Question:
Determine k so that (3k − 2), (4k − 6) and (k + 2) are three consecutive terms of an AP.
Solution:
It is given that (3k − 2), (4k − 6) and (k + 2) are three consecutive terms of an AP.
∴ (4k − 6) − (3k − 2) = (k + 2) − (4k − 6)
⇒ 4k − 6 − 3k + 2 = k + 2 − 4k + 6
⇒ k − 4 = −3k + 8
⇒ k + 3k = 8 + 4
⇒ 4k = 12
⇒ k = 3
Hence, the value of k is 3.
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