For what value of 'K', K+2, 4K-6, 3K-2 are three consecutive terms of an A.P.?

Options

A. 1
B. -1
C. 3
D. -3

Answer

To find the value of 'K' for which \( K + 2, 4K - 6, 3K - 2 \) are three consecutive terms of an Arithmetic Progression (A.P.), we can use the property of A.P. that states the difference between consecutive terms is constant.

### Step-by-Step Solution:

1. **Identify the terms**:
- Let the three terms be:
- First term \( a_1 = K + 2 \)
- Second term \( a_2 = 4K - 6 \)
- Third term \( a_3 = 3K - 2 \)

2. **Set up the A.P. condition**:
- For the terms to be in A.P., the difference between the second and first terms must equal the difference between the third and second terms:
\[ a_2 - a_1 = a_3 - a_2 \]

3. **Substitute the terms**:
- Substitute the expressions for \( a_1, a_2, \) and \( a_3 \):
\[ (4K - 6) - (K + 2) = (3K - 2) - (4K - 6) \]

4. **Simplify the left side**:
- Simplifying the left-hand side:
\[ 4K - 6 - K - 2 = 3K - 8 \]

5. **Simplify the right side**:
- Simplifying the right-hand side:
\[ 3K - 2 - 4K + 6 = -K + 4 \]

6. **Set the two sides equal**:
- Now, we have:
\[ 3K - 8 = -K + 4 \]

7. **Solve for K**:
- Add \( K \) to both sides:
\[ 3K + K - 8 = 4 \]
- Combine like terms:
\[ 4K - 8 = 4 \]
- Add 8 to both sides:
\[ 4K = 12 \]
- Divide by 4:
\[ K = 3 \]

### Final Answer:
The value of \( K \) for which \( K + 2, 4K - 6, 3K - 2 \) are three consecutive terms of an A.P. is \( K = 3 \).