Sequences

Count the rungs of a ladder, the days until an exam, the balance in a savings account each month — in every case you are looking at an ordered list of numbers, one for each position: first, second, third, and so on. Mathematics gives this idea a name, a sequence, and a precise definition: a function from the natural numbers into the real numbers. That single idea — attaching a number $a_n$ to every position $n = 1, 2, 3, \dots$ — is the doorway into the whole "Sequence and Series" chapter of CBSE Class 11 Algebra. This article stays exactly where Pass 1 left it: what a sequence is, the two ways to describe one (an explicit formula for $a_n$ , or a recursive rule linking $a_n$ to earlier terms), and how a sequence differs from a series. It adds one grounded derivation — turning a recursive rule into an explicit formula — a fresh worked example built on the same numbers as before, and a compact exam-card summary. No progressions or infinite sums yet; those are the next articles in the chapter.

The core insight

A sequence is a function $a: \mathbb{N} \to \mathbb{R}$ whose values $a_1, a_2, a_3, \dots$ can be described either by an explicit formula for $a_n$ or by a recursive rule linking each term to the one(s) before it, and the two descriptions are just different windows onto the same ordered list.

The core idea

Definition. A sequence is a function whose domain is the natural numbers $\mathbb{N} = \{1, 2, 3, \dots\}$ (or $\{1, 2, \dots, n\}$ for a finite sequence) and whose range is a set of real numbers. Writing the function's values in order gives the familiar picture of a sequence:

a_1,\ a_2,\ a_3,\ \dots,\ a_n,\ \dots

where: $a_n$ (also written $a(n)$ ) is the $n$ -th term — the value the function assigns to the input $n$ — and the whole list is the sequence itself. A sequence with a last term is finite; one that never ends is infinite.

Two languages for the same list. A sequence can be pinned down explicitly, by a formula that produces $a_n$ straight from $n$ — for instance $a_n = 2n - 1$ generates the odd numbers $1, 3, 5, 7, \dots$ with no reference to earlier terms at all. Or it can be pinned down recursively, by fixing one or more starting terms and a rule that builds each later term from the term(s) just before it — for instance $a_1 = 1,\ a_2 = 1,\ a_n = a_{n-1} + a_{n-2}$ for $n > 2$ generates the Fibonacci sequence $1, 1, 2, 3, 5, 8, \dots$ , where every term depends on remembering the two that came before. Both languages describe a genuine sequence; neither is "more correct" than the other. What changes is how easily you can jump to a distant term. An explicit formula lets you compute, say, $a_{100}$ directly by substitution. A recursive rule, as written, forces you to climb through every earlier term first — unless you can convert it into an explicit formula, which is exactly the derivation below.

Sequence versus series. Do not confuse the two: a sequence is a list, $a_1, a_2, a_3, \dots$ , separated by commas; a series is what results from adding those terms, $a_1 + a_2 + a_3 + \dots$ . The next article in this chapter studies series; here the terms are only ever listed, never summed. The figure below lines up the two descriptions of a sequence side by side, showing how an explicit rule reads straight across while a recursive rule must climb step by step.

Figure

Two views of the same sequence, an explicit formula reading straight across from n to a sub n, and a recursive rule climbing step by step from one term to the next using the previous term

Common traps

Mixing up "sequence" and "series." A sequence is a list ( $a_1, a_2, a_3, \dots$ ); a series is a sum ( $a_1 + a_2 + a_3 + \dots$ ). Writing a "+" where a comma belongs (or vice versa) is a very common slip.
Losing track of the index. $a_n$ means the $n$ -th term, not " $a$ times $n$ ." CBSE convention starts sequences at $n = 1$ ; substituting the wrong index, or starting at $n = 0$ , gives a shifted and wrong list of terms.
Forgetting a recursive rule needs a base case. A rule like $a_n = a_{n-1} + 4$ is meaningless without a known starting value $a_1$ ; always check enough starting terms are given to pin the sequence down uniquely.
Assuming every sequence is an AP or a GP. A sequence like $a_n = n^2$ looks "regular" but has neither a constant difference nor a constant ratio between consecutive terms, so it is neither an Arithmetic Progression nor a Geometric Progression — just a sequence.

Derivation ATurning a recursive rule into an explicit formula

The goal is to show that a sequence built by "start at $a_1$ and add a fixed number $d$ to get each next term" — the recursive rule $a_n = a_{n-1} + d$ for $n \ge 2$ , with $a_1$ given — always has a matching explicit formula $a_n = a_1 + (n-1)d$ , so that any term can be reached directly instead of by climbing through every earlier one; this is also the pattern that underlies the general term of an Arithmetic Progression met later in the chapter.

