To solve the problem step by step, we will use the formulas for the sum of the first n terms of an arithmetic progression (AP).
### Step 1: Write the formulas for \( S_5 \), \( S_7 \), and \( S_{10} \)
The sum of the first n terms of an AP is given by the formula:
\[
S_n = \frac{n}{2} \left(2a + (n-1)d\right)
\]
where:
- \( a \) is the first term,
- \( d \) is the common difference,
- \( n \) is the number of terms.
For \( S_5 \):
\[
S_5 = \frac{5}{2} \left(2a + 4d\right) = \frac{5}{2} (2a + 4d)
\]
For \( S_7 \):
\[
S_7 = \frac{7}{2} \left(2a + 6d\right) = \frac{7}{2} (2a + 6d)
\]
For \( S_{10} \):
\[
S_{10} = \frac{10}{2} \left(2a + 9d\right) = 5(2a + 9d)
\]
### Step 2: Set up the equations based on the given information
We know:
1. \( S_5 + S_7 = 167 \)
2. \( S_{10} = 235 \)
From the first equation:
\[
\frac{5}{2} (2a + 4d) + \frac{7}{2} (2a + 6d) = 167
\]
Combining these:
\[
\frac{5(2a + 4d) + 7(2a + 6d)}{2} = 167
\]
\[
5(2a + 4d) + 7(2a + 6d) = 334
\]
Expanding this:
\[
10a + 20d + 14a + 42d = 334
\]
Combining like terms:
\[
24a + 62d = 334 \quad \text{(Equation 1)}
\]
From the second equation:
\[
5(2a + 9d) = 235
\]
\[
2a + 9d = 47 \quad \text{(Equation 2)}
\]
### Step 3: Solve the system of equations
Now we have two equations:
1. \( 24a + 62d = 334 \)
2. \( 2a + 9d = 47 \)
We can solve these equations simultaneously. Let's first solve Equation 2 for \( a \):
\[
2a = 47 - 9d
\]
\[
a = \frac{47 - 9d}{2}
\]
Substituting \( a \) into Equation 1:
\[
24\left(\frac{47 - 9d}{2}\right) + 62d = 334
\]
Multiplying through by 2 to eliminate the fraction:
\[
24(47 - 9d) + 124d = 668
\]
Expanding:
\[
1128 - 216d + 124d = 668
\]
Combining like terms:
\[
1128 - 92d = 668
\]
Rearranging gives:
\[
-92d = 668 - 1128
\]
\[
-92d = -460
\]
\[
d = \frac{460}{92} = 5
\]
### Step 4: Substitute \( d \) back to find \( a \)
Now substitute \( d = 5 \) back into Equation 2:
\[
2a + 9(5) = 47
\]
\[
2a + 45 = 47
\]
\[
2a = 2
\]
\[
a = 1
\]
### Step 5: Write the AP
Now we have:
- First term \( a = 1 \)
- Common difference \( d = 5 \)
The AP is:
\[
1, 1 + 5, 1 + 2 \times 5, 1 + 3 \times 5, \ldots
\]
So the AP is:
\[
1, 6, 11, 16, 21, \ldots
\]
### Final Answer:
The arithmetic progression is:
\[
1, 6, 11, 16, 21, \ldots
\]
