Certainly! Let's work through this problem step-by-step. We need to find the first term ([tex]\(a\)[/tex]) and the common difference ([tex]\(d\)[/tex]) of an arithmetic series given the sum of the first 10 terms ([tex]\(S_{10} = 27.5\)[/tex]) and the 10th term ([tex]\(a_{10} = 5\)[/tex]).
### Step 1: Use the formula for the nth term of an arithmetic series
The general formula for the [tex]\(n\)[/tex]th term ([tex]\(a_n\)[/tex]) of an arithmetic series is:
[tex]\[ a_n = a + (n-1)d \][/tex]
For the 10th term:
[tex]\[ a_{10} = a + 9d \][/tex]
Given [tex]\(a_{10} = 5\)[/tex]:
[tex]\[ 5 = a + 9d \][/tex]
Rearrange to solve for [tex]\(a\)[/tex]:
[tex]\[ a = 5 - 9d \quad \text{(Equation 1)} \][/tex]
### Step 2: Use the formula for the sum of the first [tex]\(n\)[/tex] terms of an arithmetic series
The formula for the sum of the first [tex]\(n\)[/tex] terms ([tex]\(S_n\)[/tex]) is:
[tex]\[ S_n = \frac{n}{2} \left( 2a + (n-1)d \right) \][/tex]
For the sum of the first 10 terms:
[tex]\[ S_{10} = \frac{10}{2} \left( 2a + 9d \right) \][/tex]
Given [tex]\(S_{10} = 27.5\)[/tex]:
[tex]\[ 27.5 = 5 \left( 2a + 9d \right) \][/tex]
Simplify:
[tex]\[ 5.5 = 2a + 9d \quad \text{(Equation 2)} \][/tex]
### Step 3: Substitute Equation 1 into Equation 2
Substitute [tex]\(a = 5 - 9d\)[/tex] from Equation 1 into Equation 2:
[tex]\[ 5.5 = 2(5 - 9d) + 9d \][/tex]
Expand and simplify:
[tex]\[ 5.5 = 10 - 18d + 9d \][/tex]
[tex]\[ 5.5 = 10 - 9d \][/tex]
Rearrange to solve for [tex]\(d\)[/tex]:
[tex]\[ -4.5 = -9d \][/tex]
[tex]\[ d = 0.5 \][/tex]
### Step 4: Substitute [tex]\(d\)[/tex] back into Equation 1 to find [tex]\(a\)[/tex]
Now, substitute [tex]\(d = 0.5\)[/tex] back into Equation 1:
[tex]\[ a = 5 - 9(0.5) \][/tex]
[tex]\[ a = 5 - 4.5 \][/tex]
[tex]\[ a = 0.5 \][/tex]
### Summary
The first term [tex]\(a\)[/tex] is [tex]\(0.5\)[/tex], and the common difference [tex]\(d\)[/tex] is [tex]\(0.5\)[/tex].
