已知等差数列{an}的公差为d,求证:am-an/m-n=d

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已知等差数列{an}的公差为d,求证:am-an/m-n=d
已知等差数列{an}的公差为d,求证:am-an/m-n=d

已知等差数列{an}的公差为d,求证:am-an/m-n=d
等差公式可知am=a1+(m-1)*d,an=a1+(n-1)*d
相减得am-an=(m-n)d

证明:∵am=a1+(m-1)d an=a1+(n-1)d
∴am-an=(m-n)d
∴(am-an)/(m-n)=d
得证

am=a1+(m-1)d
an=a1+(n-1)d
(am-an)/(m-n)=d(m-n)/(m-n)=d

(am-an)/(m-n)
=(a1+(m-1)d-a1-(n-1)d)/(m-n)
=(m-n)d/(m+n)
=d