Missing Pages...

Let page numbers m to n be missing.

∴ n(n+1)/2 - m(m+1)/2 = 9808

∴ n(n+1) - m(m+1) = 9808*2

∴ n^2 + n - m^2 - m = 9808*2

∴ (n+m)(n-m) + (n-m) = 9808*2

If we take prime factors of 9808 * 2, we get 2^5 * 613.

Considering the first possibility of two factors as 32 and 613, we get the following:

(n-m)(n+m+1) = 32 * 613

∴ n – m = 32 and n + m = 612

Solving the above equations,

n = 322 & m = 290

Let the number of missing pages be n and the first missing page be p+1. Then the pages p+1 up to and including p+n are missing, and n times the average of the numbers of the missing pages must be equal to 9808:

n×( ((p+1)+(p+n))/2 ) = 9808

n×(2×p+n+1)/2 = 2×2×2×2×613

n×(2×p+n+1) = 2×2×2×2×2×613

One of the two terms n and 2p+n+1 must be even, and the other one must be odd. Moreover, the term n must be smaller than the term 2p+n+1. It follows that there are only two solutions:

n=1 and 2×p+n+1=2×2×2×2×2×613, so n=1 and p=9808, so only page 9808 is missing

n=2×2×2×2×2 and 2×p+n+1=613, so n=32 and p=290, so the pages 291 up to and including 322 are missing

Since the question is which pages (plural) are missing, the solution is that the pages 291 up to and including 322 are missing.

Let the pages number missing are ( $p+1),(p+2),(p+3),...,(p+n)$ . Th the pages number is an AP so that

the sum of the pages number is $\frac{n}{2} \times [(p+1)+(p+n)]= 9808$ .

$n \times (n+2p+1)= 19616 = 32 \times\ (32+2\times 290+1$ )

Now, we get $p=290$ and $n=32$ . Thus, the answer is $\boxed{291.322}$