Year 10Extension

Geometric Sequence

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

Worked example

e.g.

(3;\ 9;\ 27;\ 81)

Determine that there is a constant ratio if not told $r = T_{2} \div T_{1} = T_{3} \div T_{2} = 3$
Determine the first term $a = 3$
Substitute into formula $T_{n} = 3 \times 3^{n-1}$
Simplify if possible $T_{n} = 3^{n}$

For the following patterns, determine (a) the constant ratio, (b) the general formula, (c) the $8^{th}$ term:

2;\ 4;\ 8;\ 16\ldots

Answer

r = 2,\quad T_{n} = 2 \times 2^{n-1} = 2^{n},\quad T_{8} = 256

32;\ 64;\ 128;\ 256\ldots

Answer

r = 2,\quad T_{n} = 32 \times 2^{n-1},\quad T_{8} = 4096

5;\ 15;\ 45;\ 135\ldots

Answer

r = 3,\quad T_{n} = 5 \times 3^{n-1},\quad T_{8} = 10\,935

1;\ \tfrac{1}{2};\ \tfrac{1}{4};\ \tfrac{1}{8}\ldots

Answer

r = \tfrac{1}{2},\quad T_{n} = \left(\tfrac{1}{2}\right)^{n-1},\quad T_{8} = \tfrac{1}{128}

100;\ 20;\ 4;\ 0.8\ldots

Answer

r = \tfrac{1}{5},\quad T_{n} = 100 \times \left(\tfrac{1}{5}\right)^{n-1},\quad T_{8} = \tfrac{1}{781.25} = 0.00128

3;\ -6;\ 12;\ -24\ldots

Answer

r = -2,\quad T_{n} = 3 \times (-2)^{n-1},\quad T_{8} = -384

81;\ 27;\ 9;\ 3\ldots

Answer

r = \tfrac{1}{3},\quad T_{n} = 81 \times \left(\tfrac{1}{3}\right)^{n-1},\quad T_{8} = \tfrac{1}{27}

7;\ 14;\ 28;\ 56\ldots

Answer

r = 2,\quad T_{n} = 7 \times 2^{n-1},\quad T_{8} = 896

\tfrac{1}{2};\ 2;\ 8;\ 32\ldots

Answer

r = 4,\quad T_{n} = \tfrac{1}{2} \times 4^{n-1},\quad T_{8} = 8192

\text{Is } 4;\ 12;\ 36;\ 108 \text{ arithmetic or geometric?}

Answer

\text{Geometric, } r = 3. \text{ The differences } (8, 24, 72) \text{ are not constant, the ratios are.}