determine k so that k2 + 4k+8, 2k2+3k+6 and 3k2+4k+k are three consecutive terms of an AP
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Answer:
k = 4/3
Step-by-step explanation:
In AP, middle number is the mean of its surrounding number:
2b = a+c
2(2k2+3k+6) = k2 + 4k+8 + 3k2+4k+k
4k2 + 6k + 12 = 4k2 + 9k + 8
3k = 4
k=4/3
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