Rule  Description 

1  Average of two or more numbers/quantities is called the mean of these numbers, which is given by Average = \[\frac{\text{Sum of observation / quantities}}{\text{No. of observation / quantities}}\] OR Average of Numbers = \[\frac{x_1+x_2+x_3+.......+x_n}{n}\] 
2 
If the given observations (x) are occurring with certain frequency (A) then, Average = \[\frac{A_1x_1+A_2x_2+.......+A_nx_n}{x_1+x_2+.......+x_n}\] where, \[A_1, A_2, A_3, ......., A_n\] are frequencies 
3 
The average of ‘n’ consecutive natural numbers starting from 1 i.e. Average of \[1,2,3, .....n\] \[=\frac{n+1}{2}\] 
4 
The average of squares of ‘n’ consecutive natural numbers starting from 1 i.e. Average of \[1^2, 2^2, 3^2, 4^2.......x^2\] \[= \frac{(n+1)(2n+1)}{6}\] 
5 
The average of cubes of ‘n’ consecutive natural numbers i.e. Average of \[1^3, 2^3, 3^3, 4^3.......n^3\] \[= \frac{n(n+1)^2}{4}\] 
6 
The average of first ‘n’ consecutive even natural numbers i.e. Average of \[2, 4, 6, ..... 2n\] \[= (n + 1)\] 
7 
The average of first ‘n’ consecutive odd natural numbers i.e. \[1, 3, 5, ..... (2n – 1)\] \[= n\] 
8 
The average of certain consecutive numbers \[a, b, c, ......... n\] is \[\frac{a+n}{2}\] 
9 
The average of 1st ‘n’ multiples of certain numbers x is \[\frac{x(1+n)}{2}\] 
10 
If the average of ‘\[n_1\]’ numbers is \[a_1\] and the average of ‘\[n_2\]’ numbers is \[a_2\], then average of total numbers \[n_1\] and \[n_1\] is Average = \[\frac{n_1a_1+n_2a_2}{n_1+n_2}\] 
11 
If A goes from P to Q with speed x km/h and returns from Q to P with speed y km/h, then the average speed of total journey is Average speed = \[\frac{2xy}{x+y}\] = \[\frac{\text{total distance}}{\text{total time taken}}\] 
12 
If a distance is travelled with three different speeds a km/h, b km/h and c km/h, then Average speed of total journey = \[\frac{3abc}{ab+bc+ca}\] km/h 
13 
If the average of m numbers is x and out of these ‘m’ numbers the average of n numbers is y. then, the average of remaining numbers will be \[\frac{mxny}{mn}\] 
14 
If from (n + 1) numbers, the average of first n numbers is ‘F’ and the average of last n numbers is ‘L’,and the first number is ‘f’ and the last number is ‘l’ then \[f – l = n(F – L)\] 
15 
‘t’ years before, the average age of N members of a family was ‘T’ years. If during this period ‘n’ children increased in the family but average age (present) remains same, then, Present age of n children = \[n.T – N.t\] 
16 
If in the group of N persons, a new person comes at the place of a person of ‘T’ years, so that average age, increases by ‘t’ years, then, the age of the new person = \[T + N.t\] If the average age decreases by ‘t’ years after entry of new person, then, the age of the new person = \[T – N.t\] 
17 
The average age of a group of N students is ‘T’ years. If ‘n’ students join, the average age of the group increases by ‘t’ years, then Average age of new students = \[T+\left(\frac{N}{n}+1\right)t\] If the average age of the group decreases by ‘t’ years,then Average age of new students = \[T\left(\frac{N}{n}+1\right)t\] 
18 
If the average of ‘n’ observations is ‘x’ and from these the average of 1st ‘m’ observations is ‘y’ and the average of last ‘m’ observations is ‘z’ then

19 
If the average age of ‘n’ persons is x year and from them ‘m’ persons went out whose average age is ‘y’ years and same number of persons joined whose average age is ‘z’ years then, the average age of n persons = \[\left\{x\frac{m\left(yz\right)}{n}\right\}\] years 
20 
If in a group, one member is replaced by a new member, then, Age of new member = \[ \text{(age of replaced member)} \pm xn\] where,

21 
If a new member is added in a group then, age of added member = \[Average \pm x(n + 1)\] where,

22 
If a member leaves the group, then age of left member = \[Average \ income \pm x (n – 1)\] where,

23 
If average of n numbers is m, later on it was found that a number ‘a’ was misread as ‘b’. The correct average will be = \[m+\frac{ab}{n}\] 
24 
If the average of n numbers is m, later on it was found that two numbers a and b misread as p and q. The correct average will be = \[m+\frac{a+bpq}{n}\] 