Q.First 3 terms of the sequence $a_n = \dfrac{2n-3}{6}$ is

Sequence Terms Evaluation

A parking garage numbers its levels 1, 2, 3, ... and posts a rule on the wall: "Level $n$ sits $n^2$ metres below street level." You don't need to walk down to level 10 to find its depth -- you simply compute $10^2 = 100$ metres. That one substitution is the entire idea behind sequence terms evaluation: once a sequence has a rule connecting position to value, any term can be found directly, without listing every term before it.

What "Evaluating a Term" Means

A sequence assigns a number to each position $1, 2, 3, \dots$. The term at position $n$ is written $a_n$ (read "a sub n"). When the sequence is given by an explicit formula $a_n = f(n)$, evaluating the $k$-th term simply means substituting $n = k$ into $f$ and simplifying.

The Precise Statement

If $\{a_n\}$ is a sequence with $a_n = f(n)$, then for any positive integer $k$, the $k$-th term is

The entire procedure is: replace $n$ with the required index, then simplify the resulting number. No equation-solving, no guesswork -- just substitution.

Common Forms You'll Meet in Exams

Worked Example

Find the 7th term of the sequence $a_n = \dfrac{3n-1}{n^2+2}$.

1. The index we need is $k = 7$.
2. Substitute $n = 7$:

3. Check whether it simplifies -- $\gcd(20, 51) = 1$, so $\dfrac{20}{51}$ is already in lowest terms and is the final answer. …