Geometric sequences with surds and other patterns

(a) $\sqrt{2}$

1. $\frac{2}{\sqrt{2}} = \sqrt{2}$: each term is $\sqrt{2}$ times the one before.

• B1 The correct answer, $\sqrt{2}$.

(b) $16$

1. The $n$th term is $(\sqrt{2})^n$.
2. $$(\sqrt{2})^8 = 2^4 = 16$$

• M1 Writing $(\sqrt{2})^8$ or continuing the sequence: $4\sqrt{2}, 8, 8\sqrt{2}, 16$.
• A1 The correct answer, $16$.

(a) $12\sqrt{2}$

1. $$3\sqrt{2}^{5}$$
2. The sixth term is the first term multiplied by the ratio five times: 3($\sqrt{2}$)$^{5}.$
3. ($\sqrt{2}$)$^{5}$ = ($\sqrt{2}$)$^{4}$ $\times$ $\sqrt{2}$ = $4\sqrt{2}.$
4. The sixth term is $12\sqrt{2}.$

• P1 Establishing $3\sqrt{2}^{5}$ or an equivalent valid method.
• A1 Correct answer: $12\sqrt{2}$

(a) $162$

1. $$18\sqrt{3}/6$$
2. $$18\sqrt{3}\times 3\sqrt{3}$$
3. Moving from the second term to the fifth multiplies by the common ratio cubed. This multiplier is ($18\sqrt{3}$)/6 = $3\sqrt{3}.$
4. Moving from the fifth term to the eighth uses the same three-step multiplier.
5. The eighth term is $18\sqrt{3}$ $\times$ $3\sqrt{3}$ = 54 $\times$ 3 = 162.

• P1 Establishing $18\sqrt{3}/6$ or an equivalent valid method.
• P1 Establishing $18\sqrt{3}\times 3\sqrt{3}$ or an equivalent valid method.
• A1 Correct answer: $162$

(a) $27$

1. There are four ratio multiplications from the first to the fifth term.
   $$3\sqrt{3}^{4}$$
2. Therefore $27$.

• M1 There are four ratio multiplications from the first to the fifth term.
• A1 Correct answer: $27$

(b) $27\sqrt{3}$

1. Multiply the fifth term once more by the common ratio.
   $$27\sqrt{3}$$
2. Therefore $27\sqrt{3}$.

• M1 Multiply the fifth term once more by the common ratio.
• A1 Correct answer: $27\sqrt{3}$

(a) $2$

1. Three ratio multiplications join the two given terms.
   $$r^{3}=64/8$$
2. Therefore $2$.

• M1 Three ratio multiplications join the two given terms.
• A1 Correct answer: $2$

(b) $4$

1. Divide the second term by the common ratio.
   $$8/2$$
2. Therefore $4$.

• M1 Divide the second term by the common ratio.
• A1 Correct answer: $4$

(a) $320$

1. Use six multiplications by the negative common ratio.
   $$5\times (-2)^{6}$$
2. Therefore $320$.

• M1 Use six multiplications by the negative common ratio.
• A1 Correct answer: $320$

(b) $-25$

1. Add the four signed terms explicitly.
   $$5-10+20-40$$
2. Therefore $-25$.

• M1 Add the four signed terms explicitly.
• A1 Correct answer: $-25$

(a) $a = 2$, $b = 5$

1. Write the terms.
   $$a, \ b, \ a + b, \ a + 2b, \ 2a + 3b, \ 3a + 5b$$
2. Form simultaneous equations.
   $$2a + 3b = 19, \qquad 3a + 5b = 31$$
3. Multiply the first by 3 and the second by 2, then subtract.
   $$6a + 10b - (6a + 9b) = 62 - 57 \Rightarrow b = 5, \ a = 2$$

• P1 Writing the fifth and sixth terms as $2a + 3b$ and $3a + 5b$.
• P1 Forming the two equations.
• P1 Eliminating a variable to find $b = 5$ or $a = 2$.
• A1 The correct answer, $(2, 5)$.

(a) $\frac{1}{2 + \sqrt{3}} = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3}$

1. The ratio is the second term divided by the first.
   $$r = \frac{1}{2 + \sqrt{3}}$$
2. Rationalise the denominator.
   $$\frac{1}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3}$$

• M1 Writing $r = \frac{1}{2 + \sqrt{3}}$.
• M1 Multiplying top and bottom by $2 - \sqrt{3}$.
• A1 Showing the denominator is 1 to reach $2 - \sqrt{3}$.

(b) $2 - \sqrt{3}$

1. Third term $= 1 \times (2 - \sqrt{3}) = 2 - \sqrt{3}$.

• B1 The correct answer, $2 - \sqrt{3}$.

(c) $7 - 4\sqrt{3}$

1. $$(2 - \sqrt{3})^2 = 4 - 4\sqrt{3} + 3 = 7 - 4\sqrt{3}$$

• M1 Expanding $(2 - \sqrt{3})^2$ with at least three correct terms.
• A1 $7 - 4\sqrt{3}$.