Question

For what value of k,k+2,4k-6,3k-2 are three consecutive terms of an AP

Answer 1

Since given terms are in AP,
$2 \times (4k - 6) = ( k + 2) + (3k - 2) \\ 8k - 12 = 4k \\ 4k = 12 \\ k = 3$

Answer 2

Answer:k=3

Step-by-step explanation:

d1=d2

4k-6-k-2=3k-2-4k+6

3k-8=4-k

4k=12

k=3