Determine \(k\) so that \(k^2+4k+8\), \(2k^2+3k+6\), \(3k^2+4k+4\) are three consecutive terms of an AP.
\(k=0\)
Step 1: Recall the property of an Arithmetic Progression (AP).
For three numbers \(A, B, C\) to be in AP, the difference between consecutive terms must be equal:
That is, \(B - A = C - B\).
Step 2: Identify the given terms:
\(A = k^2 + 4k + 8\)
\(B = 2k^2 + 3k + 6\)
\(C = 3k^2 + 4k + 4\)
Step 3: Find \(B - A\):
\(B - A = (2k^2 + 3k + 6) - (k^2 + 4k + 8)\)
= \(2k^2 - k^2 + 3k - 4k + 6 - 8\)
= \(k^2 - k - 2\)
Step 4: Find \(C - B\):
\(C - B = (3k^2 + 4k + 4) - (2k^2 + 3k + 6)\)
= \(3k^2 - 2k^2 + 4k - 3k + 4 - 6\)
= \(k^2 + k - 2\)
Step 5: Use the AP condition \(B - A = C - B\):
\(k^2 - k - 2 = k^2 + k - 2\)
Step 6: Simplify the equation:
Cancel \(k^2\) on both sides:
\(-k - 2 = k - 2\)
Add 2 to both sides:
\(-k = k\)
Add \(k\) to both sides:
\(0 = 2k\)
So, \(k = 0\).
Final Answer: \(k = 0\).