Geometric Sequences (B) Worksheet | PDF Printable Algebra Worksheet - Free Printable

Here are the solutions to the problems on the worksheet, broken down by section.

Section A

*Write down the first four terms of the following geometric sequences.*

To find the terms, we substitute $n = 1, 2, 3,$ and $4$ into each formula.

1) $2^n$
* $n=1: 2^1 = 2$
* $n=2: 2^2 = 4$
* $n=3: 2^3 = 8$
* $n=4: 2^4 = 16$
Answer: 2, 4, 8, 16

2) $5^n$
* $n=1: 5^1 = 5$
* $n=2: 5^2 = 25$
* $n=3: 5^3 = 125$
* $n=4: 5^4 = 625$
Answer: 5, 25, 125, 625

3) $(\frac{1}{2})^n$
* $n=1: (\frac{1}{2})^1 = \frac{1}{2}$
* $n=2: (\frac{1}{2})^2 = \frac{1}{4}$
* $n=3: (\frac{1}{2})^3 = \frac{1}{8}$
* $n=4: (\frac{1}{2})^4 = \frac{1}{16}$
Answer: $\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}$

4) $(\frac{3}{10})^n$
* $n=1: (\frac{3}{10})^1 = \frac{3}{10}$ (or 0.3)
* $n=2: (\frac{3}{10})^2 = \frac{9}{100}$ (or 0.09)
* $n=3: (\frac{3}{10})^3 = \frac{27}{1000}$ (or 0.027)
* $n=4: (\frac{3}{10})^4 = \frac{81}{10000}$ (or 0.0081)
Answer: $\frac{3}{10}, \frac{9}{100}, \frac{27}{1000}, \frac{81}{10000}$

5) $0.2^n$
* $n=1: 0.2^1 = 0.2$
* $n=2: 0.2^2 = 0.04$
* $n=3: 0.2^3 = 0.008$
* $n=4: 0.2^4 = 0.0016$
Answer: 0.2, 0.04, 0.008, 0.0016

6) $(\sqrt{3})^n$
* $n=1: (\sqrt{3})^1 = \sqrt{3}$
* $n=2: (\sqrt{3})^2 = 3$
* $n=3: (\sqrt{3})^3 = 3\sqrt{3}$
* $n=4: (\sqrt{3})^4 = 9$
Answer: $\sqrt{3}, 3, 3\sqrt{3}, 9$

7) $(0.1)^{n+1}$
* $n=1: (0.1)^{1+1} = 0.1^2 = 0.01$
* $n=2: (0.1)^{2+1} = 0.1^3 = 0.001$
* $n=3: (0.1)^{3+1} = 0.1^4 = 0.0001$
* $n=4: (0.1)^{4+1} = 0.1^5 = 0.00001$
Answer: 0.01, 0.001, 0.0001, 0.00001

8) $(5\sqrt{2})^{n-1}$
* $n=1: (5\sqrt{2})^{1-1} = (5\sqrt{2})^0 = 1$
* $n=2: (5\sqrt{2})^{2-1} = (5\sqrt{2})^1 = 5\sqrt{2}$
* $n=3: (5\sqrt{2})^{3-1} = (5\sqrt{2})^2 = 25 \times 2 = 50$
* $n=4: (5\sqrt{2})^{4-1} = (5\sqrt{2})^3 = 125 \times 2\sqrt{2} = 250\sqrt{2}$
Answer: $1, 5\sqrt{2}, 50, 250\sqrt{2}$

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Section B

*Answer the questions about geometric sequences.*

1) Write the next three terms of the geometric sequences.

a) 4, 16, 64, 256
* Find the multiplier (common ratio): $16 \div 4 = 4$.
* Next term: $256 \times 4 = 1024$
* Next term: $1024 \times 4 = 4096$
* Next term: $4096 \times 4 = 16384$
Answer: 1024, 4096, 16384

b) $1, \frac{2}{3}, \frac{4}{9}$
* Find the multiplier: To get from 1 to $\frac{2}{3}$, you multiply by $\frac{2}{3}$. Check: $\frac{2}{3} \times \frac{2}{3} = \frac{4}{9}$. It works.
* Next term: $\frac{4}{9} \times \frac{2}{3} = \frac{8}{27}$
* Next term: $\frac{8}{27} \times \frac{2}{3} = \frac{16}{81}$
* Next term: $\frac{16}{81} \times \frac{2}{3} = \frac{32}{243}$
Answer: $\frac{8}{27}, \frac{16}{81}, \frac{32}{243}$

2) Write down the first four terms of the geometric sequence $0.1^{1-n}$
* $n=1: 0.1^{1-1} = 0.1^0 = 1$
* $n=2: 0.1^{1-2} = 0.1^{-1} = \frac{1}{0.1} = 10$
* $n=3: 0.1^{1-3} = 0.1^{-2} = \frac{1}{0.01} = 100$
* $n=4: 0.1^{1-4} = 0.1^{-3} = \frac{1}{0.001} = 1000$
Answer: 1, 10, 100, 1000

3) Write down the nth term of this geometric sequence: $1, \frac{2}{5}, \frac{4}{25}, \frac{8}{125}$
* First term ($a$) is 1.
* Common ratio ($r$): $\frac{2}{5} \div 1 = \frac{2}{5}$.
* Formula for nth term: $a \cdot r^{n-1}$
* Substitute values: $1 \cdot (\frac{2}{5})^{n-1}$
Answer: $(\frac{2}{5})^{n-1}$

4) Sukh invests £2000 into a bank account. Sukh’s investment increases by 2.5% each month.

a) Show that Sukh’s investment forms a geometric sequence.
A geometric sequence is formed when you multiply by the same number every time.
* Increasing by 2.5% means multiplying by $1 + 0.025 = 1.025$.
* Month 1: $£2000$
* Month 2: $£2000 \times 1.025$
* Month 3: $(£2000 \times 1.025) \times 1.025$
* Because we are multiplying by the constant factor 1.025 each month, it is a geometric sequence.

b) Write down the common ratio for the geometric sequence.
As calculated above, the multiplier is 1 plus the percentage as a decimal.
$1 + 0.025 = 1.025$
Answer: 1.025

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Extension

*Write down the first four terms of the geometric sequence $(\frac{3}{\sqrt{2}})^{1-n}$*

* $n=1: (\frac{3}{\sqrt{2}})^{1-1} = (\frac{3}{\sqrt{2}})^0 = 1$
* $n=2: (\frac{3}{\sqrt{2}})^{1-2} = (\frac{3}{\sqrt{2}})^{-1} = \frac{\sqrt{2}}{3}$
* $n=3: (\frac{3}{\sqrt{2}})^{1-3} = (\frac{3}{\sqrt{2}})^{-2} = (\frac{\sqrt{2}}{3})^2 = \frac{2}{9}$
* $n=4: (\frac{3}{\sqrt{2}})^{1-4} = (\frac{3}{\sqrt{2}})^{-3} = (\frac{\sqrt{2}}{3})^3 = \frac{2\sqrt{2}}{27}$

Answer: $1, \frac{\sqrt{2}}{3}, \frac{2}{9}, \frac{2\sqrt{2}}{27}$