An A.P. consists of 57 terms of which 7th term is 13 and the last term is 108. Find the 45th term of this A.P.

Answer

To solve the problem step by step, we will use the properties of Arithmetic Progression (A.P.).

### Step 1: Understand the given information
We have an A.P. with:
- Total terms (n) = 57
- 7th term (T7) = 13
- Last term (T57) = 108

### Step 2: Write the formula for the nth term of an A.P.
The nth term of an A.P. can be expressed as:
\[ T_n = a + (n - 1)d \]
where:
- \( a \) is the first term,
- \( d \) is the common difference,
- \( n \) is the term number.

### Step 3: Write equations for the 7th and 57th terms
Using the formula for the 7th term:
\[ T_7 = a + (7 - 1)d = a + 6d \]
Given \( T_7 = 13 \), we can write:
\[ a + 6d = 13 \quad \text{(Equation 1)} \]

For the 57th term:
\[ T_{57} = a + (57 - 1)d = a + 56d \]
Given \( T_{57} = 108 \), we can write:
\[ a + 56d = 108 \quad \text{(Equation 2)} \]

### Step 4: Subtract Equation 1 from Equation 2
To eliminate \( a \), we subtract Equation 1 from Equation 2:
\[ (a + 56d) - (a + 6d) = 108 - 13 \]
This simplifies to:
\[ 50d = 95 \]

### Step 5: Solve for \( d \)
Now, divide both sides by 50:
\[ d = \frac{95}{50} = \frac{19}{10} = 1.9 \]

### Step 6: Substitute \( d \) back into Equation 1 to find \( a \)
Using Equation 1:
\[ a + 6d = 13 \]
Substituting \( d = 1.9 \):
\[ a + 6 \times 1.9 = 13 \]
\[ a + 11.4 = 13 \]
Now, subtract 11.4 from both sides:
\[ a = 13 - 11.4 = 1.6 \]

### Step 7: Find the 45th term
Now, we can find the 45th term using the formula:
\[ T_{45} = a + (45 - 1)d = a + 44d \]
Substituting the values of \( a \) and \( d \):
\[ T_{45} = 1.6 + 44 \times 1.9 \]
Calculating \( 44 \times 1.9 \):
\[ 44 \times 1.9 = 83.6 \]
Now, adding:
\[ T_{45} = 1.6 + 83.6 = 85.2 \]

### Final Answer
The 45th term of the A.P. is **85.2**.

---