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 need to use the property that in an A.P., the difference between consecutive terms is constant.

### Step 1: Set up the equation for A.P.
In an A.P., the difference between the second term and the first term should equal the difference between the third term and the second term. Therefore, we can write the equation as:

\[ (4K - 6) - (K + 2) = (3K - 2) - (4K - 6) \]

### Step 2: Simplify the left side of the equation
Now, simplify the left side:

\[ 4K - 6 - K - 2 = 3K - 2 - 4K + 6 \]

This simplifies to:

\[ (4K - K) - (6 + 2) = (3K - 4K) + (6 - 2) \]

\[ 3K - 8 = -K + 4 \]

### Step 3: Rearrange the equation
Now, let's rearrange the equation to isolate K:

\[ 3K + K = 4 + 8 \]

\[ 4K = 12 \]

### Step 4: Solve for K
Now, divide both sides by 4 to find K:

\[ K = \frac{12}{4} = 3 \]

### Conclusion
Thus, the value of \( K \) for which \( K + 2, 4K - 6, 3K - 2 \) are three consecutive terms of an A.P. is:

\[ \boxed{3} \]

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