Class 9 Mathematics Chapter 8: Predicting What Comes Next: Exploring Sequences and Progressions

A sequence is an ordered list of numbers, and each number in it is called a term. In the notation $t_n$ the subscript is the position of the term in that list, so $t_n$ means the $n$th term; it is not $t$ multiplied by $n$.

Substituting $n = 1, 2, 3$ and $4$ gives $4(1) - 3 = 1$, $4(2) - 3 = 5$, $4(3) - 3 = 9$ and $4(4) - 3 = 13$. The first four terms are $1, 5, 9, 13$.

The 100th term is $3(100) - 2 = 298$. No earlier term is needed because an explicit rule gives the $n$th term directly from its position, so the calculation skips straight to the term wanted.

It says only how to get from one term to the next, and not where to begin. The same rule fits $1, 6, 11, 16, \ldots$ and $2, 7, 12, 17, \ldots$ and infinitely many others. A recursive rule needs its starting value stated alongside it.

Working forward one step at a time, $u_2 = 2(1) + 3 = 5$, then $u_3 = 2(5) + 3 = 13$, then $u_4 = 2(13) + 3 = 29$. The first four terms are $1, 5, 13, 29$.

An explicit rule gives the $n$th term directly in terms of its position, so that any term can be reached in one step. A recursive rule instead says how each term is built from the term or terms before it, and must also state a starting value. For the sequence $1, 4, 7, 10, \ldots$ the explicit rule is $t_n = 3n - 2$, and the recursive rule is $t_1 = 1$ with $t_n = t_{n-1} + 3$ for $n \geq 2$. Both describe the same sequence.

The sequence has explicit rule $t_n = 4n - 3$. Setting this equal to 97 gives $4n - 3 = 97$, so $4n = 100$ and $n = 25$. Since 25 is a positive whole number, 97 is indeed a term of the sequence, and it is the 25th one.

Setting the rule equal to the number gives $4n - 3 = 100$, so $4n = 103$ and $n = \dfrac{103}{4}$. That is not a whole number, and a term's position must be a positive whole number, so 100 is not a term of the sequence. Listing terms would have suggested the same but proved nothing, since one could always have stopped too soon.

It is defined by $V_1 = 1$, $V_2 = 2$ and $V_n = V_{n-1} + V_{n-2}$ for $n \geq 3$. Working forward, $V_3 = 2 + 1 = 3$, $V_4 = 3 + 2 = 5$, $V_5 = 5 + 3 = 8$ and $V_6 = 8 + 5 = 13$, so the first six terms are $1, 2, 3, 5, 8, 13$. Two starting values are needed because the rule reaches back two places: with only one term given, there would be nothing for the second term to be added to, and the sequence could not get going.

A recursive rule has to be worked through in order, so reaching the 50th term means computing the 49 before it. An explicit rule reaches any term in a single substitution, which makes distant terms as cheap as near ones. It also allows questions of membership to be answered by solving an equation rather than by listing terms and hoping to have gone far enough. The recursive rule is usually easier to spot from the pattern, but the explicit rule is what makes the sequence usable.

(i) A sequence is an ordered list of numbers. Each number in it is called a term, and the order is part of the sequence, so swapping two terms gives a different sequence.

(ii) $t_n$ means the $n$th term of the sequence. The subscript records the position of the term in the list, and is not something to multiply $t$ by.

(iii) An explicit rule gives the $n$th term directly in terms of $n$, so that any term can be found from its position alone.

(iv) Substituting $n = 1, 2, 3, 4$ gives $7, 12, 17, 22$.

(v) $t_{20} = 5(20) + 2 = 102$. An explicit rule reaches a term straight from its position, so none of the nineteen earlier terms had to be worked out.

(i) A recursive rule builds each term of a sequence from the term or terms that come before it.

(ii) Because the step alone does not say where to begin. "Add 4 each time" fits $1, 5, 9, \ldots$ and $2, 6, 10, \ldots$ and infinitely many others, so the first term must be stated as part of the rule.

