The 4th term of an A.P. is 22 and 15th term is 66. Find the first term and the common difference. Hence, find the sum of the series to 8 terms.

Answer

To solve the problem step by step, we will follow these steps:

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

### Step 2: Set up equations based on given terms
We know:
- The 4th term \( T_4 = 22 \)
- The 15th term \( T_{15} = 66 \)

Using the formula for the nth term, we can write:
1. For the 4th term:
\[ T_4 = a + (4 - 1)d \]
\[ 22 = a + 3d \] (Equation 1)

2. For the 15th term:
\[ T_{15} = a + (15 - 1)d \]
\[ 66 = a + 14d \] (Equation 2)

### Step 3: Subtract Equation 1 from Equation 2
To eliminate \( a \), we subtract Equation 1 from Equation 2:
\[ (a + 14d) - (a + 3d) = 66 - 22 \]
This simplifies to:
\[ 14d - 3d = 44 \]
\[ 11d = 44 \]

### Step 4: Solve for the common difference \( d \)
Now, divide both sides by 11:
\[ d = \frac{44}{11} = 4 \]

### Step 5: Substitute \( d \) back into Equation 1 to find \( a \)
Now that we have \( d \), we can substitute it back into Equation 1 to find \( a \):
\[ 22 = a + 3(4) \]
\[ 22 = a + 12 \]
Subtract 12 from both sides:
\[ a = 22 - 12 = 10 \]

### Step 6: Find the sum of the first 8 terms of the A.P.
The formula for the sum of the first \( n \) terms of an A.P. is:
\[ S_n = \frac{n}{2} \times (2a + (n - 1)d) \]
For \( n = 8 \):
\[ S_8 = \frac{8}{2} \times (2(10) + (8 - 1)(4)) \]
\[ S_8 = 4 \times (20 + 7 \times 4) \]
\[ S_8 = 4 \times (20 + 28) \]
\[ S_8 = 4 \times 48 \]
\[ S_8 = 192 \]

### Final Answer
The first term \( a = 10 \), the common difference \( d = 4 \), and the sum of the first 8 terms \( S_8 = 192 \).

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