If k + 2, 4k – 6 and 3k – 2 are in AP, then the value of k is
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If k + 2, 4k – 6 and 3k – 2 are in AP, then the value of k is
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k + 2, 4k – 6, 3k – 2 are in AP. => 2(4k - 6) = (k + 2) + (3k - 2) 8k – 12 = k + 2 + 3k – 2 8k – 12 = 4k 4k = 12 k = 3
For three terms to be in arithmetic progression, consecutive differences must be equal: (4k−6) − (k+2) = (3k−2) − (4k−6). This simplifies to 3k − 8 = −k + 4, giving 4k = 12, so k = 3. Substituting k=3 gives the terms 5, 6, 7 — a valid AP with common difference 1, confirming the value.