等差数列(an)中,a4+a5+a6+a7=24,a4a7=27,求通项an
问题描述:
等差数列(an)中,a4+a5+a6+a7=24,a4a7=27,求通项an
答
a4+a5+a6+a7=242(a4+a7)=24a4+a7=12a4a7=27解得a4=3,a7=9或a4=9,a7=3当a4=3,a7=9时a7=a4+3d9=3+3d3d=6d=2a4=a1+3d3=a1+3*2a1=-3an=a1+(n-1)d=-3+2(n-1)=2n-5当a4=9,a7=3时a7=a4+3d3=9+3d3d=-6d=-2a4=a1+3d9=a1+3*(-2...