On the GRE, there are two types of sequences to watch out for: arithmetic, and geometric. An arithmetic sequence occurs when there is a constant difference between terms. For example, in a sequence of 3, 5, 7, 9…, then the difference is +2. In a geometric sequence, there is a constant ratio and not a constant difference.
The common ratio is found by dividing the 1st term into the 2nd term. For example, in a sequence of 2, 4, 8, 16…, the ratio is 2, since each term is multiplied by 2 to get the next term.
The concept of sequences is fairly simple, but what to do when a question asks for an impossibly high term, such as the 149th term? There isn’t enough time to write the sequence out that far, so we’d use one of the following formulas:
For Arithmetic: an = a1 + (n – 1)d
For Geometric: an = a1 * r(n-1)
In these equations, an = nth term, a1 = first term in the sequence, d = difference, r = ratio, and n = the number of the term you want to find. For example, if we were asked to find the 33rd term in the geometric sequence above, we would plug in as follows:
an = 2 * 2(33 – 1)
an = 233
Let’s look at a practice question:
1. In the sequence of numbers, a1, a2, a3, a4, a5, each number after the first is 5 times the preceding number. If a4 – a1 is 93, what is the value of a1?
For this question, it is best to choose simple numbers to see the pattern. If a1 is 1, then we know that a2 = 5, a3 = 25 and a4 = 125, so we know that a4 will be 5*5*5 or 125 times the value a1,. No matter what we choose as a1, a4 will always be 125 times greater than a1. We need to find a value such that 125x – x = 93
124x = 93
x = 3/4