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What must be added to 3k³-5k²+ 8k to obtain 4k+11k+7
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Answer:
Let 8k+4, 6k-2 and 2k-7 be 3 consecutive terms of an AP.
= (6K-2)-(8K+4) = (2K-7) - (6K-2)
= 6K-2-8K-4= 2K-7-6K +2
= -2K-6 = -4K-5
= 2K = 1
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