(iii) $t_1 = 1$ and $t_n = t_{n-1} + 4$ for $n \geq 2$.

(iv) $s_2 = 3 \times 2 = 6$, $s_3 = 6 \times 5 = 30$ and $s_4 = 30 \times 29 = 870$, so the terms are $3, 6, 30, 870$.

(v) Because each term is defined only in terms of earlier ones. To reach the 50th, the 49 terms before it must all be worked out first, in order.

(i) The terms rise by 6 each time and start at 7, so $t_n = 6n + 1$. Checking, $n = 1$ gives 7 and $n = 2$ gives 13.

(ii) $t_1 = 7$ and $t_n = t_{n-1} + 6$ for $n \geq 2$.

(iii) $t_{15} = 6(15) + 1 = 91$.

(iv) Setting $6n + 1 = 91$ gives $6n = 90$ and $n = 15$, a positive whole number, so 91 is a term — the 15th, as part (iii) also shows.

(v) Setting $6n + 1 = 100$ gives $6n = 99$ and $n = 16.5$, which is not a whole number. So 100 is not a term of the sequence.

(i) No. The primes $2, 3, 5, 7, 11, 13, \ldots$ form a perfectly good sequence, since they come in a definite order, yet there is no simple formula for the $n$th prime. Being ordered is what makes a list a sequence; having a rule is a bonus.

(ii) $V_1 = 1$, $V_2 = 2$, and $V_n = V_{n-1} + V_{n-2}$ for $n \geq 3$.

(iii) $1, 2, 3, 5, 8, 13, 21, 34$.

(iv) Its rule reaches back two places, so it needs two terms already in place before it can produce a third. With only one seed there would be nothing for the first sum to add.

(v) It shows that a recursive description is sometimes all there is. This sequence has no simple explicit rule of the kind used for arithmetic patterns, and yet it is completely and precisely defined.

An arithmetic progression is a sequence in which the difference between consecutive terms is always the same. In the usual notation $a$ is the first term of the sequence and $d$ is that constant gap, called the common difference.

The $n$th term is $t_n = a + (n-1)d$. It uses $n - 1$ because the first term is where the sequence begins rather than somewhere a step was taken to, so reaching the $n$th term costs one step fewer than its position.

The first term is $a = 3$ and the common difference is $d = 8 - 3 = 5$. So $t_n = 3 + (n-1)5 = 3 + 5n - 5 = 5n - 2$. Checking at $n = 1$ gives 3, as required.

The differences between consecutive terms are $3$, $5$ and $7$, which are not all the same. An arithmetic progression requires every such difference to be equal, so this sequence is not one; it is the sequence of square numbers.

Using the formula $\dfrac{n(n+1)}{2}$ with $n = 20$ gives $\dfrac{20 \times 21}{2} = \dfrac{420}{2} = 210$.

The definition requires only that the difference between consecutive terms be constant; it says nothing about the terms having to grow. Here each term is 4 less than the one before, so the common difference is $d = -4$, a perfectly good constant. With $a = 11$, the $n$th term is $t_n = 11 + (n-1)(-4) = 11 - 4n + 4 = 15 - 4n$. Checking at $n = 1$ gives 11, as required.

From the third term to the seventh there are $7 - 3 = 4$ steps, and the terms rise by $33 - 13 = 20$ over those steps, so $d = \dfrac{20}{4} = 5$. Working back from the third term, $a = t_3 - 2d = 13 - 10 = 3$. As a check, the progression $3, 8, 13, 18, 23, 28, 33$ does have 13 in third place and 33 in seventh.

Here $a = 2$ and $d = 7$, so the $n$th term is $t_n = 2 + (n-1)7 = 7n - 5$. Setting this equal to 100 gives $7n - 5 = 100$, so $7n = 105$ and $n = 15$. Since 15 is a positive whole number, 100 is a term of the progression, and it stands in fifteenth place.

