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Since 3k +2, 4k + 5 and 6k - 4 are consecutive terms of an AP. Their common difference must be same, which means
a2- a1 = a3 - a2
⇒4k + 5 - 3k -2 = 6k - 4 - 4k - 5
⇒ k + 3 = 2k - 9
⇒12 = k
k= 12
a2- a1 = a3 - a2
⇒4k + 5 - 3k -2 = 6k - 4 - 4k - 5
⇒ k + 3 = 2k - 9
⇒12 = k
k= 12
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