若{an}为等差数列,Ap= q,Aq=p(p ≠q),则A(p+q)= ______

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若{an}为等差数列,Ap= q,Aq=p(p ≠q),则A(p+q)= ______
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若{an}为等差数列,Ap= q,Aq=p(p ≠q),则A(p+q)= ______
若{an}为等差数列,Ap= q,Aq=p(p ≠q),则A(p+q)= ______

若{an}为等差数列,Ap= q,Aq=p(p ≠q),则A(p+q)= ______
若{an}为等差数列,Ap= q, Aq=p(p ≠q),则A(p+q)=-(p+q)
具体证明见下面:
等差数列{an}的前n项和为Sn,Sp=q,Sq=p,p≠q,则S(p+q)=-(p+q)
证明:由题意,
q=Sp=a1+a2+...+ap=pa1+p(p-1)d/2
p=Sq=a1+a2+...+aq=qa1+q(q-1)d/2
两式相减,得到
q-p=(p-q)[a1+(p+q-1)d/2]
因为p≠q,故
a1+(p+q-1)d/2=-1
因此
S(p+q)=a1+a2+...+a(p+q)=(p+q)(a1+a(p+q))/2
=(p+q)(a1+a1+(p+q-1)d)/2
=(p+q)(a1+(p+q-1)d/2)
=(p+q)*(-1)
=-(p+q)

0

Ap= q, Aq=p(p ≠q),
可知d=1
A(p+q)= 0