Question

Two consecutive numbers from $1,2,3,.....n$ are removed. Arithmetic mean of the remaining numbers is $\frac{105}{4}$. Find $n$-(3 times of the sum of those removed numbers) ?

Answer 1

Let $p$ and $p+1$ be the removed numbers from $1,2,...,n$, then sum of remaining numbers,
$=\frac{n(n+1)}{2}-(2p+1)$
From given condition,
$\frac{105}{4}=\frac{\frac{n(n+1)}{2}-(2p+1)}{n-2}$
$\Rightarrow 2n^2-103n-8p+206=0$
Since $n$ and $p$ are integers so $n$ must be even .
Let $n=2r$, we get
$p=\frac{4r^2+103(1-r)}{4}$
Since $p$ is an integer, then $(1-r)$ must be divisible by $4$. Let $r=1+4t$, we get,
$n=2+8t$ and $p=16t^2-95t+1$.
Now, $1\le p<n$
$\Rightarrow 1\le 16t^2-95t+1<8t+2$
$\Rightarrow t=6$
$\Rightarrow n=50andp=7$
Hence, the removed numbers are $7$ and $8$.