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
.
.
.
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 = n2(2a+(n−1)d)
Number of terms in the series
Number of terms = (LastTerm−FirstTerm)d+1
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