Predicting What Comes Next - Exploring Sequences and Progressions

The chapter builds a small, sturdy toolkit in a fixed order — first the general idea of a sequence, then two ways of writing a rule for one, and finally the two most important special sequences.

• Section 8.1 defines a sequence as an ordered list of terms, introduces subscript notation $t_1, t_2, t_3, \ldots$, and separates finite from infinite sequences using natural numbers, odd numbers, triangular numbers and square numbers.
• Section 8.2 introduces the explicit rule, a formula for $t_n$ directly in $n$, letting you jump to any term or check whether a given number belongs to the sequence, and at what position.
• Section 8.3 introduces the recursive rule, building each term from the one before (or, for the Virahānka-Fibonacci sequence, the two before), contrasted with the explicit rule.
• Section 8.4 defines the arithmetic progression (AP) — constant common difference — derives $t_n=a+(n-1)d$ from a growing tile pattern, and shows its terms plot as a straight line.
• Section 8.5 proves the classical sum of the first $n$ natural numbers, $S_n=n(n+1)/2$, via the reverse-and-add trick, connected to the triangular numbers of Section 8.1.
• Section 8.6 defines the geometric progression (GP) — constant common ratio — derives $t_n=a\cdot r^{\,n-1}$, contrasts its curved plot with an AP's line, and applies it to the Sierpiński triangle and a bouncing ball.

Worked examples below reuse NCERT's own example numbers wherever the source text names them directly, so you can cross-check against your textbook.

Sequences, terms and the subscript notation

A sequence is an ordered list of numbers in which each number is called a term. Order matters — $1, 3, 5, 7$ is not the same sequence as $7, 5, 3, 1$. A sequence may be finite, like $6, 12, 24, 48, 96$, or infinite, like the natural numbers $1, 2, 3, 4, \ldots$, where the dots signal the pattern continues forever. To talk about a general term without knowing its value, mathematicians use subscript notation: $t_1$ is the first term, $t_2$ the second, and $t_n$ the term in the $n$th position.

where: the subscript matches the position of the term, never its value — $t_4 = 7$ means "the fourth term equals $7$," not a fourth power. Different sequences discussed together get different letters, $t_n$ for one and $s_n$ or $u_n$ for another. The position $n$ is always a natural number, but the term $t_n$ itself can be any real number — an integer, fraction, or negative, as in $-7, -3, 1, 5, 9, \ldots$ where $s_1=-7$.

Explicit rule for the nth term

An explicit rule gives $t_n$ directly as an expression in $n$, so any term can be found by substituting its position — no earlier terms are needed. For the odd numbers, the rule is $u_n = 2n-1$, and substituting $n=1,2,3,\ldots$ gives $u_1=1$, $u_2=3$, $u_3=5$, matching $1,3,5,7,9,\ldots$.

where: $n$ is the position and $u_n$ is its value. The power of an explicit rule shows for large positions: the $300$th term needs one substitution, $u_{300}=2(300)-1=599$, with no earlier terms written out. It also answers the reverse question — is a given number a term, and which one? To check $137$, solve $2n-1=137$, giving $n=69$; since $69$ is a natural number, $137$ is the $69$th term. If solving for $n$ gives a fraction or a negative number, the number is not a term of that sequence at all.

Recursive rule for the nth term

A recursive rule gives a term in relation to the term (or terms) immediately before it, rather than its position alone. In $1, 4, 7, 10, 13, \ldots$, every term is $3$ more than the one before, so $t_2=t_1+3$, $t_3=t_2+3$, and in general $t_n=t_{n-1}+3$ for $n\geq2$, together with the starting value $t_1=1$.

where: $a$ is the first term and $d$ is the fixed number added each step; unlike an explicit rule, this needs every earlier term — there is no shortcut to the $300$th term without first computing the $299$ before it. This describes the same sequence a different way: for $1,4,7,10,\ldots$, both $t_n=3n-2$ (explicit) and $t_1=1,\ t_n=t_{n-1}+3$ (recursive) generate the identical list, worth checking against each other for a few values of $n$.

The Virahānka-Fibonacci sequence

A recursive rule need not use only the single previous term — it can reach back two terms or more. The most celebrated example is $V_1=1,\ V_2=2$, and $V_n=V_{n-1}+V_{n-2}$ for $n\geq3$, giving $1, 2, 3, 5, 8, 13, 21, 34, \ldots$, where every term after the second is the sum of the two before it.

where: each new term needs both predecessors, not just one, so $V_5$ requires knowing $V_4=5$ and $V_3=3$, giving $V_5=5+3=8$. This sequence was first written down by the Indian scholar Virahānka in his 7th-century work on Prakrit poetic meter, studied further by Gopāla and Hemachandra, and later popularised in Europe by Fibonacci around 1200 CE — hence its double name. It reappears throughout mathematics and nature as the standard example of a recursion using more than one earlier term.

