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Question

Determine k so that k2+4k+8=0,2k2+3k+6=0,3k2+4k+4=0 are the three consecutive terms of an AP

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Solution

k2+4k+8,2k2+3k+6,3k2+4k+4 are the three consecutive terms of A.P
k2+4k+8+3k2+4k+42=2k2+3k+64k2+8k+12=4k2+6k+124k3k=0k=0

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