已知{an}是等差数列,m,n∈N*,且m≠n.
(1)若am = n,an = m,试求a1,d,Sn及an+ m;
(2)若Sm = n,Sn = m,试求a1,d及an.
參考答案:1)
am=a1+(m-1)d=n
an=a1+(n-1)d=m
二式相减可解得
d=-1
a1=m+n-1
Sn=na1+n(n-1)d/2=n*(m+n-1)+n(n-1)*(-1)/2
=n(2m-n+3)/2
an+ m=a1+(n+m-1)(-1)=0
2)
Sn=na1+n(n-1)d/2=m
Sm=ma1+m(m-1)d/2=n
d=-2(m+n)/mn
a1=(mn-m-n+nn+mm)/mn
an=a1+(n-1)d=(mn-m-n+nn+mm)/mn+(n-1)*(-2)*(m+n)/mn
=(mm-nn-3m-3n-mn)/mn