Question

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

Answer 1

It is given that (3k – 2), (4k – 6) and (k + 2) are three consecutive terms of an AP.

∴ (4k – 6) – (3k – 2) = (k + 2) – (4k – 6)

⇒ 4k – 6 – 3k + 2 = k + 2 – 4k + 6

⇒ k – 4 = –3k + 8

⇒ k + 3k = 8 + 4 

⇒ 4k = 12 

⇒ k = 3 

Hence, the value of k is 3.