The sum of the positive integers between 99 and 999 is 299500
Since a = 100, and d = 2 note that the last term will be 998 using nth tem of an Ap
Un= a+(n-1)d
998= 100+ (n-1)2
Then 998- 100= (n- 1)2
= 898= (n-1)2
449= n-1
n= 450
Sum of the terms = n/2(2a+(n-1)d)
450/2(2(100) + 449*2)
Ans= 247050
=
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