Question

Determine k so that 4k+ 8,2k2 + 3k+6 and 3k2+4k+4 are three consecutive term of an ap

Answer 1

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Answer 2

Answer:

Step-by-step explanation:

let 4K+8 be A1

2K2+3K+6 be A2

3K2+4K+4 be A3

find the common difference where

d=A2-A1=A3-A2

  d=2K2+3K+6-(4K+6)=3K2+4K+4-(2K2+3K+6)

 it will result to

2K2-K-2=K2+K+2

Collect the like terms

2K2-K2=K+K-2+2

IT WILL result to; K2=2K

devide by k both sides which results to K=2