Deriving the Sum of the Arithmetic Sequence

We have discussed arithmetic progression or arithmetic sequence and you have learned how to find its nth term. In the previous post, you have also learned how to find the sum of the first n positive integers. Notice that first n positive integers 1, 2, 3, 4, 5, all the way up to n is also an arithmetic sequence with first term 1 and constant difference 1.

Now, the question that comes to mind is, how do we find the sum of an arithmetic sequence?

sum of first n integers

Recall the method that we used in “Finding the sum of the first n positive integers.” We added the integers twice with the order of the terms reversed as shown above. Clearly, we can use this method to find the sum of the arithmetic sequence 3, 7, 11, 15, 19, 23, 27.  Continue reading

Sum of the First n Positive Integers: Another Solution

Late last month, we have seen how Gauss had possibly calculated the sum of the first n positive integers.  In this post, we explore another solution. As we have done before, let us  first start with a specific example, and then generalize later. What is the sum of the first 6 positive integers?

If we let S be the sum of the first 6 positive integers, then S = 1 + 2 + 3 + 4 + 5 + 6. Since the order of the addends does not change the sum, S = 6 + 5 + 4 + 3 + 2 + 1. Adding the two equation, we have

sum-of-integers

so, 2S = 6(7) = 42 and S = 21.

We can generalize the addition above as shown in the following image.

sum-of-1-up-to-n

As we can see, each sum is equal to n + 1 and there are n of them. Therefore,

$latex 2S = n(n+1)$

which implies that

$latex S = \displaystyle\frac{n(n+1)}{2}$