Progression01: Sum and nth term of a series

Any series a, a+d, a+2d, a+3d, ..........

where a --> First term
          d --> difference between the terms
          n --> number of terms

nth term of the series 
First term -> a
Second term -> a+1d
Third term -> a+2d
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.
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nth term -> a+(n-1)d

n'th term of the series = a + (n-1)d

Sum of the first n terms

S = a+a+d+a+2d+a+3d+.........+a+(n-1)d
   = na+d+2d+3d+.....(n-1)d
   = na+d(1+2+3+......+(n-1))
   = na+d((n-1)(n-1+1)/2)
   = n(a+(n-1)d/2)

Sum of the first n terms = \(\dfrac{n}{2}\left(2a+\left(n-1\right)d\right)\)

Number of terms in the series

Number of terms = \(\dfrac{\left(LastTerm-FirstTerm\right)}{d}+1\)