The sum of the page numbers after a page is removed from a

Sum of n numbers in arithmetic progression is N x [2A + D(n-1)]/2, where
N= Number of terms....(we dont know this from in given problem)
A= 1st term (has to be 1 as its the 1st page of the book)
D= constant difference (1 as numbers must be in sequence

After simplifying above formula, we get sum of terms = N/2.(N+1).
Now we know the sum of page numbers without missing pages is 1000. Hence sum of terms must be 1000 + sum of missing number.
Now we plug the values in given options we can conclude that after only option C yields integer value of N. Hence C

Looking at the answer choices, we notice that the sum of page numbers removed from the book could be as small as 5 + 6 = 11 or as large as 21 + 22 = 43. Therefore, the actual sum of the pages before any pages were removed should be between 1000 + 11 = 1011 and 1000 + 43 = 1043.

Let’s test some numbers for the pages of books until we find a number between 1011 and 1043. As the formula for the sum of consecutive integers from 1 to n is n(n + 1)/2 and the square root of 1000 is around 30, a good place to start is:

(30 * 31)/2 = 465 - too low

(50 * 51)/2 = 1275 - too high

(40 * 41)/2 = 820 - too low

(45 * 46)/2 = 1035 - success

Since n = 45 results in a number between 1011 and 1043, the book can have a total of 45 pages. All we have to do now is to find an answer choice which adds up to 1035 - 1000 = 35.

A) 7 + 8 = 15

B) 10 + 11 = 21

C) 17 + 18 = 35

So, we know that the book had 45 pages and the sum of the page numbers was 1035 before any page was removed. After the page with page numbers 17 and 18 was removed, the sum of the remaining numbers was 1035 - (17 + 18) = 1000. The answer is C.

Answer: C

sum first positive integers: \(n(n+1)/2\) and \(n\) must be an integer value
minimum sum of page numbers without those removed: 1000
maximum sum of page numbers including those removed: 1000+21+22=1043

range: \(1043>\frac{n(n+1)}{2}>1000\)

\(n=50…\frac{n(n+1)}{2}=\frac{50(51)}{2}=1275\) too big try a smaller \(n\)
\(n=45…\frac{n(n+1)}{2}=\frac{45(46)}{2}=1035\) perfect (anything greater is outside range)

\(1035-1000=35=(17+18)\)

Answer (C)

\(\frac{n(n+1)}{2} = 1000+\)

The sum of all pages in the book is a bit above 1000.
So product of two consecutive integers n and (n+1) is a bit above 2000.

The closest square that I can think of immediately is 1600 = 40^2 and since 2500 = 50^2,
2000 would lie somewhere in the middle.
45^2 = 2025 so let's try 45*46 = 2070

This means n must be 45 and the sum of all pages must be half of 2070 i.e. 1035.

Since the new sum is 1000, a sum fo 35 is removed so the two pages must be 17 and 18 (must be consecutive).

Answer (C)

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