determine k so that k + 2, 4k - 6, 3k - 2 are the three consecutive terms of an AP.

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Solution

Let a₁= k + 2 ; a₂= 4k - 6 ; a₃= 3k - 2
Since a₁, a₂and a₃are three consecutive terms of an AP, then
a₂- 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

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