If k+2, 2k-5 and k+8 are in A.P., find the value of k.

Answer

To find the value of \( k \) such that \( k + 2 \), \( 2k - 5 \), and \( k + 8 \) are in Arithmetic Progression (A.P.), we can follow these steps:

### Step 1: Set up the condition for A.P.
In an A.P., the difference between consecutive terms is constant. Therefore, we can write the equation:
\[ (2k - 5) - (k + 2) = (k + 8) - (2k - 5) \]

### Step 2: Simplify the left-hand side
Now, simplify the left-hand side of the equation:
\[ (2k - 5) - (k + 2) = 2k - 5 - k - 2 = k - 7 \]

### Step 3: Simplify the right-hand side
Next, simplify the right-hand side of the equation:
\[ (k + 8) - (2k - 5) = k + 8 - 2k + 5 = -k + 13 \]

### Step 4: Set the two sides equal
Now we have:
\[ k - 7 = -k + 13 \]

### Step 5: Solve for \( k \)
To solve for \( k \), first add \( k \) to both sides:
\[ k + k - 7 = 13 \]
This simplifies to:
\[ 2k - 7 = 13 \]

Next, add 7 to both sides:
\[ 2k = 13 + 7 \]
This gives:
\[ 2k = 20 \]

Finally, divide both sides by 2:
\[ k = \frac{20}{2} = 10 \]

### Conclusion
The value of \( k \) is \( 10 \).

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