Question

If 2k, 3k + 1, and 5k − 1 are three consecutive terms of an A.P., find the value of k.

Answer 1

Since 2k, 3k + 1, and 5k − 1 are in A.Р.

∴ (3k + 1) − 2k = (5k − 1) − (3k + 1)    ...(∵ T₂ − T₁ = T₃ − T₂)

⇒ (k + 1) = (2k − 2)

⇒ 1 + 2 = 2k − k

⇒ 3 = k

The value of k = 3