determine k so that k + 2, 4k - 6, 3k - 2 are the three consecutive terms of an AP.
Let a1 = k + 2 ; a 2 = 4k - 6 ; a3 = 3k - 2
Since a1 , a2 and a3 are three consecutive terms of an AP, then
a 2 - a1 = a3 - a2 [ as common difference will be same]
⇒ (4k - 6) - (k + 2) = (3k - 2) - (4k - 6)
⇒ 3k - 8 = - k + 4
⇒ 4k = 12
⇒ k = 3