help!

so the mean is (0+99)*100/2=4950 then 4950/100=49.5 so the mean is 49.5

First find the sum of the numbers.

Also, since 0 has no value, I'll take it out for now.

The number series 1, 2, 3, 4, . . . . , 98, 99.
The first term a = 1
The common difference d = 1
Total number of terms n = 99

apply the input parameter values in the formula
Sum = n/2 x (a + Tn)
= 99/2 x (1 + 99)
= 9900/2
1 + 2 + 3 + 4 + . . . . + 98 + 99 = 4950

Therefore, 4950 is the sum of positive integers from 1 to 99.

Next find the average.

The amount of numbers = 100

4950/100 = 49.5

So your final answer is  49.5

I think he/she means: 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 1 + 0(for 10) + 1 +1(for 11) + 1 + 2(for 12)......etc.

If you sum up ALL the digits from 0 to 99 = 900. The number of ALL these digits =190

900 / 190 =4.7368 - This is mean of the 190 digits.