急求!等差数列{an}中,a1+a2+a3=12,a1a2a3=48,公差大于0,则首项a1=
问题描述:
急求!等差数列{an}中,a1+a2+a3=12,a1a2a3=48,公差大于0,则首项a1=
答
a1+a2+a3=12
3a2=12
a2=4
a1a2a3=48
a1a3=12
(a2-d)(a2+d)=12
(4-d)(4+d)=12
16-d²=12
d²=4
d=2
a1=4-2=2