Given that m2+m+1, m2+2m+11,and m2+3m+28 form three consecutive terms of a geometric sequence.find m
m2+3m+28 = m2+2m+11
m2+2m+11 m2+m+1
(m2+3m+28)( m2+m+1)= (m2+2m+11)2
m4+m3+m2+3m3+3m2+3m+28m2+28m+28=m4+2m3+11m2+2m3+4m2+22m+11m2+22m+121
m4-m4+m3+3m2-2m3+m2+28m2-11m2-4m2-11m2+3m+28m-22m+28-121=0
6m2-13m-93=0
m=31/6
m=-3
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