Determine k so that k + 2, 4k - 6 and 3k - 2 are three consecutive terms of an A.P

Answer

To determine the value of \( k \) such that \( k + 2 \), \( 4k - 6 \), and \( 3k - 2 \) are three consecutive terms of an arithmetic progression (A.P.), we can follow these steps:

### Step 1: Understand the property of A.P.
In an arithmetic progression, the difference between consecutive terms is constant. Therefore, we can set up the equation:
\[ a_3 - a_2 = a_2 - a_1 \]
where \( a_1 = k + 2 \), \( a_2 = 4k - 6 \), and \( a_3 = 3k - 2 \).

### Step 2: Substitute the terms into the equation
Substituting the values of \( a_1 \), \( a_2 \), and \( a_3 \) into the equation gives:
\[ (3k - 2) - (4k - 6) = (4k - 6) - (k + 2) \]

### Step 3: Simplify both sides
Now, simplify both sides of the equation.

**Left Side:**
\[ 3k - 2 - 4k + 6 = -k + 4 \]

**Right Side:**
\[ 4k - 6 - k - 2 = 3k - 8 \]

So, we have:
\[ -k + 4 = 3k - 8 \]

### Step 4: Rearrange the equation
Now, let's rearrange the equation to isolate \( k \):
\[ -k - 3k = -8 - 4 \]
\[ -4k = -12 \]

### Step 5: Solve for \( k \)
Dividing both sides by -4 gives:
\[ k = \frac{-12}{-4} = 3 \]

### Conclusion
Thus, the value of \( k \) that makes \( k + 2 \), \( 4k - 6 \), and \( 3k - 2 \) consecutive terms of an A.P. is:
\[ \boxed{3} \]