Arithmetic progressions and their nth term

Consider a growing tile pattern where every stage adds four squares to the corners of the last: $1, 5, 9, 13, 17, \ldots$. Rewriting each term as the first term plus a whole number of jumps, $1, 1+1\times4, 1+2\times4, 1+3\times4, \ldots$, reveals $t_n=1+(n-1)\times4$. Any sequence built this way — adding the same fixed number at every step — is an arithmetic progression, or AP.

where: $a$ is the first term and $d$ is the common difference, the constant amount added each step (found as $t_2-t_1$, or any $t_{k+1}-t_k$). It can be negative — in $11, 7, 3, -1, -5, \ldots$, $a=11$ and $d=-4$, since the terms decrease. Not every "patterned" sequence is an AP: primes $2, 3, 5, 7, 11, \ldots$ have no constant difference at all, so they form a sequence but not a progression. The recursive rule for an AP is simply $t_1=a,\ t_n=t_{n-1}+d$ for $n\geq2$.

Visualising an AP: points on a straight line

Plot each term of an AP against its position, the ordered pairs $(n, t_n)$, and a striking fact appears: every point lies on one straight line. For $1, 5, 9, 13, 17$, the pairs $(1,1), (2,5), (3,9), (4,13), (5,17)$ line up exactly, because $t_n=4n-3$ is a linear function of $n$, and the graph of a linear function is always a straight line.

where: the slope of the line is exactly the common difference $d$, since moving one step along $n$ always moves $d$ units along $t_n$ — a fast diagnostic: if plotted points do not lie on a straight line, the sequence is not an AP, however regular it looks.

Sum of the first n natural numbers

Before the full derivation below, the idea is worth stating on its own: the sum of the first $n$ natural numbers has a closed form avoiding adding them one by one.

where: $S_n$ is the sum of the first $n$ natural numbers. This is also the explicit rule for the triangular numbers $1, 3, 6, 10, 15, \ldots$ from Section 8.1, since the $n$th triangular number is exactly this sum. The formula turns a sum of a hundred or a million terms into one multiplication and one division.

Geometric progressions and their nth term

Now consider a growing pattern where the count of green squares doubles every stage: $3, 6, 12, 24, \ldots$. Rewriting each term as the first term times a whole number of doublings, $3,\ 3\times2,\ 3\times2^2,\ 3\times2^3,\ \ldots$, reveals $t_n=3\times2^{\,n-1}$. Any sequence built this way — multiplying by the same fixed number every step — is a geometric progression, or GP.

where: $a$ is the first term and $r$ is the common ratio, found by dividing any term by the one before it, $r=t_{k+1}/t_k$, and it must be the same for every consecutive pair — checking two or three ratios, not one, confirms a genuine GP. The ratio can be a fraction less than $1$, so terms shrink, as in a bouncing ball's heights; it can also be negative, as in $1, -1, 1, -1, \ldots$ with $r=-1$. The recursive rule is $t_1=a,\ t_n=r\cdot t_{n-1}$ for $n\geq2$ — multiplying, not adding, the essential contrast with an AP.

Visualising a GP and the Sierpiński fractals

Plot a GP's terms against position and the points curve away from any straight line — slowly if $r$ is close to $1$, dramatically otherwise. For $3, 6, 12, 24, 48, \ldots$, the pairs $(1,3), (2,6), (3,12), (4,24), (5,48)$ rise faster and faster, since $t_n=3\times2^{\,n-1}$ is exponential in $n$, not linear.

where: the curve bends upward more sharply as $n$ grows because each step multiplies, so the gaps between terms keep growing, unlike an AP where every gap is the fixed $d$. The same exponential behaviour explains the Sierpiński triangle, a fractal built by repeatedly removing the central triangle from every black triangle remaining: black triangles at stage $n$ number $t_n=3^{\,n}$ (a GP, $a=1, r=3$), while the area of the black region shrinks as a GP with $a=1, r=\tfrac{3}{4}$ — triangles multiply explosively even as their total area shrinks toward zero.

Board level - Checking membership using an explicit rule

Consider the sequence generated by the explicit formula $s_n = 5n-2$. Write its first six terms, find its $100$th term, and check whether $308$ and $471$ are terms of this sequence (this reuses NCERT's own Example 2).

Step 1. Substitute $n=1,2,\ldots,6$ into $s_n=5n-2$: the first six terms are $3, 8, 13, 18, 23, 28$.

Step 2. Substitute $n=100$ directly, with no need for earlier terms: $s_{100}=5(100)-2=498$.

Step 3. To check $308$, solve $5n-2=308$: this gives $5n=310$, so $n=62$, a natural number, confirming $308$ is the $62$nd term.

Step 4. To check $471$, solve $5n-2=471$: this gives $5n=473$, so $n=94.6$, not a natural number, so $471$ is not a term. First six terms: $3, 8, 13, 18, 23, 28$; $s_{100}=\mathbf{498}$; $308$ is the $\mathbf{62}$nd term; $471$ is not a term.