Write $S = 1 + 2 + 3 + \cdots + 9 + 10$, and then write the same sum backwards underneath it as $S = 10 + 9 + 8 + \cdots + 2 + 1$. Adding the two lines column by column, every column gives the same total: $1 + 10 = 11$, $2 + 9 = 11$, $3 + 8 = 11$, and so on. There are ten columns, so $2S = 11 \times 10 = 110$, and therefore $S = 55$.

Draw a staircase of circles with columns of 1, 2, 3 and so on up to $n$; the number of circles in it is exactly the sum wanted. Now take a second staircase identical to the first, turn it about, and slot it into the gaps of the first. The two fit together with nothing left over and nothing overlapping, forming a rectangle $n$ circles across and $n + 1$ down. That rectangle holds $n(n+1)$ circles, and it is made of two copies of the sum, so the sum itself is $\dfrac{n(n+1)}{2}$.

(i) An arithmetic progression is a sequence in which the difference between consecutive terms is always the same. That constant is the common difference.

(ii) $t_n = a + (n-1)d$, taking $a$ as the entry it starts from and $d$ as the constant gap.

(iii) Because the first term is where the sequence begins rather than somewhere a step was taken to. The second term costs one step, the third two, and the $n$th costs $n - 1$.

(iv) Here $a = 7$ and $d = 6$, so $t_n = 7 + (n-1)6 = 6n + 1$.

(v) $t_{12} = 6(12) + 1 = 73$.

(i) Work out the differences between consecutive terms and see whether they all agree.

(ii) The differences are $5, 5, 5$, all the same, so it is an AP with $d = 5$.

(iii) The differences are $2, 4, 8$, which do not agree, so it is not an AP. Its ratios are all 2, which makes it a geometric progression instead.

(iv) Because any two consecutive terms have some difference between them, whatever the sequence. It is the agreement of all the differences that makes a sequence an AP, and $1, 4, 9, 16$ shows how a first difference can be perfectly ordinary while the sequence is not an AP at all.

(v) Its differences are all $-5$. Nothing in the definition requires the terms to grow, only that the difference be constant, and a constant may be negative.

(i) $a = 2$ and $d = 9 - 2 = 7$.

(ii) $t_n = 2 + (n-1)7 = 7n - 5$.

(iii) $t_{15} = 7(15) - 5 = 105 - 5 = 100$.

(iv) Setting $7n - 5 = 100$ gives $7n = 105$ and $n = 15$, a positive whole number, so 100 is a term — the fifteenth, as part (iii) confirms.

(v) Setting $7n - 5 = 90$ gives $7n = 95$, and 95 does not divide by 7, so $n$ would not be a whole number. Therefore 90 is not a term of this progression.

(i) Write $S = 1 + 2 + \cdots + n$ and, underneath it, the same sum reversed: $S = n + (n-1) + \cdots + 1$. Adding column by column, every column gives $n + 1$, and there are $n$ columns, so $2S = n(n+1)$.

(ii) $1 + 2 + \cdots + n = \dfrac{n(n+1)}{2}$.

(iii) $\dfrac{50 \times 51}{2} = \dfrac{2550}{2} = 1275$.

(iv) A staircase of circles with columns $1, 2, \ldots, n$ holds the sum. A second identical staircase, turned about and slotted in, fills a rectangle $n$ across and $n+1$ down, holding $n(n+1)$ circles in all. The sum is therefore half of that.

(v) The triangular numbers are $1, 3, 6, 10, 15, \ldots$, whose differences are $2, 3, 4, 5, \ldots$ These grow rather than staying constant, so the sequence fails the test for an AP. The running totals of an AP need not themselves form an AP.

A geometric progression is a sequence in which the ratio of consecutive terms is always the same. In the usual notation $a$ is the first term of the sequence and $r$ is that constant multiplier, called the common ratio.

The $n$th term is $t_n = a r^{,n-1}$. The power is one less than the position because the first term is where the sequence begins rather than somewhere it multiplied to, so reaching the $n$th term takes $n - 1$ multiplications.

