求一道很简单的等差数列题
已知数列an是等差数列,且a1+a6=12,a14=64.设a6与a14的等差数列中项为x,a6与x的等差中项为y,x与a14的等差中项为z,求x+y+z
人气：338 ℃ 时间：2020-08-22 04:29:33
解答
2*a1 + 5 d = 12
a1 + 13 d = 64
联立求出：a1 = -164/21, d = 116/21
x + y + z = a10 + a8 + a12
= 3 a10 = 3*(a1+9d)
= 880/7
也不算很简单啊.
推荐
猜你喜欢
