已知等差数列{an}的前n项和为Sn,且满足a2+a4=14,S7=70设bn=（2Sn+48）/n,求数列bn的最小项是第几项?_其他解答

已知等差数列{an}的前n项和为Sn,且满足a2+a4=14,S7=70
设bn=（2Sn+48）/n,求数列bn的最小项是第几项?并求出该项的值

人气：259 ℃　时间：2019-08-17 23:25:18

解答

an=a1+(n-1)da2+a4=142a1+4d =14a1+2d =7 (1)S7=70(a1+3d)7 =70a1+3d =10 (2)(2)-(1)d=3a1=1an = 1+(n-1)3 = 3n-2bn = (2Sn+48)/n= [(3n-1)n+48]/n= (3n^2-n+48)/nletf(x) = (3x^2-x+48)/xf'(x) = 3- 48/x^2 =03x^2-...