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S = (a) + (a + d) +(a+2d) +.... + (a+(n-1)d)
S = (a+a+a+a+....n)+(d+2d+3d+.........+(n-1)d)
S = na +d(1 + 2 +3 + .....+n-1)
S= na + d((n-1)(n)/2) 0Σrn-1 = (n-1)(n)/2
S = n/2 (2a + (n-1)d)
let L be the last term
L = a + (n-1)d
S = n/2(a + L)
Now a + L /2 is the arithmetic mean = M
Therefore
S = nM