Step 1. Write out what the recursive rule says for the first few terms, replacing each earlier term as soon as it is known. $a_2 = a_1 + d$ . Then $a_3 = a_2 + d = (a_1 + d) + d = a_1 + 2d$ . Then $a_4 = a_3 + d = (a_1 + 2d) + d = a_1 + 3d$ .

Step 2. Spot the pattern in how many copies of $d$ have been added. To reach $a_2$ from $a_1$ , one $d$ was added. To reach $a_3$ , two copies of $d$ were added. To reach $a_4$ , three copies of $d$ were added. In general, reaching $a_n$ from $a_1$ requires adding $d$ exactly $(n-1)$ times, since there are $(n-1)$ steps between position $1$ and position $n$ .

Step 3. Write the general claim. $a_n = a_1 + (n-1)d$ for every $n \ge 1$ . Check it reduces correctly at $n = 1$ : $a_1 = a_1 + (1-1)d = a_1 + 0 = a_1$ , which is trivially true, confirming the formula is consistent with the given starting term.

Step 4. Confirm the formula actually satisfies the original recursive rule, so the two descriptions genuinely describe the same sequence. Using the formula for consecutive terms, $a_n - a_{n-1} = [a_1 + (n-1)d] - [a_1 + (n-2)d] = d$ , which is exactly $a_n = a_{n-1} + d$ — the recursive rule is recovered. So $a_n = a_1 + (n-1)d$ is the explicit formula equivalent to the recursive rule $a_1$ given, $a_n = a_{n-1} + d$ .

Worked example

A sequence is defined recursively by $a_1 = 3$ and $a_n = a_{n-1} + 4$ for $n \ge 2$ (the same rule as Pass 1's Example 2, where $a_6 = 23$ was found by climbing term by term). Use the explicit formula just derived to find $a_{15}$ directly, and confirm it agrees with the earlier value of $a_6$ .

Step 1. Identify the constants for this sequence. Here $a_1 = 3$ and the fixed amount added at each step is $d = 4$ .

Step 2. Substitute into the explicit formula $a_n = a_1 + (n-1)d$ derived above. $a_n = 3 + (n-1)(4) = 3 + 4n - 4 = 4n - 1$ .

Step 3. Use this explicit formula to jump straight to $a_{15}$ , with no need to compute any of the terms in between. $a_{15} = 4(15) - 1 = 60 - 1 = 59$ .

Step 4. Cross-check against the term already known from the recursive climb. Pass 1 found $a_6 = 23$ by adding $4$ five times starting from $3$ . The explicit formula gives $a_6 = 4(6) - 1 = 24 - 1 = 23$ , exactly matching.

Answer: $a_n = 4n - 1$ , so $a_{15} = 59$ , and the formula correctly reproduces $a_6 = 23$ .

Competitive stretch

Using the same explicit formula $a_n = 4n - 1$ , find which term of the sequence equals $79$ . Set $a_n = 79$ : $4n - 1 = 79 \Rightarrow 4n = 80 \Rightarrow n = 20$ . So $79$ is the $20$ th term, $a_{20} = 79$ — found without listing a single one of the nineteen terms before it.

Ask the tutor: if you have a recursive rule and want to see it turned into an explicit formula step by step for your own numbers, ask the AI tutor to walk through the conversion with you.

If a sequence can be written both explicitly and recursively, which form should I use when solving a problem?

Use whichever form the question hands you as a starting point, but convert to the explicit formula whenever you need a term far down the list, since a recursive rule forces you to compute every earlier term first. If the recursive rule adds (or multiplies by) a fixed amount each step — as in this article's derivation — turning it into an explicit formula is quick and worth doing immediately; you will meet exactly this shortcut again as the general term formulas for Arithmetic and Geometric Progressions.

Why does CBSE insist sequences start at n = 1 instead of n = 0?

It is a convention, not a mathematical necessity — a sequence is still a valid function even if you chose to index it from $0$ . CBSE fixes $n = 1$ as the first term simply so that "the first term," "the second term," and so on line up exactly with everyday counting and so that every student's answer for $a_1, a_2, \dots$ is comparable. Always follow the indexing given in a problem; if a question explicitly starts a sequence at $n = 0$ , use that instead, but assume $n = 1$ unless told otherwise.

Is every sequence either an Arithmetic Progression or a Geometric Progression?

No, and this is one of the most common misconceptions. A sequence only qualifies as an AP if consecutive terms differ by the same constant amount every time, and only as a GP if consecutive terms share the same constant ratio every time. Plenty of perfectly valid sequences, such as $a_n = n^2$ or $a_n = \frac{n^2+1}{2}$ , satisfy neither condition — they are still sequences, generated by a legitimate explicit formula, just not members of either special family you meet next in this chapter.

Sequences

The core idea

Common traps

Worked example

Competitive stretch

Before this

Where this leads