Find the value of k, if k2+ 4k + 8, 2k^2 + 3k + 4 and 3k^2 + 4k + 4 are 3 consecutive terms of an AP.
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Answer: since they are in AP,
The condition is 2b=a+c
2(2k²+3k+4)=k²+4k+8+3k²+4k+4
4k²+6k+8=4k²+8k+12
6k+8=8k+12
2k=-4
k=-2
Step-by-step explanation:
Hope it helps you frnd......
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