Question

If k + 2,4k – 6, 3k – 2 are the 3 consecutive terms of an A.P, then the value of K is …

(1) 2 

(2) 3 

(3) 4 

(4) 5

Answer 1

(2) 3

Here, a = k + 2, b = 4k - 6, c = 3k - 2 

We know that a, b, c are in A.P. 

b – a = c – b ⇒ 2b = a + c 

2(4k – 6) = k + 2 + 3k – 2 

8k – 12 = 4k ⇒ 4k = 12 

K = 12/4 = 3