Determine k so that (3k – 2), (4k – 6) and (k+2) are three consecutive
terms of an AP.
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Answer:
Step-by-step explanation:
(4k-6)-(3k-2)=(k+2)-(4k-6)
4k-6-3k+2=k+2-4k+6
K-4=-3k+8
K+3k=8+4
4k=12
K=3
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