(n) A.Ms between (a) and (b): (A_1,A_2,\ldots,A_n)

Common difference (d=\dfrac{b-a}{n+1})

[ A_k=a+kd ]

Sum of A.Ms: [ \sum_{k=1}^{n}A_k=na+d\sum k=na+d\cdot\dfrac{n(n+1)}{2}=na+\dfrac{b-a}{n+1}\cdot\dfrac{n(n+1)}{2} ] [ =na+\dfrac{n}{2}(b-a)=\dfrac{n}{2}(2a+b-a)=\dfrac{n}{2}(a+b)=n\cdot\dfrac{a+b}{2} ]

which is (n) times the single A.M. between (a) and (b).

Proved.