Determine k so that (3k − 2), (4k − 6) and (k + 2) are three consecutive terms of an AP.
Question:

Determine k so that (3− 2), (4− 6) and (+ 2) are three consecutive terms of an AP. 

Solution:

It is given that (3− 2), (4− 6) and (+ 2) are three consecutive terms of an AP.

∴ (4− 6) − (3− 2) = (+ 2) − (4− 6)

⇒ 4− 6 − 3k + 2 = + 2 − 4+ 6

⇒ − 4 = −3+ 8

⇒ + 3k = 8 + 4

⇒ 4k = 12

⇒ k = 3

Hence, the value of k is 3.

 

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