Solution:
(3k – 2), (4k – 6) and (k + 2) are three consecutive terms of an AP.
(4k – 6) – (3k – 2) = (k + 2) – (4k – 6)
⇒ 2(4k – 6) = (k + 2) + (3k – 2)
⇒ 8k – 12 = 4k + 0
⇒ 8k – 4k = 0 + 12
⇒ 4k = 12
k = 3
Determine k so that (3k -2), (4k – 6) and (k +2) are three consecutive terms of an AP.
Determine k so that (3k -2), (4k – 6) and (k +2) are three consecutive terms of an AP.