26 AUG

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

A \ = \frac{a+b}{2}

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 average of that quantities

 Arithmetic \ Mean \ = \ \frac{Sum \ of \ terms}{Number \ of \ terms}

i.e If r1, r2, r3, r4, . . . . .  rn 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,+. . . . .  + rn ]

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

Here common difference of above A.P is d \ = \frac{b-a}{n+1}

r1 + r2 + r3 + r4,+. . . . .  + rn   = n \ \left [ \frac{a +b}{2} \right ]

r_{1} = a + \ \left [ \frac{b -a}{n+1} \right ]         r_{2} = a + \ \left [ \frac{2(b -a)}{n+1} \right ]    similarly r_{n} = a + \ \left [ \frac{n(b -a)}{n+1} \right ]

2. Selection of the terms in an Arithmetic Progression

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

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

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

If number of terms is 6 then assume them as  ” a-5d, a-3d, a-d, a+d ,  a+3d & a+5d ”  and 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 terms of A.M is, a, b, & c

According to 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 Arithmetic mean of numbers 1/2 and 1/3

Solution: Arithmetic \ Mean \ = \ \frac{Sum \ of \ terms}{Number \ of \ terms}

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

Example – 3: Find the second term of the given A.M terms \frac{1}{log_{3}2} , X  , \frac{1}{log_{12}2}

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 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 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

1 \ + \frac{b+c}{a} , \ \ \ 1 \ + \frac{a+c}{b} , \ \ \ \ 1 \ + \frac{a+b}{c} , \   are in A.P

Add ” -1″ to all terms then

\frac{b+c}{a} , \ \ \ \frac{a+c}{b} , \ \ \ \ \frac{a+b}{c}   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

1 \ + \frac{b+c}{a} , \ \ \ 1 \ + \frac{a+c}{b} , \ \ \ \ 1 \ + \frac{a+b}{c} , \   are in A.P

Add ” -2″ to all terms then

\frac{b+c}{a} -1 , \ \ \ \frac{a+c}{b}-1, \ \ \ \ \frac{a+b}{c} -1  are in A.P

\frac{b+c}{a} - \frac{a}{a} , \ \ \ \frac{a+c}{b} - \frac{b}{b}, \ \ \ \ \frac{a+b}{c} - \frac{c}{c} are in A.P

\frac{b+c-a}{a} , \ \ \ \frac{a+c-b}{b}, \ \ \ \ \frac{a+b-c}{c} 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 Arithmetic mean formula

a + c = 2 b . . . . . . . .  ( i )

Take the terms b+c, c+a , a+b 

Add 1st and 3rd term of the above

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

Now substitute equation  (i) in above

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

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

Concept, formula with examples for Arithmetic Mean | practice questions for arithmetic mean

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My self Sivaramakrishna Alluri. Thank you for watching my blog friend

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