Exam practice

Up by 1: 5, 6.

Up by 2: 12, 14.

Up by 5: 30, 35.

Each term is double the one before: 32, 64.

Down by 2: 12, 10.

Up by 2: −4 + 2 = −2, then 0.

Look at the differences: are they all the same?

The differences are 3, 4, 5, 6, growing by 1. Next come differences of 7 and 8: 20 + 7 = 27, 27 + 8 = 35.

The numerators go up by 3 (1, 4, 7, 10) and the denominators by 2 (3, 5, 7, 9). Next: $\frac{13}{11}$ and $\frac{16}{13}$, that is $\frac{13}{11}, \frac{16}{13}$.

Up by 2: compare with 2n (2, 4, 6, 8); each term is 2 more. $t_n = 2n + 2$.

Up by 4: compare with 4n (4, 8, 12, …); each term is 3 more. $t_n = 4n + 3$ (Example 2).

Up by 4: compare with 4n (4, 8, 12, …); each term is 2 less. $t_n = 4n - 2$.

Down by 3 each time, so compare with −3n (−3, −6, −9, …); each term is 28 more (25 = −3 + 28). $t_n = 28 - 3n$. Check: t₄ = 28 − 12 = 16 ✓.

Find a rule for the numerators and one for the denominators.
Numerators 1, 4, 7, 10: up by 3, so 3n − 2. Denominators 3, 5, 7, 9: up by 2, so 2n + 1.
$t_n = \frac{3n - 2}{2n + 1}$.

Numerators 2, 5, 8, 11: 3n − 1. Denominators 7, 8, 9, 10: n + 6.
$t_n = \frac{3n - 1}{n + 6}$.

Down by 2: compare with −2n (−2, −4, …); each term is 42 more. $t_n = 42 - 2n$.

Numerators 2, 4, 6, 8: 2n. Denominators 5, 8, 11, 14: 3n + 2.
$t_n = \frac{2n}{3n + 2}$.

n = 1, 2, 3, 4, 5: 6, 8, 10, 12, 14, …

2, 5, 8, 11, 14, …

$3^1, 3^2, 3^3, 3^4, 3^5$: 3, 9, 27, 81, 243, …

1 − 1, 4 − 1, 9 − 1, 16 − 1, 25 − 1: 0, 3, 8, 15, 24, …

$(-1)^n$ is −1 when n is odd and +1 when it is even, so the signs take turns.
−1, + 4, −9, + 16, −25: −1, 4, −9, 16, −25, …

$$\begin{aligned} t_1 &= 1 + 2 + 3 = 6 \\ t_2 &= 4 + 4 + 3 = 11 \\ t_3 &= 9 + 6 + 3 = 18 \\ t_4 &= 16 + 8 + 3 = 27 \\ t_5 &= 25 + 10 + 3 = 38 \end{aligned}$$
6, 11, 18, 27, 38, …

3 − 5, 12 − 5, 27 − 5, 48 − 5, 75 − 5: −2, 7, 22, 43, 70, …

Each figure adds one triangle to the strip: one new dot and two new sticks.

Sticks: 1, 3, 5, 7, 9, up by 2, so $t_n = 2n - 1$. (Counting dots instead: 2, 3, 4, 5, 6, so $t_n = n + 1$.)

Each stroke of the N gets one more stick, so each figure has 3 more sticks and 3 more dots than the one before.

Dots: 4, 7, 10, 13, 16, up by 3, so $t_n = 3n + 1$. (Counting sticks instead: 3, 6, 9, 12, 15, so $t_n = 3n$.)

Each figure joins the midpoints of the newest middle triangle, which adds 3 dots and 4 small triangles.

Triangles: 1, 5, 9, 13, 17, up by 4, so $t_n = 4n - 3$. (Counting dots instead: 3, 6, 9, 12, 15, so $t_n = 3n$.)
Whatever you count, say what it is: “the number of triangles in figure n is 4n − 3”.

A sequence is a list of numbers written in order by a fixed rule, each number being a term. Example: 3, 6, 9, 12, …, the multiples of 3, whose general term is $t_n = 3n$.

A sequence: terms separated by commas.

A series: terms joined by plus signs.

A sequence.

A series: Σ means add, so it is 4 + 7 + 10.

A series.

A sequence.

n = 0, 1, 2, 3, 4: −1 + 1 + 3 + 5 + 7 = 15.

n = 2 to 6: 8 + 11 + 14 + 17 + 20 = 70.

2 + 5 + 10 + 17 + 26 = 60.

$n^2 + 2n + 1 = (n + 1)^2$, so the terms are $2^2, 3^2, 4^2$: 4 + 9 + 16 = 29.

5 + 10 + 15 + ⋯ + 50 = 5(1 + 2 + ⋯ + 10) = 5 × 55 = 275.

