Question

If the mean deviation of the numbers 1, 1 + d, 1 + 2d, …….. , 1 + 100d from their mean is 255, then the value of d is equal to: -
(a) 10.0
(b) 20.0
(c) 10.1
(d) 20.2

Answer 1

Hint: Apply the formula for mean deviation given as: - Mean deviation = \[\dfrac{1}{n}\sum\limits_{i=1}^{n}{\left| \overline{x}-{{x}_{i}} \right|}\], where ‘n’ is the number of terms, ‘\[\overline{x}\]’ is the mean of the given numbers and ‘\[{{x}_{i}}\]’ is the given number where i = 1, 2, 3, …. , n. To find the mean, \[\overline{x}\] add the given numbers by using the formula of sum of ‘n’ terms of A.P. given as: - \[{{S}_{n}}=\dfrac{n}{2}\left[ {{1}^{st}}\text{term}+ \text{last term} \right]\], and divide this sum by ‘n’. Substitute all the given values and the founded value in the mean deviation formula to find the value of ‘d’.

Complete step-by-step solution:
We have been given the numbers 1, 1 + d, 1 + 2d, … , 1 + 100d and their mean deviation is given as 255. We have to find the value of ‘d’.
Now, we know the formula for Mean deviation is: -
Mean deviation = \[\dfrac{1}{n}\sum\limits_{i=1}^{n}{\left| \overline{x}-{{x}_{i}} \right|}\], where ‘n’ is the number of terms, ‘\[\overline{x}\]’ is the mean or average of the given numbers, and ‘\[{{x}_{i}}\]’ are the numbers given, where i = 1, 2, 3, ….. , n.
The value of the Mean deviation is given as 255. Therefore, substituting in the above formula, we get,
\[\Rightarrow 255=\dfrac{1}{n}\sum\limits_{i=1}^{n}{\left| \overline{x}-{{x}_{i}} \right|}\] - (1)
Now, let us find the value of \[\overline{x}\].
The given numbers are 1, 1 + d, 1 + 2d, ….. , 1 + 100d. These can be written as 1 + 0.d, 1 + d, 1 + 2d, ….. , 1 + 100d. So, clearly, we can see that the co – efficient of ‘d’ is starting from 0 and ending at 100. Therefore, counting from 0 to 100, we will get 101 terms. So, n = 101.
\[\Rightarrow \overline{x}=\dfrac{1+\left( 1+d \right)+\left( 1+2d \right)+......+\left( 1+100d \right)}{n}\]
\[\Rightarrow \overline{x}=\dfrac{1+\left( 1+d \right)+\left( 1+2d \right)+......+\left( 1+100d \right)}{101}\] - (2)
Clearly, the numerator in the above relation is an A.P. with 101 terms. So, the sum is given as: -
\[\begin{align}
  & \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ {{1}^{st}} \text{term}+ \text{last term} \right] \\
 & \Rightarrow {{S}_{101}}=\dfrac{101}{2}\left[ 1+\left( 1+100d \right) \right] \\
 & \Rightarrow {{S}_{101}}=\dfrac{101}{2}\left[ 2+100d \right] \\
 & \Rightarrow {{S}_{101}}=\dfrac{101}{2}\times 2\times \left[ 1+50d \right] \\
\end{align}\]
Cancelling the common terms we get,
\[\Rightarrow {{S}_{101}}=101\times \left[ 1+50d \right]\]
Substituting the value of \[{{S}_{101}}\] in equation (2), we get,
\[\begin{align}
  & \Rightarrow \overline{x}=\dfrac{101\times \left[ 1+50d \right]}{101} \\
 & \Rightarrow \overline{x}=1+50d \\
\end{align}\]
Now, substituting the value of \[\overline{x}\] in equation (1), we get,
\[\begin{align}
  & \Rightarrow 255=\dfrac{1}{101}\times \sum\limits_{i=1}^{101}{\left| \left( 1+50d \right)-{{x}_{i}} \right|} \\
 & \Rightarrow 255=\dfrac{1}{101}\times \left\{ \left| 1+50d-{{x}_{1}} \right|+\left| 1+50d-{{x}_{2}} \right|+.......+\left| 1+50d-{{x}_{101}} \right| \right\} \\
 & \Rightarrow 255=\dfrac{1}{101}\times \left\{ \left| 1+50d-1 \right|+\left| 1+50d-\left( 1+d \right) \right|+......+\left| 1+50d-\left( 1+100d \right) \right| \right\} \\
\end{align}\]
Solving the terms inside the bracket, we get,
\[\Rightarrow 255=\dfrac{1}{101}\times \left\{ \left| 50d \right|+\left| 49d \right|+.....\left| 0d \right|+\left| -d \right|+\left| -2d \right|+......+\left| -50d \right| \right\}\]
Removing the modulus sign, we get,
\[\begin{align}
  & \Rightarrow 255=\dfrac{1}{101}\times \left\{ 50d+49d+.....d+0+d+2d+......+50d \right\} \\
 & \Rightarrow 255=\dfrac{1}{101}\times 2\left\{ d+2d+3d+......+50d \right\} \\
\end{align}\]
\[\Rightarrow 255=\dfrac{1}{101}\times 2d\left\{ 1+2+3+......+50 \right\}\] - (3)
Now, \[\left( 1+2+3+......+50 \right)\] is the sum of 50 terms of A.P.
\[\begin{align}
  & \Rightarrow {{S}_{50}}=\dfrac{50}{2}\left[ 1+50 \right] \\
 & \Rightarrow {{S}_{50}}=\dfrac{50}{2}\times 51 \\
\end{align}\]
Substituting the value of \[{{S}_{50}}\] in equation (3), we get,
\[\begin{align}
  & \Rightarrow 255=\dfrac{1}{101}\times 2d\times \dfrac{50\times 51}{2} \\
 & \Rightarrow d=\dfrac{255\times 101\times 2}{2\times 50\times 51} \\
 & \Rightarrow d=\dfrac{5\times 101\times 2}{100} \\
 & \Rightarrow d=\dfrac{101}{10} \\
 & \Rightarrow d=10.1 \\
\end{align}\]
Hence, option (c) is the correct answer.

Note: One may note that the modulus of a given number is always positive, it doesn’t matter, whether the number is positive or negative inside the modulus sign. You can see that the term of ‘d’ first goes from 50 to 0 and then again goes from 1 to 50. So, as a result, it becomes twice the sum of numbers from 1 to 50. These small observations helped us in applying the formula of A.P.