等差数列{an} ,a2+a7+a10=17,a4+a5+a6+...+a14=77,ak=13,则k=
问题描述：
等差数列{an} ,a2+a7+a10=17,a4+a5+a6+...+a14=77,ak=13,则k=
答
设等差数列{an}的通项为：an = a1 +(n - 1)d,其中a1为首项,d为公差,由a2+a7+a10 = (a1 + d) + (a1 + 6d) + (a1 + 9d) = 3a1 + 16d,得：3a1 + 16d = 17-----（1）由a4+a5+a6+...+a14 = 1/2*(a4 + a14)*11 = 11*(a1 + ...