设数列an,bn满足:bn=(a1+a2+a3+a4+...+an)/n,若bn是等差数列,求证an也是等差数列

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设数列an,bn满足:bn=(a1+a2+a3+a4+...+an)/n,若bn是等差数列,求证an也是等差数列

首先等差数列的通项公式是关于n的一次式
bn是等差数列,设bn=A*n+B
则:a1+a2+a3+a4+...+an=n(A*n+B)=A(n^2)+Bn
a1+a2+a3+a4+...+a(n-1)=A((n-1)^2)+B(n-1)
相减得:an=A(2n-1)+B=A*n+(B-A)
故an是等差数列

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