Sum of all the numbers from 1 to n = n(n+1) / 2
Assuming the erased number is x
After erasing one number remaining numbers = (n-1)
∴ Sum - x = 35 7/17 (n-1)
∴ {n(n+1) / 2} - x = 35 7/17 *(n-1)
As 1 to n all are natural numbers & x is also a natural number between 1 to n
On a series of natural numbers, after erasing one natural number the maximum change in average can be only 0.5
So, post-erasing
the average will be in the vicinity of 35 (≈ 35)
Before erasing the average is in the vicinity of 35 (≈ 35)
We know that the average is the middle term of any sequence
So, the middle term is also in the vicinity of 35 (≈ 35)
∴ The number of terms n has to be in the vicinity of 70 [35*2]
∴ (n-1) also has to be in the vicinity of 70
And
post-erasing
the series will be 1 to (n-1)
∴ Sum / (n-1) = 35 7/17
∴ (n-1) has to be a multiple of 17
and as (n-1) is in the vicinity of 70
∴ (n-1) = 17 * 4 = 68
∴ n= 69
Now, we'll put the value of n in 1)
{n(n+1) / 2} - x = 35 7/17 *(n-1)
69 *70 / 2 - x = [ {(35 *17) + 7} * 68 ] / 17
2415 - x = 40936 / 17
x = 2415 - 2408
∴ x = 7
∴ 15 is erased from the sequence or series of natural numbers from 1 to 69