设Bn=(A1+A2+...+An)/n,若{an}等差数列,且公差为d,问｛Bn}是否为等差数列
an=a1+(n-1)d
sn=na1+n(n-1)d/2
bn=(A1+A2+...+An)/n=sn/n=a1+(n-1)d/2
所以bn是首项为a1,公差为d/2等差数列.