Finding the general term of a sequence

Solution 1:

$$-\frac{1}{6},\,\frac{2}{7},\,\frac{5}{8},\,\frac{8}{9},\,\frac{11}{10},\,\ldots \quad a_n = \;\;?$$

Look for patterns:

• Numerator: each term increases by $3$, starting at $-1$ when $n = 0\;\; \rightarrow \;\;3\cdot n - 1\;$?

• Denominator: each term increases by $1$, starting at $6$ when $n = 0 \;\; \rightarrow \;\;1\cdot n+6\;$?

Note that both the numerators and the denominators form arithmetic progressions.

Test it:

$$\{a_n\},\; n\in \{0, 1, 2, 3, ...\};\;\;a_n\;=\; \dfrac{3n - 1}{n+6} \quad = \quad -\frac{1}{6},\,\frac{2}{7},\,\frac{5}{8},\,\frac{8}{9},\,\frac{11}{10},\,\frac{14}{11},\, \frac{17}{12},\,\ldots$$

If testing it fails, try to "tweak it" the expression of the general term (reevaluate to ensure correct term for $a_0$, ensure that the numerator and denominators do, in fact, represent arithmetic progressions...etc.)

Not applicable to your particular sequence, but a useful tool to have in your toolbox of strategies: Another progression you'll see arise frequently in sequences is the geometric progression, another "good to know" progression to learn so you can spot it when it occurs, and make use of how to form the general expression for a term in such a sequence.