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Using this method, one can show that the sum
 

Tn = 1 + 2 + 3 + 4 + ... + n

of the first n positive integers is n(n + 1)/2. Some values of Tn are given in the table which follows.
 

n 1 2 3 4 5 6 ...
Tn 1 3 6 10 15 21 ...
 
 
The sequence Tn may be defined for all positive integers n by
 
T1 = 1, T2 = T1 + 2, T3 = T2 + 3,
T4 = T3 + 4, ..., Tn+1 = Tn + (n + 1), ... .

The values 1, 3, 6, 10, 15, ... of Tn are called triangular numbers because they give the number of objects in triangular arrays of the type shown in Figure 2.

An arithmetic progression may have a negative common difference d. One with a = 7/3, d = -5/3, and n = 8 is:
 

7/3, 2/3, -1, -8/3, -13/3, -6, -23/3, -28/3.

The average (or arithmetic mean) of n numbers is their sum divided by n. For example, the average of 1, 3, and 7 is 11/3. If each of the terms of a sum is replaced by the average of the terms, the sum is not altered. We note that the average of all the terms of an arithmetic progression is the average of the first and last terms, and that the average is the middle term when the number of terms is odd, that is, whenever there is a middle term.

If a is the average of r and s, then it can easily be seen that r, a, s are consecutive terms of an arithmetic progression. (The proof is left to the reader.) This is why the average is also called the arithmetic mean.
 

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Sunday, March 15, 1998