Question

Two consecutive numbers from $1,2,3,........................,n$ are removed. The arithmetic mean of the remaining number is $\dfrac{{105}}{4}$, then the value of ‘$n$’ is,
A. 48
B. 49
C. 50
D. 51

Answer 1

Hint: First of all, find the sum of the remaining terms and then their arithmetic mean. As ‘$p$’ and ‘$n$’ are integers, so ‘$n$’ must be even. Then find the value of $n$ by substituting it with another variable which is always even (like $2r$). So, use this concept to reach the solution of the problem.

Complete step-by-step answer:

Given the arithmetic mean of the remaining terms when two consecutive terms are removed is $\dfrac{{105}}{4}$

Let $p,p + 1$ be the removed numbers from $1,2,3,........................,n$ then the remaining terms are $n - 2$.

Sum of the $1,2,3,........................,n$ terms $= \dfrac{{n\left( {n + 1} \right)}}{2}$

Now sum of the remaining terms = \dfrac{{n\left( {n + 1} \right)}}{2} - \left\\{ {p + \left( {p + 1} \right)} \right\\}

The arithmetic mean of remaining terms is $\dfrac{{\dfrac{{n\left( {n + 1} \right)}}{2} - \left( {2p + 1} \right)}}{{n - 2}}$ which equals to $\dfrac{{105}}{4}$.

$$\Rightarrow \dfrac{{\dfrac{{n\left( {n + 1} \right)}}{2} - \left( {2p + 1} \right)}}{{n - 2}} = \dfrac{{105}}{4} \\\ \Rightarrow \dfrac{{n\left( {n + 1} \right) - 2\left( {2p + 1} \right)}}{2} = \dfrac{{105\left( {n - 2} \right)}}{4} \\\ \Rightarrow {n^2} + n - 4p - 2 = \dfrac{{105n - 210}}{2} \\\ \Rightarrow 2\left( {{n^2} + n - 4p - 2} \right) = 105n - 210 \\\ \Rightarrow 2{n^2} + 2n - 8p - 4 - 105n + 210 = 0 \\\ \Rightarrow 2{n^2} - 103n - 8p + 206 = 0 \\\$$

Since ‘$p$’ and ‘$n$’ are integers, so ‘$n$’ must be even.

$$\Rightarrow 8p = 2{n^2} - 103n + 206 \\\ \therefore p = \dfrac{1}{8}\left( {2{n^2} - 103n + 206} \right) \\\$$

Let $n = 2r$ then

$$\Rightarrow p = \dfrac{{2{{\left( {2r} \right)}^2} - 103\left( {2r} \right) - 206}}{8} \\\ \Rightarrow p = \dfrac{2}{8}\left[ {4{r^2} - 103r - 103} \right] \\\ \Rightarrow p = \dfrac{{4{r^2} + 103\left( {1 - r} \right)}}{4} \\\$$

Since ‘$p$’ is an integer so $\left( {1 - r} \right)$ must be divisible by 4.
Let $r = 4t + 1$ then $n = 2\left( {4t + 1} \right) = 8t + 2$ and here ‘$t$’ is positive

\Rightarrow p = \dfrac{{4{{\left( {4t + 1} \right)}^2} + 103\left\\{ {1 - \left( {4t + 1} \right)} \right\\}}}{4} \\\ \Rightarrow p = \dfrac{{4\left( {16{t^2} + 8t + 1} \right) - 4\left( {103t} \right)}}{4} \\\ \Rightarrow p = \dfrac{4}{4}\left( {16{t^2} + 8t - 103t + 1} \right) \\\ \Rightarrow p = 16{t^2} - 95t + 1 \\\ \therefore p = 16{t^2} - 95t + 1 \\\

Here $1 \leqslant p < n$ as $p,p + 1$ are consecutive terms in $1,2,3,........................,n$

$$\Rightarrow 1 \leqslant 16{t^2} - 95t + 1 < n \\\ \Rightarrow 1 \leqslant 16{t^2} - 95t + 1 < 8t + 2{\text{ }}\left[ {\because n = 8t + 2} \right] \\\$$

Splitting the terms, we have

$$\Rightarrow 1 \leqslant 16{t^2} - 95t + 1{\text{ }} \\\ \Rightarrow 0 \leqslant 16{t^2} - 95t{\text{ }} \\\ \Rightarrow 16{t^2} \geqslant 95t \\\ \Rightarrow t \geqslant \dfrac{{95}}{{16}} \\\ \therefore t \geqslant 5.9375 \\\$$

And the other term is

$$\Rightarrow 16{t^2} - 95t + 1 < 8t + 2 \\\ \Rightarrow 16{t^2} - 103t - 1 < 0 \\\$$

By using the formula $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ we have

$$\Rightarrow t = \dfrac{{103 \pm \sqrt {{{\left( { - 103} \right)}^2} - 4\left( {16} \right)\left( { - 1} \right)} }}{{2\left( {16} \right)}} \\\ \Rightarrow t = \dfrac{{103 \pm \sqrt {10609 + 64} }}{{32}} \\\ \Rightarrow t = \dfrac{{103 \pm 103.31}}{{32}} \\\ \Rightarrow t = \dfrac{{103 + 103.31}}{{32}},\dfrac{{103 - 103.31}}{{32}} \\\ \Rightarrow t = \dfrac{{206.31}}{{32}},\dfrac{{ - 0.31}}{{32}} \\\ \Rightarrow t = 6.45, - 0.0097 \\\$$

Since ‘$t$’ is positive we have $t < 6.45$

So, from $t \geqslant 5.9375{\text{ and }}t < 6.45$ we get $t = 6$

As we have $n = 8t + 2 = 8 \times 6 + 2 = 48 + 2 = 50$

Therefore, the value of ‘$n$’ is $50$

Thus, the correct option is C.

Note: As ‘$t$’ is an integer we have considered the value which is satisfying the inequalities of $t$. In the equation $p = \dfrac{{4{r^2} + 103\left( {1 - r} \right)}}{4}$ , $4{r^2}$ is divisible by 4. As $p$ is an integer $\left( {1 - r} \right)$ must be divisible by 4.