25 + 36 + 49 + 64 + 81 + 100 = 355.

7 + 14 + 23 + 34 + 47 + 62 = 187.

The terms are $\frac{3}{1}, \frac{5}{2}, \frac{7}{3}, \frac{9}{4}, \frac{11}{5}$. Over the common denominator 60:
$$\frac{180 + 150 + 140 + 135 + 132}{60} = \frac{737}{60}$$
$\frac{737}{60}$ (about 12.28).

The terms are $0, \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}$. Over 60:
$$\frac{0 + 30 + 40 + 45 + 48}{60} = \frac{163}{60}$$
$\frac{163}{60}$ (about 2.72).

Up by 2, so $t_n = 2n + 3$. The last term: 2n + 3 = 21 gives n = 9.
$\sum_{n=1}^{9} (2n + 3)$

$\sum_{n=1}^{6} 2n$

Down by 5, so $t_n = 35 - 5n$. The last term: 35 − 5n = 5 gives n = 6.
$\sum_{n=1}^{6} (35 - 5n)$

Up by 4, so $t_n = 4n - 3$; six terms.
$\sum_{n=1}^{6} (4n - 3)$

$\sum_{n=1}^{4} n^2$

The power of b is one less than the term’s position: $t_n = ab^{n-1}$.
$\sum_{n=1}^{5} ab^{n-1}$

$\dfrac{t_2}{t_1} = \dfrac{2}{4} = \dfrac{1}{2}$, $\dfrac{t_3}{t_2} = \dfrac{1}{2}$, $\dfrac{t_4}{t_3} = \dfrac{1}{2} \div 1 = \dfrac{1}{2}$: the ratio of consecutive terms is the same, so it is geometric, with $r = \frac{1}{2}$.

10 − 4 = 6, 16 − 10 = 6: the difference is the same, so it is arithmetic, with d = 6.

$\frac{ab}{a} = b$, $\frac{ab^2}{ab} = b$: the same ratio, so it is geometric, with r = b.

$\frac{12}{6} = \frac{24}{12} = 2$: geometric (a geometric series), with r = 2.

27 − 30 = −3, 24 − 27 = −3: arithmetic (an arithmetic series), with d = −3.

Write both with a + (n − 1)d and subtract.

t₃ = a + 2d = 9 … (i); t₉ = a + 8d = 33 … (ii).
(ii) − (i): 6d = 24, so d = 4; then a = 9 − 8 = 1.
t₅ = a + 4d = 1 + 16 = 17.

d = 6 − 2 = 4. $t_{10} = a + 9d = 2 + 36 =$ 38.

a + 3d = 20 … (i); a + 6d = 35 … (ii).
(ii) − (i): 3d = 15, d = 5; a = 20 − 15 = 5.
$t_{10} = a + 9d = 5 + 45 =$ 50.
(Quicker: from the 7th to the 10th is three more steps of 5, 35 + 15 = 50.)

In an arithmetic sequence, t₂ − t₁ = t₃ − t₂.

$$\begin{aligned} (x + 2) - (x - 1) &= 3x - (x + 2) \\ 3 &= 2x - 2 \\ x &= \tfrac{5}{2} \end{aligned}$$
$x = \frac{5}{2}$. The terms are $\frac{3}{2}, \frac{9}{2}, \frac{15}{2}$, so a = $\frac{3}{2}$ and d = 3.
$t_5 = a + 4d = \frac{3}{2} + 12 =$ $\frac{27}{2}$.

a + 4d = 17, a + 7d = 26: subtracting, 3d = 9, d = 3, and a = 17 − 12 = 5.
$$\begin{aligned} 5 + (n - 1) \times 3 &= 44 \\ (n - 1) \times 3 &= 39 \\ n - 1 &= 13 \end{aligned}$$
n = 14: 44 is the 14th term.

$r = \frac{6}{12} =$ $\frac{1}{2}$.
$t_{10} = ar^9 = 12 \times \left(\frac{1}{2}\right)^9 = \frac{12}{512} =$ $\frac{3}{128}$.

ar = 6, $ar^4 = 162$: dividing, $r^3 = 27$, r = 3, and a = 2 (as in Example 4).
$t_9 = ar^8 = 2 \times 3^8 = 2 \times 6561 =$ 13,122.

In a geometric sequence, $\dfrac{t_2}{t_1} = \dfrac{t_3}{t_2}$.

$$\begin{aligned} \frac{z}{z + 6} &= \frac{z - 3}{z} \\ z^2 &= (z + 6)(z - 3) \\ z^2 &= z^2 + 3z - 18 \\ 3z &= 18 \end{aligned}$$
z = 6. The terms are 12, 6, 3, so a = 12 and $r = \frac{1}{2}$.
$t_5 = ar^4 = 12 \times \frac{1}{16} =$ $\frac{3}{4}$.