The sum of the `5 ^(th) and the 7^(th)` terms of an A.P. is 52 and then `10^(th)` term is 46. Find the A.P..

Answer

To find the Arithmetic Progression (A.P.) given the conditions, we can follow these steps:

### Step 1: Define the terms of the A.P.
Let the first term of the A.P. be \( a \) and the common difference be \( d \).

The \( n^{th} \) term of an A.P. can be expressed as:
\[ T_n = a + (n-1)d \]

### Step 2: Write the expressions for the 5th, 7th, and 10th terms.
The 5th term \( T_5 \) is:
\[ T_5 = a + 4d \]

The 7th term \( T_7 \) is:
\[ T_7 = a + 6d \]

The 10th term \( T_{10} \) is:
\[ T_{10} = a + 9d \]

### Step 3: Set up the equations based on the problem statement.
According to the problem:
1. The sum of the 5th and 7th terms is 52:
\[ T_5 + T_7 = 52 \]
Substituting the expressions:
\[ (a + 4d) + (a + 6d) = 52 \]
This simplifies to:
\[ 2a + 10d = 52 \quad \text{(Equation 1)} \]

2. The 10th term is 46:
\[ T_{10} = 46 \]
Substituting the expression:
\[ a + 9d = 46 \quad \text{(Equation 2)} \]

### Step 4: Solve the equations.
From Equation 1:
\[ 2a + 10d = 52 \]
Dividing the entire equation by 2:
\[ a + 5d = 26 \quad \text{(Equation 3)} \]

Now we have:
- Equation 3: \( a + 5d = 26 \)
- Equation 2: \( a + 9d = 46 \)

### Step 5: Subtract Equation 3 from Equation 2.
\[ (a + 9d) - (a + 5d) = 46 - 26 \]
This simplifies to:
\[ 4d = 20 \]
Dividing by 4:
\[ d = 5 \]

### Step 6: Substitute \( d \) back into Equation 3 to find \( a \).
Substituting \( d = 5 \) into Equation 3:
\[ a + 5(5) = 26 \]
This simplifies to:
\[ a + 25 = 26 \]
Subtracting 25 from both sides:
\[ a = 1 \]

### Step 7: Write the A.P.
Now that we have \( a = 1 \) and \( d = 5 \), we can write the A.P.:
- First term: \( a = 1 \)
- Second term: \( a + d = 1 + 5 = 6 \)
- Third term: \( a + 2d = 1 + 10 = 11 \)
- Fourth term: \( a + 3d = 1 + 15 = 16 \)
- Fifth term: \( a + 4d = 1 + 20 = 21 \)
- Sixth term: \( a + 5d = 1 + 25 = 26 \)
- Seventh term: \( a + 6d = 1 + 30 = 31 \)
- Eighth term: \( a + 7d = 1 + 35 = 36 \)
- Ninth term: \( a + 8d = 1 + 40 = 41 \)
- Tenth term: \( a + 9d = 1 + 45 = 46 \)

Thus, the A.P. is:
\[ 1, 6, 11, 16, 21, 26, 31, 36, 41, 46 \]