Question

等差数列中,已知A2+A5+A9=9,a3a5a8=-21,求数列的通项公式

Answer 1

a2=a1+d

a5=a1+4d

a9=a1+8d

a2+a5+a9=(a1+d)+(a1+4d)+(a1+8d)=3a1+13d=9

a3=a1+2d

a8=a1+7d

a3a5a8=(a1+2d)(a1+4d)(a1+7d)=a1^3+13da1^2+50a1d^2+56d^3

a1=3—13/3d

把a1=3—13/3d带入a1^3+13da1^2+50a1d^2+56d^3中,求出d

再把d的值代入a1=3—13/3d中求出a1

an=a1+(n—1)d