The first term is $a = 2$ and the common ratio is $r = \dfrac{6}{2} = 3$, found by dividing rather than subtracting. So $t_n = 2 \times 3^{,n-1}$, and at $n = 1$ that gives $2 \times 3^0 = 2$, as required.

Its ratios are $\dfrac{4}{2} = 2$, then $\dfrac{6}{4} = \dfrac{3}{2}$, then $\dfrac{8}{6} = \dfrac{4}{3}$, and these do not agree. A geometric progression needs every such ratio to be the same. This sequence is an arithmetic progression instead, since its differences are all 2.

Here $a = 3$ and $r = 2$, so $t_8 = 3 \times 2^7 = 3 \times 128 = 384$.

The definition asks only that the ratio of consecutive terms be constant; it says nothing about the terms having to grow. If that constant ratio lies between 0 and 1, each term is a fraction of the one before and the sequence shrinks. For example $64, 32, 16, 8, 4, \ldots$ has ratio $\dfrac{32}{64} = \dfrac{1}{2}$ throughout, so it is a geometric progression with $t_n = 64 \left(\dfrac{1}{2}\right)^{n-1}$.

Dividing the second term by the first gives $\dfrac{15/4}{5} = \dfrac{15}{20} = \dfrac{3}{4}$. Dividing the third by the second gives $\dfrac{45}{16} \times \dfrac{4}{15} = \dfrac{180}{240} = \dfrac{3}{4}$, and the fourth by the third gives $\dfrac{135}{64} \times \dfrac{16}{45} = \dfrac{3}{4}$ as well. All the ratios agree, so the sequence is a geometric progression with $a = 5$ and $r = \dfrac{3}{4}$, and its $n$th term is $t_n = 5\left(\dfrac{3}{4}\right)^{n-1}$.

Multiplying by a negative number flips the sign, so every step changes the sign of the term. The result is that the terms alternate: those in odd positions keep the sign of the first term and those in even positions take the opposite sign. This follows from the formula, since a negative $r$ raised to an even power is positive and to an odd power is negative. For example $2, -6, 18, -54, \ldots$ has $r = -3$: the size grows by a factor of 3 each time while the sign flips at every step.

An arithmetic progression has a constant difference between consecutive terms, while a geometric progression has a constant ratio. For $1, 5, 9, 13$ the differences are $4, 4, 4$, all equal, so it is an arithmetic progression; its ratios are $5$, $\dfrac{9}{5}$ and $\dfrac{13}{9}$, which disagree, so it is not geometric. For $3, 6, 12, 24$ the ratios are $2, 2, 2$, all equal, so it is geometric; its differences are $3, 6, 12$, which disagree, so it is not arithmetic.

An arithmetic progression adds the same amount at every step, so its growth is steady and its total after $n$ steps rises in proportion to $n$. A geometric progression multiplies at every step, so the amount added grows along with the sequence itself, and the total rises far faster. The two sequences begin only two apart: at the first term they stand at 1 and 3. By the fifth term they are 17 and 48, and by the tenth 37 and 1536 — the geometric one is then more than forty times the other. This is why compound interest and populations grow so much faster than intuition expects.

(i) A geometric progression is a sequence in which the ratio of consecutive terms is always the same. That constant is called the common ratio.

(ii) $t_n = a r^{,n-1}$, where $a$ is the first term and $r$ the common ratio.

(iii) The power is one less than the position because the first term is where the sequence begins rather than somewhere it multiplied to. The second term costs one multiplication, the third two, and the $n$th costs $n - 1$.

(iv) Here $a = 2$ and $r = \dfrac{6}{2} = 3$, so $t_n = 2 \times 3^{,n-1}$.

(v) $t_6 = 2 \times 3^5 = 2 \times 243 = 486$.

(i) An arithmetic progression has a constant difference between consecutive terms; a geometric progression has a constant ratio.

(ii) Its differences are $4, 4, 4$, all equal, so it is arithmetic. Its ratios are $5$, $\dfrac{9}{5}$ and $\dfrac{13}{9}$, which disagree, so it is not geometric.

