9d\)[/tex]:
[tex]\[
a = 5 - 9 \times (-0.0143596746906912)
\][/tex]
[tex]\[
a = 5.12923707221622
\][/tex]
### Conclusion
The first term [tex]\(a\)[/tex] is approximately [tex]\(5.12923707221622\)[/tex] and the common difference [tex]\(d\)[/tex] is approximately [tex]\(-0.0143596746906912\)[/tex].