Board level - Generating and checking terms with a recursive rule

Find the first four terms of the sequence given by the recursive rule $u_1=1,\ u_n = 2u_{n-1}+3$ for $n \geq 2$. Is $133$ a term of this sequence (this reuses NCERT's own Example 3)?

Step 1. Compute term by term, since a recursive rule needs the previous value: $u_2=2u_1+3=2(1)+3=5$.

Step 2. Continue: $u_3=2u_2+3=2(5)+3=13$, then $u_4=2u_3+3=2(13)+3=29$. So the first four terms are $1, 5, 13, 29$.

Step 3. To check $133$, keep computing forward, since there is no explicit shortcut here: $u_5=2(29)+3=61$, $u_6=2(61)+3=125$, $u_7=2(125)+3=253$.

Step 4. The sequence jumps from $u_6=125$ straight to $u_7=253$, stepping over $133$ entirely, and every later term only grows larger. The first four terms are $1, 5, 13, 29$, and $133$ is $\mathbf{not}$ a term of this sequence.

Higher-order thinking (HOTS / olympiad-style) - An AP from two given terms

An AP consists of $50$ terms in which the $3$rd term is $12$ and the last ($50$th) term is $106$. Find the $29$th term (this reuses NCERT's own Exercise Set 8.2, Question 4).

Step 1. Translate both given facts into the AP formula $t_n = a+(n-1)d$. The $3$rd term gives $a+2d=12$, and the $50$th (last) term gives $a+49d=106$ — two linear equations in the two unknowns $a$ and $d$.

Step 2. Subtract the first equation from the second to eliminate $a$: $(a+49d)-(a+2d) = 106-12$, giving $47d = 94$, so $d=2$.

Step 3. Substitute $d=2$ back into $a+2d=12$: $a+4=12$, so $a=8$. The AP is $8, 10, 12, 14, \ldots$, and it is worth checking: the $50$th term should be $a+49d = 8+98=106$, which matches.

Step 4. Find the $29$th term using the same formula: $t_{29} = a+28d = 8+28(2) = 8+56 = 64$. The $29$th term is $\mathbf{64}$. Timing note: about three minutes. The whole problem is two simultaneous linear equations disguised inside AP language — the moment you see two terms given by position, translate both into $a+(n-1)d$ immediately rather than trying to guess $a$ and $d$ by inspection. Always sanity-check by re-substituting into both original conditions (the $3$rd and $50$th terms), not just one, since a value of $d$ that satisfies only one equation is worthless.

• Treating any "patterned" list as an AP or a GP. Primes $2, 3, 5, 7, 11, \ldots$ have no constant difference or ratio; always compute two or three consecutive differences (or ratios) before naming a sequence an AP or a GP.
• Forgetting the $(n-1)$, not $n$, in both nth-term formulas. Both $t_n=a+(n-1)d$ and $t_n=a\,r^{\,n-1}$ use $(n-1)$ jumps to reach position $n$ from position $1$; writing $a+nd$ or $a\,r^n$ shifts every term by one position.
• Accepting a fractional or negative position as an answer. If solving $t_n=$ (given number) for $n$ gives anything but a natural number, that number is simply not a term of the sequence.
• Confusing addition and multiplication between AP and GP. An AP's recursive rule adds $d$; a GP's multiplies by $r$. Mixing the two up is the commonest slip when a problem states both kinds together.
• Checking only one pair of terms for a common ratio. A GP's ratio must match for every consecutive pair; one matching pair followed by a mismatch means it is not a GP.
• Mixing up stage numbering with term numbering in fractal problems. The Sierpiński triangle often starts at Stage $0$, so $t_n=3^n$ uses the stage number directly — check whether numbering starts at $0$ or $1$ before applying a formula.

• Define a sequence and a term, and distinguish a finite sequence from an infinite one.
• Read and write subscript notation such as $t_n$, $s_n$, $u_n$, matching the subscript to the term's position.
• Use an explicit rule to find any term directly, and check whether a given number is a term (and at which position).
• Use a recursive rule, including one depending on two previous terms like the Virahānka-Fibonacci sequence, to build up terms from a starting value.
• Identify an AP, find its first term $a$ and common difference $d$, and write its explicit and recursive rules.
• Recognise an AP visually: its terms, plotted against position, always lie on a straight line.
• Derive and apply $S_n = \dfrac{n(n+1)}{2}$ for the sum of the first $n$ natural numbers, and connect it to the triangular numbers.
• Identify a GP, find its first term $a$ and common ratio $r$, and write its explicit and recursive rules.
• Recognise a GP visually: its terms curve away from a straight line, as in fractals and bouncing-ball heights.
• Decide, given an unfamiliar sequence, whether it is an AP, a GP, or neither, by testing for a constant difference or ratio.

Ask the tutor: Want to practice spotting whether a mixed batch of sequences are APs, GPs, or neither, or work through a few more "which term is this number" problems until the natural-number check becomes automatic?