Observe the following pattern in an arithmetic progression {eq}a_1, a_2, a_3, \cdots {/eq}
{eq}a_2 = a_1 + d\\ a_3 = a_2 + d = (a_1 + d) + d = a_1 +2 d\\ a_4 = a_3 + d = (a_1 + 2 d) + d {/eq}
So, if we want the n-th term of the sequence, it will be given by $$a_n = a_1 + (n - 1)d, $$ where {eq}a_1 {/eq} is the first term of the sequence, {eq}d {/eq} is the common difference and {eq}n {/eq}, the position of the term we want to evaluate.
Let {eq}a_1, a_2, \cdots {/eq} be an arithmetic sequence with {eq}a_1 = 5 {/eq} and {eq}d = 3 {/eq}. Find out the general term of the sequence.
We have {eq}a_n = 5 + (n - 1)3 = 5 + 3 n - 3 = 2 + 3 n {/eq}. Therefore, the general term is given by {eq}a_n = 2 + 3 n {/eq}
Example 2: How to Write the General Term of an Arithmetic Sequence
The first two terms of an arithmetic sequence are 6 and 2. Write an explicit formula for the general term.
Since {eq}a_1 = 6 {/eq} and {eq}a_2 = 2 {/eq}, it follows that {eq}d = 2 - 6 = -4 {/eq}. Plugging the values in the formula gives {eq}a_n = 6 + (n - 1) (- 4) = 6 - 4 n + 4 = 10 - 4 n {/eq}
Now that we saw the explicit formula, we are going to introduce the recursive formula. In an arithmetic progression, we have
{eq}a_1\\ a_2 = a_1 + d\\ a_3 = a_2 + d\\ \vdots {/eq}
So, in general, we have the relation a_n = a_{n-1} + d, which is called a recursive formula because it provides a way of knowing the general term knowing the previous one and the common difference.
Example 3: How to Solve Arithmetic Sequence Problems
In the fifteen-ball pool game, 15 balls are displayed in a triangle with the number of balls in each row forming an arithmetic sequence. The first row contains 5 balls, the last, 1 ball and, from one row to the next, the number of balls is decreased by 1 unity. Find the number of rows in the sequence.
The following is given: {eq}a_1 = 5 {/eq}, {eq}a_n = 1 {/eq} and {eq}d = - 1 {/eq}. We want to find {eq}n {/eq}. We are going to use the explicit formula to solve this problem.
{eq}1 = 5 + (n - 1)(- 1) \implies - 4 = - n + 1 \implies n = 5 {/eq} Therefore, there are 5 rows, as depicted below.
Scheme that represents the balls in the fifteen-ball pool game
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Example 4: How to Do Arithmetic Sequence Problems
The first and eleventh terms of a sequence are equal to 4 and 19, respectively. Find the common difference and the twenty-eighth term of the arithmetic progression.
Since {eq}a_1 = 4 {/eq} and {eq}a_{1 1} = 19 {/eq}, we have {eq}1 9 = 4 + 1 0 d \implies d = \displaystyle \frac{3}{2} {/eq}. With the common difference we can find the twenty-eighth term: {eq}a_{28} = 4 + 27 \displaystyle \frac{3}{2} = \displaystyle \frac{89}{2} {/eq}.
Sum of Finite Terms of an Arithmetic Sequence
Suppose you are given an arithmetic sequence (finite or infinite) for which you want to determine the sum of a finite number of consecutive terms. Because we are interested in a finite number of terms of the sequence, we can relabel them to {eq}a_1, a_2, a_3, \cdots, a_{n} {/eq} in case the terms we are interested in are not the first terms of the sequence. Assume that, in the given sequence, the common difference is {eq}d {/eq}. We can rewrite the terms as follows:
$$a_1\\ a_2 = a_1 + d\\ a_3 = a_1 + 2 d\\ \vdots\\ a_{n-2} = a_1 + (n-3)d\\ a_{n-1} = a_1 + (n - 2)d\\ a_n = a_1 + (n-1)d\\ $$
If we add the first and last terms, second and second to last, and so on, we are going to notice a pattern. We are going to do the first three of those sums.
{eq}a_1 + a_n = a_1 + a_1 + (n - 1)d = 2 a_1 + (n - 1)d\\ a_2 + a_{n-1} = a_1 + d + a_1 + (n - 2)d = 2 a_1 + (n - 1)d\\ a_3 + a_{n-2} = a_1 + 2 d + a_1 + (n - 3)d = 2 a_1 + (n - 1)d {/eq}
With the first three sums, we can observe that each one of them is equal to {eq}a_1 + a_n {/eq}. From {eq}a_1 {/eq} to {eq}a_n {/eq} there are {eq}n {/eq} terms and because we are grouping these terms in pairs, it turns out we have {eq}\displaystyle \frac{n}{2} {/eq} summands. Therefore,
$$a_1 + a_2 + \cdots + a_n = (a_1 + a_n) \displaystyle \frac{n}{2} $$
Let's use the above formula to evaluate the sum of the first 30 terms of a sequence with {eq}a_1 = 3 {/eq} and {eq}d = - 2 {/eq}.
We need to find {eq}a_{3 0} {/eq} before we can plug the values in the sum formula. The general term formula is {eq}a_n = a_1 + (n - 1)d {/eq}. So, we have {eq}a_{3 0} = 3 + 29 (- 2) = - 5 5 {/eq}. Now, substituting the values in the sum formula gives us {eq}a_1 + a_2 + \cdots + a_{3 0} = \displaystyle \frac{(3 - 5 5) 3 0}{2} = - 7 8 0 {/eq}
Lesson Summary
We saw that an arithmetic sequence (or progression) is a sequence where the difference between two consecutive terms is a constant. The explicit formula for the general term is {eq}a_n = a_1 + (n - 1) d {/eq}, and the recursive formula is {eq}a_n = a_{n - 1} + d {/eq}, but, in this case, we also need to be given the term {eq}a_1 {/eq}. We finished the lesson presenting the formula that gives the sum of a finite number of terms of an arithmetic sequence: {eq}a_1 + a_2 + \cdots + a_n = (a_1 + a_n) \displaystyle \frac{n}{2} {/eq}.
