Question

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

• A. $6,7$
• B. $7,8$
• C. $8,9$
• D. $9,10$

Answer 1

The correct option is B $7,8$

Let $p,p+1$ be 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$16t^2-95t+1<8t+2$
or$16t^2-103t-1<0$ 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, $n=2+8t=2+8\times 7=2+48=50$ and
$p=16t^2-95t+1=166^2-95\times 6+1=7$
$\therefore n=50$ and $p=7$
Hence the removed numbers are $7,8$ (from above)