Determine k so that (3k -2), (4k-6) and (k +2) are three consecutive terms of an AP.

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Solution

It is given that (3k -2), (4k – 6) and (k + 2) are three consecutive terms of an A.P.
Therefore,
(4k–6)–(3k−2)=(k+2)–(4k–6)
4k–6–3k+2=k+2–4k+6
⇒k–4=−3k+8
⇒k+3k=4+8
⇒4k=12
⇒k=3
Hence the value of k is 3.

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