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.