For what value of 'k', k-3, 2k+1 and 4k+3 are in A.P.?

Answer

To find the value of \( k \) such that \( k-3 \), \( 2k+1 \), and \( 4k+3 \) are in Arithmetic Progression (A.P.), we can use the property of A.P. that states the middle term is equal to the average of the other two terms. This means:

\[ 2k + 1 = \frac{(k - 3) + (4k + 3)}{2} \]

Now, let's solve this step by step.

### Step 1: Set up the equation
We start with the equation from the definition of A.P.:

\[ 2k + 1 = \frac{(k - 3) + (4k + 3)}{2} \]

### Step 2: Simplify the right side
Combine the terms in the numerator:

\[ k - 3 + 4k + 3 = 5k \]

Thus, the equation becomes:

\[ 2k + 1 = \frac{5k}{2} \]

### Step 3: Eliminate the fraction
To eliminate the fraction, multiply both sides by 2:

\[ 2(2k + 1) = 5k \]

This simplifies to:

\[ 4k + 2 = 5k \]

### Step 4: Rearrange the equation
Now, rearranging the equation to isolate \( k \):

\[ 4k + 2 - 5k = 0 \]

This simplifies to:

\[ -k + 2 = 0 \]

### Step 5: Solve for \( k \)
Now, solve for \( k \):

\[ -k = -2 \implies k = 2 \]

### Conclusion
The value of \( k \) for which \( k-3 \), \( 2k+1 \), and \( 4k+3 \) are in A.P. is:

\[ \boxed{2} \]