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