Year 10Extension

Arithmetic Sequence

Sequences and series are not part of the Year 10 core curriculum. Offered as preparation for senior mathematics.

Know your formulas.

Know your formulas

Arithmetic sequence

T_{n} = a + (n-1)d

$T_{n}$ = the $n^{th}$ term, $a$ = first term, $d$ = constant difference, $n$ = position of the term.

Geometric sequence

T_{n} = ar^{n-1}

$r$ = constant ratio.

e.g.

(8;\ 15;\ 22;\ 29)

Determine that there is a constant difference if not told $d = T_{2} - T_{1} = T_{3} - T_{2} = 7$
Determine the first term $a = 8$
Substitute into formula $T_{n} = 8 + (n-1)7$
Simplify if asked $T_{n} = 7n + 1$

For the following patterns, determine (a) the constant difference, (b) the general formula, (c) the $15^{th}$ term:

8;\ 10;\ 12;\ 14\ldots

Answer

d = 2,\quad T_{n} = 8 + (n-1)2 = 2n + 6,\quad T_{15} = 36

1;\ \tfrac{1}{2};\ 0;\ -\tfrac{1}{2}\ldots

Answer

d = -\tfrac{1}{2},\quad T_{n} = 1 + (n-1)(-\tfrac{1}{2}),\quad T_{15} = -6

-0.9;\ 0.8;\ 2.5;\ 4.2\ldots

Answer

d = 1.7,\quad T_{n} = -0.9 + (n-1)1.7,\quad T_{15} = 22.9

-29;\ 171;\ 371;\ 571\ldots

Answer

d = 200,\quad T_{n} = -29 + (n-1)200,\quad T_{15} = 2771

\tfrac{1}{7};\ \tfrac{17}{21};\ \tfrac{31}{21};\ \tfrac{15}{7}\ldots

Answer

d = \tfrac{2}{3},\quad T_{n} = \tfrac{1}{7} + (n-1)\tfrac{2}{3},\quad T_{15} = \tfrac{199}{21}

-14.1;\ -12.2;\ -10.3;\ -8.4\ldots

Answer

d = 1.9,\quad T_{n} = -14.1 + (n-1)1.9,\quad T_{15} = 12.5

-12;\ -22;\ -32;\ -42\ldots

Answer

d = -10,\quad T_{n} = -12 + (n-1)(-10),\quad T_{15} = -152

20;\ 30;\ 40;\ 50;\ 60\ldots

Answer

d = 10,\quad T_{n} = 20 + (n-1)10,\quad T_{15} = 160

2;\ \tfrac{8}{3};\ \tfrac{10}{3};\ 4\ldots

Answer

d = \tfrac{2}{3},\quad T_{n} = 2 + (n-1)\tfrac{2}{3},\quad T_{15} = \tfrac{34}{3}

3x+1;\ 2x;\ 3x-7\ldots \text{ are the first three terms of a linear pattern. Find } x

Answer

x = 3. \text{ Constant difference: } 2x-(3x+1) = (3x-7)-2x \Rightarrow -x-1 = x-7 \Rightarrow x = 3