Question

If k+2, 4k-6 and 3k-2 are 3 consecutive terms of an A.P. Then value of k is
(A) 0
(B) 1
(C) 2
(D) 3​

Answer 1

Answer:

If k+2 , 4k-6 and 3k-2 are in AP then,

(4k-6) - (k -2) = d

4k - k -6 +2 = d

3k - 4 = d

Also,

(3k - 2) - ( 4k-6) = d

3k - 4k -2 + 4 = d

2 -k = d

Putting d = 3k-4

2 - k = 3k - 4

4k = 8

k = 2

Hence, Option (C) 2 is correct.

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