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The arithmetic mean between two numbers is defined to be the sum of numbers divided by the quantity of numbers. Likewise, in a collection, the arithmetic mean is defined to be the sum of the collection of numbers divided by the number of numbers present in the collection.
Another term that is closely associated with arithmetic mean is arithmetic progression. While arithmetic mean is abbreviated as A.M., arithmetic progression is abbreviated as A.P. If three terms are in A.P., then the middle term is called the arithmetic mean (A.M.) between the other two. Mathematically, we can say that
If a, b, c are in A.P. then b = (a+c)/2 is the A.M. of a and c.
As discussed above, in order to find the arithmetic mean between a and b, we just need to divide the sum of terms by 2. Hence, in the same manner, mathematically, it can be said that if a_{1}, a_{2}, ……, a_{n} are n numbers, then the arithmetic mean of these numbers is
A.M = 1/n (a_{1} + a_{2} + ….. + a_{n}).
Hence, if in particular we wish to find the arithmetic mean between 2a and 2b then it is simply a+b.
Arithmetic mean is an important measure of central tendency. Some of its basic properties are listed below:
* The numbers lying to the left of mean and those lying to the right of mean balance each other. It is the unique number for which the deviations sum to zero. Mathematically, we can write it as,
If x_{1}, x_{2}, …….., x_{n} have mean as X, then by this property
(x_{1} –X) + (x_{2} – X) + …….. (x_{n} – X) = 0.
* The arithmetic mean is the best possible number to represent a group of values. The reason behind this is that it minimizes the sum of squared deviations from the particular value. Mathematically, the squared deviations imply the value (xi – x)^{2}.
v The n numbers A_{1}, A_{2}, ……, A_{n} are said to be A.M.’s between the numbers a and b if a, A_{1}, A_{2}, ……, A_{n}, b are in A.P. If d is the common difference of this A.P., then b = a + (n + 2 – 1)d ⇒ d = b–a/n+1.
⇒ A_{1} = a + b–a/n+1 = na+b
A_{2} = a + 2(b–a)/n+1
…..
A_{n} = a + n(b–a)/n+1 = a+nb/n+1.
* When the number of items is multiplied by the arithmetic mean, it gives the total of items.
* If X_{1} and X_{2} are the arithmetic mean of two samples of sizes n_{1} and n_{2}, then the arithmetic mean of the distribution considering both the samples is given by
X = (n_{1}X_{1 }+ n_{2}X_{2}) / (n_{1}+ n_{2})
* Suppose the mean of ‘n’ items is x. then if each observation is increased by ‘t’ then the mean of the new set of observations is (x+t).
We shall now move on to certain illustrations based on the concept of arithmetic mean:
Illustration: The average marks of three groups of students having 70, 50 and 30 students respectively are 50, 55 and 45. Find the average marks of all the 150 students, taken together.
Solution: let x be the required average marks of all 150 students taken together. Further, let us assume that x_{1}, x_{2} and x_{3} denote the mean or average of groups 1, 2 and 3 respectively and likewise, n_{1}, n_{2} and n_{3} denote the number of students in each group.
Group 1
Group 2
Group 3
X_{1}
X_{2}
X_{3}
Average Mean
50
55
45
Number of students
70
30
x = (n_{1}x_{1} + n_{2}x_{2} + n_{3}x_{3})/ (n_{1} + n_{2} + n_{3})
= (70.50 + 50.55 + 30.45) / (70+50+30)
= 7600/ 150
= 50.67
Hence the average marks of all 150 students taken together are 50.67 marks.
Remark: Though the concepts of geometric mean and harmonic mean will be discussed in later sections but it is important to recall the relation between the three types of means as it is helpful in solving numerical.
A.M > G.M > H.M
Illustration: If 1^{st} and 2^{nd} terms of an A.P. are 1 and –3 respectively, find the nth term and the sum of the first n terms.
Solution: Let us assume the first term to be ‘a’ and the common difference is‘d’.
Then,
1^{st} term = a, 2^{nd} term = a + d where a = 1 and a + d = –3,
⇒ d = –4(common difference of A.P.)
⇒ a_{n} = a + (n – 1)d = 1 + (n–1)(–4) = 5 – 4n.
and S_{n} = n/2{a + a_{n}} = n/2{1 + 5 – 4n} = n(3 – 2n).
Illustration: If 6 arithmetic means are inserted between 1 and 9/2, find the 4^{th} arithmetic mean.
Solution: Let a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6} be six arithmetic means
⇒ 1, a_{1}, a_{2}, ……, a_{6}, 9/2 will be in A.P.
Now, 9/2 = 1 + 7d ⇒ 7/2 = 7d ⇒ d = 1/2.
Therefore, the fourth arithmetic mean is 3.
Illustration: The product of n positive numbers is unity. Then their sum is
1. A positive integer 2. Divisible by n
3. Equal to n + 1/n 4. Never less than n
Solution: We know that A.M. ≥ G.M.
Hence, 1/n [r_{1} + r_{2} + r_{3} + …… + r_{n}] ≥ ( r_{1}r_{2}r_{3} …… r_{n})^{1/n }≥ 1.
Where r_{1}, r_{2}, r_{3}, ..…… , r_{n} are positive numbers. Hence this clearly shows that the sum of n positive numbers is never less than 1.
askIITians offers extensive study material which covers all the topics of IIT JEE Mathematics. The topic of arithmetic mean in A.P. or finding arithmetic mean between two numbers has also been explained in detail along with several solved examples. It is important to have a strong base in order to remain competitive in the JEE.
Click here for the Detailed Syllabus of IIT JEE Mathematics.
Look into the Previous Year Papers with Solutions to get a hint of the kinds of questions asked in the exam.
You can get the knowledge of Recommended Books of Mathematics here.
You may also like to refer Geometric Mean and Harmonic Mean.
To read more, Buy study materials of Sequences and Series comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Mathematics here.
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