If 3k - 2, 4k - 6 and k + 2 are these consecutive terms of A.P, then find the value of k.

Answer

To solve the problem, we need to find the value of \( k \) such that \( 3k - 2 \), \( 4k - 6 \), and \( k + 2 \) are consecutive terms of an Arithmetic Progression (A.P).

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

### Step 2: Set up the equation.
From the property of A.P., we have:
\[ (4k - 6) - (3k - 2) = (k + 2) - (4k - 6) \]

### Step 3: Simplify both sides.
1. Left side:
\[ 4k - 6 - 3k + 2 = k - 4 \]
This simplifies to:
\[ k - 4 \]

2. Right side:
\[ k + 2 - 4k + 6 = -3k + 8 \]
This simplifies to:
\[ -3k + 8 \]

### Step 4: Set the two sides equal to each other.
Now we have:
\[ k - 4 = -3k + 8 \]

### Step 5: Solve for \( k \).
1. Add \( 3k \) to both sides:
\[ k + 3k - 4 = 8 \]
This simplifies to:
\[ 4k - 4 = 8 \]

2. Add 4 to both sides:
\[ 4k = 12 \]

3. Divide by 4:
\[ k = 3 \]

### Conclusion:
The value of \( k \) is \( 3 \).

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