Question

Two consecutive numbers from $1,2,3,...,n$ are removed. The Arithmetic mean of the remaining number is $\frac{105}{4}$, Then value of ${}^{\prime }{n}^{\prime }$ is,

• A. $48$
• B. $49$
• C. $50$
• D. $51$

Answer 1

The correct option is C $50$

Let $p,p+1$ be the removed numbers from $1,2,3,...n$ then
sum of the remaining numbers $=\frac{n(n+1)}{2}-(2p+1)$

$\therefore \frac{105}{4}=\frac{n(n+1)}{2}-(2p+1)$

On evaluation we get $2n^2-103n-8p+206=0$

Since $n$ and $p$ are integers so ${}^{\prime }{n}^{\prime }$ must be even.

Let $n=2r,$ we get $p=\frac{4r^2+\left(103\right)(1-r)}{4}$

Since ${}^{\prime }{p}^{\prime }$ is an integer so $(1-r)$ must be divisible by $4$.

Let $r=1+4t$ we get $n=2+8t$ and $p=16t^2-95t+1$

Now $1\le p<n$
$\Rightarrow 1\le 16t^2-95t+1<8t+2$

On solving the equation$16t^2-95t+1<8t+2$
$\Rightarrow 16t^2-103t-1<0$

Using the formula $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, we get

$t=6.18$ or $-0.25$

As $t$ is an integer we take $t=6$

From above, $n=2+8t=2+8\times 7=2+48=50$
$\therefore n=50$