Answer :

shivam2000
1+2+3+4+5+6+7+.....................+200

Here first term , a = 1
Common difference, d = a₂ - a₁
                                 = 2 - 1
                                 =1
Last term or an = 200
Number of terms = 200

Sn = n/2 [ 2a + (n-1)d]
Sn = 100[2 + 199]
Sn = 100 × 201
Sn = 20100
birendrak1975
The formula for finding 1+2+3+4+5+.......+n is n(n+1)/2
So in this case n=200
200(200+1)/2=200 x 201/2=40200/2=20100
therefore sum of1+2+3+4+5+6+7+...........................+200 is 20100

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