Question

If the mean deviation of the numbers 1, 1+ d, ..., 1+ 100d $1,1 + d,1 + 2d...........1 + 100d$ from their mean is $255$ , then a value of d is:
 A $10.1$
 B $20$
 C $10$
 D $20.2$

Answer 1

Hint: As we know that the mean of a data set is found by adding all numbers in the data and then dividing by the number of values in the set and mean deviation is the mean of deviation of the given values . In this first calculate the mean of the data then mean deviation and compare it to the given value. Mean deviation = $\dfrac{{\sum\limits_{i = 1}^n {{{\left| {({X_i} - \bar X)} \right|}^{}}} }}{n}$ .

Complete step-by-step answer:
In this question first we have to calculate the mean of the given data that is equal to the sum of all the data divided by the total numbers of values in it .
As we see the total number of data is $101$
 Hence Mean = $\dfrac{{1 + 1 + d + 1 + 2d...........1 + 100d}}{{101}}$
 The above term is in AP with the first term is $1$ and the last term is $1 + 100d$
 Sum of this series = Number of term (First term + last term ) divided by $2$
 Therefore , on putting values
 $\dfrac{{\dfrac{{101}}{2}(1 + 1 + 100d)}}{{101}}$
$\Rightarrow$ \[\dfrac{{\dfrac{{101}}{2}(2 + 100d)}}{{101}}\]
Taking $2$ outside the bracket and dividing by $101$
we get
Mean = $(1 + 50d)$
Now we have to find the deviation of series as we know that
Mean deviation = $\dfrac{{\sum\limits_{i = 1}^n {{{\left| {({X_i} - \bar X)} \right|}^{}}} }}{n}$
= $\dfrac{{(1 + 50d - 1) + (1 + 50d - 1 - d) + ....... + (1 + 100d - 1 - 50d)}}{n}$
After further solving ;
 = $\dfrac{{50d + 49d....... + 0 + d + ....... + 50d}}{{101}}$
= \[\dfrac{{2d(1 + ....... + 50)}}{{101}}\]

= \[\dfrac{{2d(50)(51)}}{{101}}\] Since $1 + 2 + .......n = \dfrac{{n(n + 1)}}{2}$
It is given that the mean deviation = $255$
$\Rightarrow$ \[255 = \dfrac{{2d(50)(51)}}{{101}}\]
$\Rightarrow$ $d = \dfrac{{101 \times 255}}{{50 \times 51 \times 2}}$
After solving we get d = $10.1$
Hence option A is the correct answer .

Note: As in the we use the summation of A.P terms = Number of term (First term + last term ) divided by $2$. We can also use instead of this is $\dfrac{n}{2}(2a + (n - 1)d)$ where n = number of term a= first term d = common difference of the A.P .
Deviation of the term always being positive .