急.急.求一道很简单的等差数列题,
问题描述：
急.急.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
答
设公差是M
an=a1+(n-1)M
a1+a6=2a1+5M=12
a14=a1+13M=64
3a1+18M=76
M=116/21
a6与a14的等差数列中项为a10
a6与a10的等差数列中项为a8
a10与a14的等差数列中项为a12
a8+a10+a12=a1+7M+a1+9M+a1+11M=3a1+27M=76+9M=76+9*116/21=76+348/7