Sequences and series problems can seem complicated at first, but they are simple with enough practice. Let’s look at an example:

• There are 50 soccer players lined up in a row. Their jerseys follow the pattern: blue, red, blue, red … Their soccer balls follow the pattern: brown, pink, yellow, brown, pink, yellow. How many soccer players have a blue jersey and a pink ball?

This is a sequences and series problem. We can identify these problems because the problem contains patterns. First, let’s think about this problem in a more orderly manner.

The first player has a blue jersey, the second has a red one, and the third has a blue one. We can see that every odd player has a blue jersey. The first player has a brown ball, the second player has a pink ball, the third has a yellow ball, the fourth has a brown ball, and the fifth has a pink ball. Since the length of the pattern is 3, every third player will have a pink ball, starting with the second player. The 2nd, 5th, 8th, … players will have pink balls.

For a player to have a blue jersey and a pink ball, the player number needs to be odd and align with the above pattern. Let’s write out the pattern above to visualize it better: 2,5,8,11,14,17,20,23 …

Let’s focus on only the odd terms: 5,11,17,23 …

We can see that the values in the term are added by 6, starting at 5. Thus, the final sequence is: 5,11,17,23,29,35,41,47. There are 8 terms in this sequence, so the final answer is 8.

However, sometimes the values are too large to write out. Thus, we need a systematic way of solving these equations. Let’s see what we can do.

Let’s go back to the step where we found the pattern for the final sequence being terms added by 6, starting at 5. Let’s see if we can find an equation to calculate the value of each term of the sequence.

To do this, let’s use a variable, t. This variable stands for the term number in a sequence. For the first term, t equals 1 and for the fifth term, t equals 5. Since this is an arithmetic sequence, the equation will follow this template: Value = Initial + t x Common Difference. The common difference is the difference between a term and its previous term, or what the term is being added by. In this case, the common difference is 6. So, the equation is: Value = 5 + 6t. Let’s see if this works.

When we plug this back in, we see that the first term should be 11, but the actual value is 5. What went wrong? In this problem, we are counting the first term as the initial term. To combat this, we can subtract 1 from t. The new equation will be: Value = 5 + 6(t - 1). If we plug in values for t, this equation works

To get the final answer, we need to set this equation to the upper limit, which is 50. Then, we need to solve for t, which would be the final term number. Let’s do it!

50 = 5 + 6(t-1). By the distributive property: 50 = 5 + 6t - 6, or 50 = 6t - 1.

Adding 1 to both sides, we get: 51 = 6t. If we divide both sides by 6, we get t = 8r3.

Since we cannot have half of a term, we need to round the value. We must round down as t = 9 would be above the upper limit. Thus, t = 8. This is our final answer, as term numbers start at 1 and go until 8, each number representing a different person.