1.
设an=a1+(n-1)d
2n=a1+a(2n-1)=a1+a1+(2n-2)d
n=a1+(n-1)d=an
an=n;
2.
1/an=1/n
sn=1/1+1/2++1/3+……+1/n
s(2n)=1/1+1/2++1/3+……+1/n+1/(n+1)+……+1/(2n-1)+1/(2n)
f(n)=S2n-Sn
=1/(n+1)+……+1/(2n-1)+1/(2n)
≥1/(n+1)+……+1/(n+1)+1/(n+1)……（n项）
=n/(n+1)
=1-1/(n+1)
≥1-1/2=1/2
f(n)的最小值为1/2；
3.
g(x)=log2x-12[1/(n+1)+……+1/(2n-1)+1/(2n)]
=log2x-12f(n)
≤log2x-12*（1/2）
=log2x-6
＜0
log2x＜6
1＜x＜2^6=64
所以1＜a＜b＜64.