Question : how do we find sum of an arithmetic progression?
Solution :
n terms of an arithmetic progression is a sequence of the form a, a+d, a+2d, a+3d, a+4d,…….a+(n-1)d
It’s sum is
\[S_n = a+a+d+a+2d+a+3d+ a+4d+……+a+(n-1)d \]
\[S_n = na+d+2d+3d+4d+…..+(n-1)d\]
\[S_n = na+(1+2+3+4+…..+(n-1))d\]
\[S_n = na+((n-1)(n-1+1)/2)d\]
\[S_n = (n/2) [2a + (n – 1)× d]\]
Thus we found formula to calculate the sum of n terms of an arithmetic progression
\[S_n = (n/2) [2a + (n – 1)× d]\]
Example : Find the sum of 100 terms of the AP 1+3+5+7+9+…….
Solution :
First term a = 1 Common difference d = 3-1 = 2
Number of terms to get the sum is n =100.
Formula of Sum of n Terms of AP
\[S_n =n/2 [2a + (n – 1)× d]\]
Putting a =3, d =2, n =100
\[S_100 =100/2 [2*1 + (100 – 1)× 2]\]
=50[2 + 99× 2] =50[2 + 198]
=50[200]
=10000
Hence, the sum of the first 100 terms of AP is 10000.