(iii) Its ratios are $2, 2, 2$, all equal, so it is geometric. Its differences are $3, 6, 12$, which disagree, so it is not arithmetic.

(iv) Both. Its difference is 0 and its ratio is 1, and each of those is constant, so the sequence satisfies both definitions at once. A constant sequence is the only kind that does.

(v) Because any two consecutive terms have some difference and some ratio between them, whatever the sequence. What distinguishes the two families is that all the differences, or all the ratios, agree, and that cannot be seen from a single pair.

(i) $a = 64$ and $r = \dfrac{32}{64} = \dfrac{1}{2}$.

(ii) $t_n = 64 \left(\dfrac{1}{2}\right)^{n-1}$.

(iii) $t_5 = 64 \times \left(\dfrac{1}{2}\right)^4 = \dfrac{64}{16} = 4$.

(iv) $t_7 = 64 \times \left(\dfrac{1}{2}\right)^6 = \dfrac{64}{64} = 1$.

(v) The definition requires only that the ratio of consecutive terms be constant, and says nothing about the terms having to grow. A ratio lying between 0 and 1 makes each term a fraction of the one before, so the sequence shrinks while remaining perfectly geometric.

(i) $t_1 = a$ and $t_n = r,t_{n-1}$ for $n \geq 2$. The starting value must be stated, since the step alone fits infinitely many sequences.

(ii) $t_1 = 3$ and $t_n = 2t_{n-1}$ for $n \geq 2$.

(iii) The points of an arithmetic progression lie on a straight line, because a fixed amount is added at every step. The points of a geometric progression curve upwards ever more steeply, because each term is multiplied, so the amount added grows along with the sequence.

(iv) The tenth term of $1, 5, 9, \ldots$ is $1 + 9 \times 4 = 37$, while the tenth term of $3, 6, 12, \ldots$ is $3 \times 2^9 = 1536$. The two began only two apart, and the geometric one is now more than forty times the other.

(v) Because many real quantities grow geometrically rather than arithmetically: money under compound interest, a population, the spread of a rumour or an infection. Each multiplies rather than adds, and so reaches sizes that steady addition would take vastly longer to reach.

A sequence is an ordered list of numbers, and each number in it is called a term. In the notation $t_n$ the subscript is the position of the term in that list, so $t_n$ means the $n$th term; it is not $t$ multiplied by $n$.

Substituting $n = 1, 2, 3$ and $4$ gives $4(1) - 3 = 1$, $4(2) - 3 = 5$, $4(3) - 3 = 9$ and $4(4) - 3 = 13$. The first four terms are $1, 5, 9, 13$.

The 100th term is $3(100) - 2 = 298$. No earlier term is needed because an explicit rule gives the $n$th term directly from its position, so the calculation skips straight to the term wanted.

An arithmetic progression is a sequence in which the difference between consecutive terms is always the same. In the usual notation $a$ is the first term of the sequence and $d$ is that constant gap, called the common difference.

The $n$th term is $t_n = a + (n-1)d$. It uses $n - 1$ because the first term is where the sequence begins rather than somewhere a step was taken to, so reaching the $n$th term costs one step fewer than its position.

The first term is $a = 3$ and the common difference is $d = 8 - 3 = 5$. So $t_n = 3 + (n-1)5 = 3 + 5n - 5 = 5n - 2$. Checking at $n = 1$ gives 3, as required.

A geometric progression is a sequence in which the ratio of consecutive terms is always the same. In the usual notation $a$ is the first term of the sequence and $r$ is that constant multiplier, called the common ratio.

The $n$th term is $t_n = a r^{,n-1}$. The power is one less than the position because the first term is where the sequence begins rather than somewhere it multiplied to, so reaching the $n$th term takes $n - 1$ multiplications.

The first term is $a = 2$ and the common ratio is $r = \dfrac{6}{2} = 3$, found by dividing rather than subtracting. So $t_n = 2 \times 3^{,n-1}$, and at $n = 1$ that gives $2 \times 3^0 = 2$, as required.