Q16 of 40 Page 1

If 3k – 2, 4k – 6 and k + 2 are three consecutive terms of A.P., then find the value of k.

If three terms a1, a2 and a3 are in AP then

a3 – a2 = a2 – a1


⇒ a1 + a3 = 2a2


⇒ 3k – 2 + k + 2 = 2(4k – 6)


⇒ 4k = 8k – 12


⇒ 4k = 12


⇒ k = 3


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