Arithmetic Mean Formula | Practice Questions for Arithmetic Mean

In this article explained about Definition, Properties, Formula and Examples with Solutions of Arithmetic Mean

Basic Concept, formula with examples for Arithmetic Mean

The arithmetic mean “A” of any two quantities of ” a” and ” b”. Then

Here a, A, b are in A.P . We much have b – A = A- a ; Each being equal to the common difference

For any given two quantities it is always possible to insert any number of terms such that the whole series thus formed shall be in A.P. Then terms thus inserted are called the Arithmetic mean.

Properties of Arithmetic Mean

1. Mathematically Arithmetic Mean is also called the average of quantities

 

i.e If r₁, r₂, r₃, r₄, . . . . .  r_n be an finite A.P with n terms. Then arithmetic mean “A” of these numbers is given by

A = (1/n) [ r₁ + r₂ + r₃ + r₄,+. . . . .  + r_n ]

1. The “n” numbers r₁, r₂, r₃, r₄, . . . . .  r_n  are said to be an arithmetic mean between “a” and “b” if a, r₁, r₂, r₃, r₄, . . . . .  r_n, b  are in A.P

Here, the common difference of the above A.P is

r₁ + r₂ + r₃ + r₄,+. . . . .  + r_n   

similarly 

2. Selection of the terms in an Arithmetic Progression

If the number of terms is 3, then assume them as ” a-d, a & a+d” and the common difference is  “d”

If the number of terms is 4, then assume them as  ” a-3d, a-d, a+d & a+3d ”  and the common difference is  “2d”

If the number of terms is 5, then assume them as  ” a-2d, a-d, a, a+d & a+2d ”  and the common difference is  “d”

If the number of terms is 6, then assume them as  ” a-5d, a-3d, a-d, a+d ,  a+3d & a+5d ”  and the common difference is  “2d”

Arithmetic Mean Examples

Example-1: In an A.M, the sum of three consecutive terms is -3, and their product is 8. Then, find the terms.

Solution: Let the terms of A.M is, a, b, & c

According to the A.M formula

a + c = 2b  . . . . . .  ( i )

a + b + c = -3 . . . . . .  ( ii )

abc = 8 . . . . . .  ( iii )

From equation ( i ) & ( ii )  b = -1

From equation ( ii ) , ( iii )  & b = -1

a = 2  and  c = -4

So the sequence is 2, -1, -4

Example- 2: Find the Arithmetic mean of numbers 1/2 and 1/3

Solution: 

Arithmetic mean = [ (1/2) + (1/3) ] / 2 = 5/16

Example – 3: Find the second term of the given A.M terms  , X  , 

Solution: The terms can be rewritten as log₂ 3 , X,   log₂ 12

⇒ 2X = log₂ 3 +  log₂ (4 x 3) = log₂ 3 +  log₂ 4 + log₂  3

⇒ 2 X = 2 ( log₂ 3) +   2 log₂ 2

⇒ 2 X = 2 ( log₂ 3) +   (2 x 1 )

⇒ X =  log₂ 3 +   1  = log₂ 3 + log₂ 2 = log₂ 6

Example – 4: If  10-r , a , r +2  are the consecutive terms of an A.P then find the value of ” a” 

Solution: Given consecutive terms are 10-r , a , r +2 

According to the formula of Arthritic mean

⇒ (10-r ) + ( r +2 ) = 2a

⇒ 12 = 2a ⇒ a = 6

Example – 5: If (a+2), (4a -6) & (3a -2) are the consecutive terms of an A.P, then find the value of ” a”

Solution: Given consecutive terms are (a+2), (4a -6) & (3a -2)

According to the formula of Arthritic mean

⇒ (a+2) + (3a -2) = (4a -6)

⇒ 4a = 12 ⇒ a = 3

Example – 6: If  1/a , 1/b and 1/c are in A.P then bc, ca & ab are also in A.P

Solution: Given consecutive terms are 1/a , 1/b and 1/c are in A.P

Multiplying the above terms with ” abc”

Then abc/a , abc/b and abc/c  are in A.P

So bc , ca & ab are in A.P

Example – 7: If  a,b & c are in A.P then 1/bc, 1/ca & 1/ab are also in A.P

Solution: Given consecutive terms are a, b & c are in A.P

Multiplying the above terms with “1/abc”

Then a/abc , b/abc  &  c/abc  are in A.P

So 1/bc, 1/ca & 1/ab are in A.P

Example -8: If  1/a,1/b  & 1/c terms are in arithmetic progression then the terms of  b+c/a, c+a/b & a+b/c are also in A.P

Solution: Given consecutive terms are 1/a , 1/b & 1/c are in A.P

Multiplying the above terms with ” a+b+c”

 a+b+c/a , a+b+c/b &  a+b+c/c are also in A.P

  are in A.P

Add ” -1″ to all terms then

  are in A.P

Example – 9: If  1/a,1/b,1/c terms are in arithmetic progression then the terms of  b+c-a/a, c+a-b/b, a+b-c/c are also in A.P

Solution: Given consecutive terms are 1/a , 1/b &  1/c are in A.P

Multiplying the above terms with ” a+b+c”

 a+b+c/a , a+b+c/b and a+b+c/c are also in A.P

  are in A.P

Add ” -2″ to all terms then

  are in A.P

  are in A.P

  are also in A.P

Example -10: If a, b & c  terms are in arithmetic progression, then the terms of  b+c, c+a , a+b are also in A.P.

Solution: The given terms are a, b & c in A.P

According to the Arithmetic mean formula

a + c = 2 b . . . . . . . .  ( i )

Take the terms b+c, c+a , a+b 

Add the 1st and 3rd terms of the above

i.e (b + c )+ (a +b) = (a +c) + 2b

Now substitute equation  (i) in the above

(a +c) + 2b  = (a +c) + (a +c)  =  2(a +c)

So The terms b+c, c+a , a+b are in A.P.

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