Question

Prove that in a arithmetic sequence mean and median are equal......

Answer 1

The following theorem can be proved by using the facts about the terms of an AP and there formula

Let an arithmetic progression of n terms with common difference d

The AP would be like :

a- nd, a-(n-1)d, a-(n-2)d,............ a-d, a , a+d ,................ a+(n-2)d, a+(n-1)d, a+nd

The total number of terms would be 2n+1

Here, we can apprehend that any arithmetic progression can be written in this form.

The mean of this data would be

[a- nd +a-(n-1)d +a-(n-2)d+............ a-d+ a + a+d +................+ a+(n-2)d+ a+(n-1)d+ a+nd] / 2n+1

a*(2n+1)/(2n+1)  = a

The sequence would already be sorted in increasing order, hence median would be middle most term i.e. a

Answer 2

Answer:

Step-by-step explanation:

Arithmetic sequences of numbers have the nifty property that the median and mean of the set are the same number. Furthermore the mean of an arithmetic sequence is just the average of the first and last term, and since the median is equal to the mean, the median is also the average of the first and last term.

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