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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?

  • 1
    Who says the two removed numbers are neighbors? – 2017-01-28
  • 1
    Do the numbers removed need to be consecutive? – 2017-01-28
  • 0
    Yes i forgot to mention they are consecutive – 2017-01-28

2 Answers 2

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Hint:
After getting $2n^2-103n+206=8k$, notice that $1\leq k\leq n-1$
So $$8\leq 2n^2-103n+206\leq 8(n-1)$$ Then you get only few possible values of n.

  • 2
    Just to add: For the inequality on the left you get that $n\leq 2$ or $n\geq 50$. For the other one you get that $2\leq n \leq 53$. Hence, the values of $n$ that satisfy both inequalities are $50 \leq n\leq 53 $ – 2017-01-28
  • 0
    So then $n=50$ and the two numbers removed are $7$ and $8$. – 2017-01-28
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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.

  • 0
    It should be 103 instead of 102 in the calculation. Also, the OP already reached at this stage and asks how to proceed from this. – 2017-01-28