Question

数学的等差数列 (17 10:40:42)

Answer 1

二元一次方程组

设等差数列通项公式为an=a1+(n-1)d

则

a4+a7+a10=3a1+(4+7+10-3)d=3a1+18d=17

a4+a5+a6+a7+a8+a9+a10+a11+a12+a13+a14

=(14-4+1)a1+[(3+13)×(14-4+1)/2]d

=11a1+88d=77

由a1+8d=7

则

3a1+18d=17

a1+8d=7

联立解得：a1=5/3,d=2/3

则ak=a1+(k-1)d=(5/3)+(k-1)×(2/3)=13

解得k=18