Question

sum of all natural numbers from 100 to 200 with process

Answer 1

n
$n(n + 1) \div 2$

Answer 2

100, 101, 102, .......... 200
a= 100
d= 1
Tn =200
=> a+(n-1) d = 200
=> 100 + (n-1) 1 = 200
=> n-1 = 100
=> n = 101

No. of terms = 101

S101 = 101[2*100 + (101-1) 1]/2
S101 = 101 (200+100)/2
S101 = 101*300/2
S101 = 101 * 150
S101 = 15150

Sum of all natural numbers from 100 to 200 is 15150