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
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...
405
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