Pages: [1]



Author

Topic: Arithmetic Mean (Read 4231 times)

wiki kid
Full Member
Karma: +0/0
Offline
Posts: 148

Find 4 Arithmetic mean between 2 and 23.
How can i solve this question
Thanks wiki kid



Logged




The Edifier
Full Member
Karma: +0/0
Offline
Posts: 184
Ask. We will try helping you...

A series is in AP if each subsequent number is obtainable by adding the same number to the prior one.
x_{1}, x_{2}, x_{3}, .... are in AP if
x_{2} = x_{1} + d
x_{3} = x_{2} + 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
x_{3}  x_{2} = x_{2}  x_{1} =  d 
x_{1}  x_{2} = x_{2}  x_{3} =  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 n^{th} term of a serires is in AP is obtainable from the relation.
T_{n} = a + (n 1)d
Every number in a series in AP, is the arithmetic mean of the two numbers on either side of it.
x_{1}, x_{2}, x_{3}, x_{4}
x_{2} is the AM of x_{1} and x_{3}
x_{3} is the AM of x_{2} and x_{4}
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 => T_{1} = 2 => a = 2 the second one the last term of the series.
=> T_{n} = 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, T_{n} = 33 => T_{6} = 33
T_{6} = 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 ..



Logged

Regards Team FutureAccountant at krishbhavara



wiki kid
Full Member
Karma: +0/0
Offline
Posts: 148

=> 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



Logged




The Edifier
Full Member
Karma: +0/0
Offline
Posts: 184
Ask. We will try helping you...

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
Check



Logged

Regards Team FutureAccountant at krishbhavara



Pages: [1]





 