等差数列{an}中,若a10等于0,则有 a1+a2+a3+…+an=a1+a2+…a19-n(n

设公差为d
a2-a1=d
a3-a2=d
.
an-a(n-1)=d
an=(n-1)d+a1
a10=9d+a1
a1=-9d
an=(n-1)d-9d=(n-10)d
a19-n=(19-n-10)d=(9-n)d
a1+a2+.+an=1/2(a1+an)n=1/2n(-9d+(n-10)d)=1/2n(n-19)d
a1+a2+.+a19-n=1/2(a1+a19-n)(19-n)=1/2(-9d+(9-n)d)(19-n)=1/2(-nd)(19-n)=1/2nd(n-19)
所以得证命题

公比为q,q不等于0,
b2/b1=q
bn=b1q^(n-1)
b9=b1q^8=1
要q=1作为特例求解

∵{an}是等差数列,a10=0,
∴a(n+1)+a(19-n)=a(n+2)+a(18-n)=···=0，
∴a1+a2+···+an+a(n+1)+a(n+2)+···+a(19-n)
=a1+a2+···+an+[a(n+1)+a(19-n)]
+[a(n+2)+a(18-n)]+···
=a1+a2+······+an.
在等比数列{bn}中，若b9=1，则
b1·b2····bn=b1·b2·····b(17-n).(n<17，n∈N*)