Find the sum of the first 40 terms of the A.P. whose 4th term is 8 and 6th term is 14.

Answer

To find the sum of the first 40 terms of the arithmetic progression (A.P.) whose 4th term is 8 and 6th term is 14, we can follow these steps:

### Step 1: Define the terms of the A.P.
The nth term of an A.P. can be expressed as:
\[ a_n = a + (n-1)d \]
where \( a \) is the first term and \( d \) is the common difference.

### Step 2: Write equations for the given terms.
From the problem, we know:
- The 4th term \( a_4 = 8 \):
\[ a + 3d = 8 \] (Equation 1)

- The 6th term \( a_6 = 14 \):
\[ a + 5d = 14 \] (Equation 2)

### Step 3: Subtract Equation 1 from Equation 2.
To eliminate \( a \), we can subtract Equation 1 from Equation 2:
\[ (a + 5d) - (a + 3d) = 14 - 8 \]
This simplifies to:
\[ 2d = 6 \]
So, we find:
\[ d = 3 \]

### Step 4: Substitute \( d \) back into one of the equations to find \( a \).
Now, we can substitute \( d = 3 \) back into Equation 1:
\[ a + 3(3) = 8 \]
This simplifies to:
\[ a + 9 = 8 \]
Thus:
\[ a = 8 - 9 = -1 \]

### Step 5: Use the formula for the sum of the first \( n \) terms of an A.P.
The formula for the sum of the first \( n \) terms \( S_n \) of an A.P. is given by:
\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]
For our case, we want to find \( S_{40} \):
\[ S_{40} = \frac{40}{2} \times (2(-1) + (40-1)(3)) \]

### Step 6: Simplify the expression.
Calculating further:
\[ S_{40} = 20 \times (2(-1) + 39 \times 3) \]
\[ = 20 \times (-2 + 117) \]
\[ = 20 \times 115 \]
\[ = 2300 \]

### Final Answer:
Thus, the sum of the first 40 terms of the A.P. is:
\[ \boxed{2300} \]

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