Question

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

Answer 1

To be term of an A.P. the difference between two consecutive terms must be the same.

So, if 3k – 2, 4k – 6 and k + 2 are terms of an A.P.,

Then, 4k – 6 – (3k – 2) = k + 2 – (4k – 6)

⇒ 4k – 6 – 3k + 2 = k + 2 – 4k + 6

⇒ k – 4 = 8 – 3k

⇒ 4k = 12

⇒ k = 3

Hence, the value of k is 3.