Let the total pages are 'n'. We have
1 + 2 + 3 + .........+n = 1097
⇒ n(n+1)/2 = 1096
n (n+1) = 2194
The square nearest to 2194 is 2209.This is the square of 47. Hence, we will take n= 63.
Therefore, sum of the pages= n(n+1)/2 = 47*48/2= 1128
Missing number = 1128-1097 =31.
Now, we have to think of 2 consecutive numbers whose sum is 31.
x+x+1=31 ⇒ ×= 15, x+1= 16