Let 'n' be the total no. of pages in the book.
And, let the page with page no. 'p' be the one which was mistakenly added twice
Hence, the sum by including page 'p' twice would be:-
\(sum=\frac{n(n+1)}{2}+p=1000\)Therefore,\(\frac{n(n+1)}{2}=(1000-p)\)Now, examining the options carefully:-
If p=10 then \(\frac{n(n+1)}{2}=990\)or n(n+1)=2x990=2x9x2x5x11=(2x2x11).(9x5)=44x45
Hence, n=44
We see that by using p=10 we obtain a proper integral value of 'n'.
Therefore. page no. 10 was added twice
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If you shut your door to all errors, truth will be shut out.