已知等差数列{an}的前n项和为Sn,且满足a2+a4=14,S7=70

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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

an=a1+(n-1)d
a2+a4=14
2a1+4d =14
a1+2d =7 (1)
S7=70
(a1+3d)7 =70
a1+3d =10 (2)
(2)-(1)
d=3
a1=1
an = 1+(n-1)3 = 3n-2
bn = (2Sn+48)/n
= [(3n-1)n+48]/n
= (3n^2-n+48)/n
let
f(x) = (3x^2-x+48)/x
f'(x) = 3- 48/x^2 =0
3x^2-48=0
x= 4
f''(x) = 96/x^3
f''(4)>0 (min)
min bn at n=4
min bn = b4
= (48-4+48)/4
= 92/4
=23

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