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sum of all even number between 1 and 1000 is 2+4+6.......................................998
an = a+ (n-1)d
That's the sum of all terms of the arithmetic progression, with
a=2 and d=2. last term is 998
nth term = 998
therefore sum of 499 terms = 249500
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@shrishti, I think you made a slight error while calculating 996/2, which should equal 498, instead you have 488.
There are 500 even numbers
first is 2 last is 1000
second is 4 last but one is 998
You will see (just like Gauss did ) that they both sum to 1002
So as there are 250 of these 'sums' ...
250 x 1002 = 250 500
This is a correct answer
Another way of seeing this is sum from k =1 to k = 500 (2k) sorry about the lack of mathematics text.
hmmm still not good is it?
1002 * 250 = 250,500
jacobmridul check your answer again the nth term is 1000
@gianfranco, I'm afraid the question is sum of all even numbers "BETWEEN" 1 and 1000, in my humble understanding it excludes 1000. so the number of terms is 499 not 500.