For what value of k, the terms 2k-1, 7 and 3k are in A.P.?

Answer

To find the value of \( k \) such that the terms \( 2k - 1 \), \( 7 \), and \( 3k \) are in Arithmetic Progression (A.P.), we can follow these steps:

### Step 1: Understand the condition for 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 \( a_1 = 2k - 1 \), \( a_2 = 7 \), and \( a_3 = 3k \).

### Step 2: Set up the equation using the A.P. condition
Substituting the values into the A.P. condition gives us:
\[ 7 - (2k - 1) = (3k - 7) \]

### Step 3: Simplify the left-hand side
Simplifying the left-hand side:
\[ 7 - 2k + 1 = 8 - 2k \]
So, we have:
\[ 8 - 2k = 3k - 7 \]

### Step 4: Rearrange the equation
Now, we can rearrange the equation to isolate \( k \):
\[ 8 + 7 = 3k + 2k \]
This simplifies to:
\[ 15 = 5k \]

### Step 5: Solve for \( k \)
Dividing both sides by 5 gives:
\[ k = 3 \]

### Conclusion
Thus, the value of \( k \) for which the terms \( 2k - 1 \), \( 7 \), and \( 3k \) are in A.P. is:
\[ \boxed{3} \]