16.答案:[6,12]依题意,2S。=a十am,,站,一=+。当n=1时,2S1=2a1=a12+a1,因am>0,解得a1=1;当n≥2时,2S-1=a-1十am-1,两式相减可得,2a.=a:十an一a?-1一am-1,化简可得,(a,十a-1)(am一a。-1-1)=0,故a.-a,-1=1,则an=n,=+片所以6.-6,=a+)一(6+)=a-3》-c)故.-“3n①当n≤2时,b,≥b,即3n≤c,所以c≥6;②当m=3时,6,=b,满足题意;③当m≥4时,b.≥b,恒成立,故3n≥c,则c≤12:女灵:十音项,背故c的取值范围为[6,12].