Suppose a sequence \(a_1,\,a_2,\,a_3,\,\cdots,a_n,\,\cdots\) has the property that the same number is being added (or subtracted) to get the next term of the sequence. In other words, \(a_k=a_{k-1}+d\), from a recursive format. When this occurs, the sequence is called an arithmetic sequence, and the number \(d\) is called the common difference.
Consider the sequence given by \(2, -3, -8, -13, \cdots\).
To check for a common difference, since \(a_k=a_{k-1}+d\), solving for \(d\): \(d=a_k-a_{k-1}\). This means, start with a term and subtract the term IN FRONT of it. $$\begin{array}{l} d=-3-2 & \implies d=-5 \\ d=-8-(-3) & \implies d=-5 \\ d=-13-(-8) & \implies d=-5 \end{array}$$ The common difference is \(d=-5\) and this sequence is arithmetic.
Since each term is "adding" \(d=-5\), a pattern can be found to help generate an explicit formula for the general term \(a_n\): $$\begin{array}{l} a_1=2 & & \implies a_1=2+(-5)(0) \\ a_2=-3 & \implies a_2=2+(-5) & \implies a_2=2+(-5)(1) \\ a_3=-8 & \implies a_3=-3+(-5) & \implies a_3=2+(-5)(2) \\ a_4=-13 & \implies a_4=-8+(-5) & \implies a_4=2+(-5)(3) \end{array}$$ The pattern emerges as starting at \(a_1=2\) and adding \(-5\) a multiple that is one less than the term number. For example, \(a_4\) starts at \(a_1=2\) and adds \(-5\) three times, \(a_3\) starts at \(a_1=2\) and adds \(-5\) two times, \(a_2\) starts at \(a_1=2\) and adds \(-5\) one time. To finish the pattern, \(a_1\) starts at \(2\) and adds \(-5\) zero times. This leads to the explicit formula: \(a_n=2+(-5)(n-1)\).
This example shows how to create a more general formula for all arithmetic sequences. Once the common difference \(d\) has been found, each term starts at \(a_1\) and adds \(d\;\;n-1\) times. This is expressed in the explicit formula below:
For an arithmetic sequence with first term \(a_1\) and common difference \(d\), the general term is given by: $$a_n=a_1+d(n-1)$$
Consider the series formed from an arithmetic sequence: \(\displaystyle{S_n=\sum_{k=1}^n\left(a_1+d(n-1)\right)}\), the \(S_n\) is a quick short-hand notation for this sum. Since the sequence is arithmetic, a formula to calculate this sum can be obtained that depends on only the first and last terms of the sequence. To illustrate how the formula is derived, consider an arithmetic sequence of just 5 terms:
| Want the sum: | \(S_5=a_1+a_2+a_3+a_4+a_5\) |
| Applying arithmetic sequence formula: | \(\begin{array}{l} S_5=a_1+(a_1+d)+(a_1+2d)+(a_1+3d)+(a_1+4d) \\ \implies S_5=5\cdot a_1+10\cdot d \end{array}\) |
| Clever trick of sequence "backwards": | \(S_5=a_5+a_4+a_3+a_2+a_1\) |
| Now subtracting common difference in formula: | \(\begin{array}{l} S_5=a_5+(a_5-d)+(a_5-2d)+(a_5-3d)+(a_5-4d) \\ \implies S_5=5\cdot a_5-10\cdot d \end{array}\) |
| Now add these two equations together: | \(\begin{array}{l} S_5=5\cdot a_1+10\cdot d \\ \underline{+S_5=5\cdot a_5-10\cdot d} \\ 2S_5=5a_1+5a_5 \end{array}\) |
| Solve for \(S_5\): | \(\begin{array}{l} 2S_5=5\left(a_1+a_5\right) \\ \displaystyle{S_5=\frac{5}{2}\left(a_1+a_5\right)}\;\color{green}{\checkmark} \end{array}\) |
This example shows how the following general formula for an arithmetic series is derived:
For an arithmetic series with first term \(a_1\) and last term \(a_n\) , the sum \(S_n\) is given by: $$S_n=\frac{n}{2}\left(a_1+a_n\right)$$