<s>[INST] Two consecutive numbers removed from first $n$ natural numbers If two consecutive numbers are removed from first  $n$ natural numbers, and the arithmetic mean of remaining numbers is$ \frac {105}{4}$, find $n $.
Let $k $ and $k+1$ be two removed numbers, then we have :
$$ 1+2+3+\cdots +n = 2k+1+\frac{105n-210}{4}. $$ 
Rearranging I got quadratic as  below
$$ 2n^2-103n+206-8k=0.$$
So:
$$ 4n=103 \pm \sqrt {8961+64k}. $$
How do I proceed from here?
 [/INST] The sum of the first n natural number is $\frac{n(n+1)}{2}$.  If the arithmetic mean of these remaining n-2 numbers is 105/4 then their sum is $\frac{(n- 2)105}{4}$.  The difference in sums is the sum of those two missing numbers: $\frac{n(n+1)}{2}- \frac{105(n- 2)}{4}= \frac{2n^2+ 2n- 105n+ 210}{4}= \frac{2n^2- 103n+ 210}{4}$.  Calling the two consecutive numbers that were removed "k" and "k+ 1", then we must have $k+ (k+ 1)= 2k+ 1= \frac{2n^2- 103n+ 210}{4}$.  $2n^2- 102n+ 210= 8k+ 4$ so $2n^2- 102n+ 206- 8k= 0$.  Solve for n using the quadratic formula.
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