# If n A.M.s are inserted between two numbers, prove that the sum of the means equidistant from the beginning and the end is constant.

$Let\;a\;and\;b\;be\;the\;first\;and\;last\;terms\\* The\;series\;be\;a,\;A-1,A_2,A_3,.........,A_n,b\\* We\;know,\;Mean\;=\frac{(a+b)}{2}\\* Mean\;of\;A_1\;and\;A_n=\frac{(A_1+A_n)}{2}\\* \therefore\;A_1=a+d\\*A_n=a-d\\* So,\;AM=\frac{(a+d+b-d)}{2}=\frac{(a+b)}{2}\\* AM\;between\;A_2\;and\;A_{n-1}=\frac{a+2d+b-2d)}{2}=\frac{(a+b)}{2}\\* Similarly,\;\frac{(a\;+\;b)}{2}\;is\;constant\;for\;all\;such\;numbers\\* Hence,\;AM=\frac{(a+b)}{2}$

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