This can be found out by using the formula for sum of n terms of an arithmetic progression.
here,
n is not known.
a=2
d=2 (since it's even)
nth term=98 (last even number less than 100)
using formula for nth term,
nth term=a+(n-1)d
98=2+(n-1)2
therefore, n=49
so, sum of n terms=n/2[2*a+(n-1)d]
putting n=49, a=2, and d=2,
sum=2450
There are no positive integers less than any negative ones.
Given any positive odd integer x the number of positive even integers less than x is given by (x-1)/2.
3, 2, 1
6
7 of them.
The set of positive odd integers.
Should be 50! Every odd integer is 1 less than the corresponding even integer and there are 50 of each in 100...
Five of them.
Assuming you refer to positive integers, the answer is 10:9
The sum of the first positive odd integers less than 101 is 10,000.
The positive even integers are: 0 2 4 6 and 8
the negative integers are below 0, for example -6.
There are 44 positive integers less than 2,010 that have an odd number of factors.
Positive integers are greater than zero. Negative integers are less than zero.
666 integers.
49
The sum of all the digits of all the positive integers that are less than 100 is 4,950.