The sum of the 4th and the 8th terms of an A.P. is 24 and the sum of the 6th and the 10th terms of the same A.P. is 44. Find the first three terms of the A.P.

Answer

To solve the problem, we need to find the first three terms of an arithmetic progression (A.P.) given two conditions about the sums of certain terms. Let's break down the solution step by step.

### 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 the equations based on the given conditions.
From the problem, we know:
1. The sum of the 4th and 8th terms is 24:
\[ a_4 + a_8 = 24 \]
Using the formula for the nth term:
\[ (a + 3d) + (a + 7d) = 24 \]
Simplifying this gives:
\[ 2a + 10d = 24 \quad \text{(Equation 1)} \]

2. The sum of the 6th and 10th terms is 44:
\[ a_6 + a_{10} = 44 \]
Again using the formula for the nth term:
\[ (a + 5d) + (a + 9d) = 44 \]
Simplifying this gives:
\[ 2a + 14d = 44 \quad \text{(Equation 2)} \]

### Step 3: Solve the equations.
Now we have a system of two equations:
1. \( 2a + 10d = 24 \)
2. \( 2a + 14d = 44 \)

We can subtract Equation 1 from Equation 2:
\[ (2a + 14d) - (2a + 10d) = 44 - 24 \]
This simplifies to:
\[ 4d = 20 \]
Dividing both sides by 4 gives:
\[ d = 5 \]

### Step 4: Substitute \( d \) back to find \( a \).
Now, substitute \( d = 5 \) back into Equation 1:
\[ 2a + 10(5) = 24 \]
This simplifies to:
\[ 2a + 50 = 24 \]
Subtracting 50 from both sides gives:
\[ 2a = 24 - 50 \]
\[ 2a = -26 \]
Dividing by 2 gives:
\[ a = -13 \]

### Step 5: Find the first three terms of the A.P.
Now that we have \( a \) and \( d \):
- The first term \( a_1 = a = -13 \)
- The second term \( a_2 = a + d = -13 + 5 = -8 \)
- The third term \( a_3 = a + 2d = -13 + 2(5) = -13 + 10 = -3 \)

Thus, the first three terms of the A.P. are:
\[ \text{First three terms: } -13, -8, -3 \]