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Arithmetic Mean
« on: February 24, 2011, 03:03:08 PM »

Find 4 Arithmetic mean between -2 and 23.

How can i solve this question

Thanks
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Re: Arithmetic Mean
« Reply #1 on: February 25, 2011, 08:55:28 AM »

A series is in AP if each subsequent number is obtainable by adding the same number to the prior one.


x1, x2, x3, .... are in AP if

        x2 = x1 + d

        x3 = x2 + d

The common number is called the common difference ....

as the series can be interpreted as

The absolute difference of every two successive numbers taken in the same order would be the same

        x3 - x2 = x2 - x1  = | d |

        x1 - x2 =  x2 - x3  = | d |

A series can be completely built up or any number of the series can be found out if we know the values of the first number of the series and the common difference.

Where a is the first number and d the common difference of a series in AP, the series in its general form can be written as

              a, a + d, a + 2d , a + 3d , ..................

The nth term of a serires is in AP is obtainable from the relation.

      Tn = a + (n -1)d


Every number in a series in AP, is the arithmetic mean of the two numbers on either side of it.

x1, x2, x3, x4

         x2 is the AM of x1 and x3

         x3 is the AM of x2 and x4



Finding arithmetic means between two numbers
=> inserting numbers between the two given numbers such that the series would be in AP


Finding 4 arithmetic means between -2 and 23
=> inserting 4 numbers between -2 and 23 theryby making the series an AP

Let d be the common difference of the so formed AP

Therefore the 4 numbers to be inserted would be

                 -2 + d,  -2 + 2d, -2 + 3d, -2 + 4d

The series would therefore be

                -2, -2 + d,  -2 + 2d, -2 + 3d, -2 + 4d, 33

Of the two numbers given,
              the first one would form the first term of the series and
                  => T1 = -2
                  => a = -2
              the second one the last term of the series.

                  => Tn = 33

There would be a total of 6 terms in the series after inserting the 4 numbers
       => n = 6
       => The last term of the series obtained is the 6th term of the series.

       Thus, Tn = 33
                 => T6 = 33

 T6 = 33

=> -2 + (6 - 1) x d = 33    [a = -2 and n = 6]

=> -2 + 5 d = 33   



Find the value of 'd' and thereby write down the numbers ..



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Re: Arithmetic Mean
« Reply #2 on: February 25, 2011, 01:16:29 PM »



=> -2 + (6 - 1) x d = 33    [a = -2 and n = 6]

=> -2 + 5 d = 33   

=> 5d = 35

=> d = 7

Taking the series 2, -2 + d,  -2 + 2d, -2 + 3d, -2 + 4d, 33

=> 2, -2 + 7, -2 + 3(7), -2 + 4(7), 33
 
=> 2, 5, 19, 26, 3

This is how it should be  Huh
 

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Re: Arithmetic Mean
« Reply #3 on: February 26, 2011, 06:07:17 PM »

     -2,    -2 + d,    -2 + 2d,        -2 + 3d,         -2 + 4d,         33

=> -2,   -2 + 7,                        -2 + 3(7),       -2 + 4(7),      33
 
=>   2,       5,                               19,                  26,             3

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