how do we find sum of an arithmetic progression?

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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.


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