# IB Math Analysis & Approaches Question Bank – Geometric sequences and series -SL Topic: SL 1.3 Paper 1

### Question

Filicia baked a very large apple pie that she cuts into slices to share with her friends. The smallest

slice is cut first. The volume of each successive slice of pie forms a geometric sequence.

The second smallest slice has a volume of 30 cm3. The fifth smallest slice has a volume

of 240 cm3.

1. Find the common ratio of the sequence. [2]

2. Find the volume of the smallest slice of pie. [2] The apple pie has a volume of 61 425 cm3.

3. Find the total number of slices Filicia can cut from this pie. [2]

Ans:Â

(a)

$$u_{1}r= 30 \:and\: u_{1}r^{4}$$ = 240,

OR $$30r^{3}= 240 (r^{3}$$=8 )

r=2

(b)

$$u_{1}\times 2= 30 \: OR \: u_{1} \times 2^{4}$$= 240

$$u_{1}$$= 15

(c)

$$\frac{15 (2^{n}-1)}{2-1}$$= 61425

n= 12 (slices)

## Question

The first three terms of a geometric sequence are $$\ln {x^{16}}$$, $$\ln {x^8}$$, $$\ln {x^4}$$, for $$x > 0$$.

Find the common ratio.

[3]
a.

Solve $$\sum\limits_{k = 1}^\infty Â {{2^{5 – k}}\ln x = 64}$$.

[5]
b.

## Markscheme

correct use $$\log {x^n} = n\log x$$ Â  Â  A1

eg$$\,\,\,\,\,$$$$16\ln x$$

valid approach to find $$r$$ Â  Â  (M1)

eg$$\,\,\,\,\,$$$$\frac{{{u_{n + 1}}}}{{{u_n}}},{\text{ }}\frac{{\ln {x^8}}}{{\ln {x^{16}}}},{\text{ }}\frac{{4\ln x}}{{8\ln x}},{\text{ }}\ln {x^4} = \ln {x^{16}} \times {r^2}$$

$$r = \frac{1}{2}$$ Â  Â  A1 Â  Â  N2

[3 marks]

a.

recognizing a sum (finite or infinite) Â  Â  (M1)

eg$$\,\,\,\,\,$$$${2^4}\ln x + {2^3}\ln x,{\text{ }}\frac{a}{{1 – r}},{\text{ }}{S_\infty },{\text{ }}16\ln x +Â \ldots$$

valid approach (seen anywhere) Â  Â  (M1)

eg$$\,\,\,\,\,$$recognizing GP is the same as part (a), using their $$r$$ value from part (a), $$r = \frac{1}{2}$$

correct substitution into infinite sum (only if $$\left| r \right|$$ is a constant and less than 1) Â  Â  A1

eg$$\,\,\,\,\,$$$$\frac{{{2^4}\ln x}}{{1 – \frac{1}{2}}},{\text{ }}\frac{{\ln {x^{16}}}}{{\frac{1}{2}}},{\text{ }}32\ln x$$

correct working Â  Â  (A1)

eg$$\,\,\,\,\,$$$$\ln x = 2$$

$$x = {{\text{e}}^2}$$ Â  Â  A1 Â  Â  N3

[5 marks]

b.

## Question

Consider the arithmetic sequence $$2{\text{, }}5{\text{, }}8{\text{, }}11{\text{,}} \ldots$$ .

Find $${u_{101}}$$ .

[3]
a.

Consider the arithmetic sequence $$2{\text{, }}5{\text{, }}8{\text{, }}11{\text{,}} \ldots$$ .

Find the value of n so that $${u_n} = 152$$ .

[3]
b.

## Markscheme

$$d = 3$$Â Â Â Â Â (A1)

evidence of substitution into $${u_n} = a + (n – 1)d$$Â Â Â  Â (M1)

e.g. $${u_{101}} = 2 + 100 \times 3$$

$${u_{101}} = 302$$Â Â Â  Â A1 Â  Â  N3

[3 marks]

a.

correct approachÂ Â Â Â  (M1)

e.g. $$152 = 2 + (n – 1) \times 3$$

correct simplificationÂ Â Â Â  (A1)

e.g. $$150 = (n – 1) \times 3$$ , $$50 = n – 1$$ , $$152 = – 1 + 3n$$

$$n = 51$$Â Â Â  Â A1 Â  Â  N2

[3 marks]

b.

## Examiners report

Candidates probably had the most success with this question with many good solutions which were written with the working clearly shown. Many used the alternate approach of $${u_n} = 3n – 1$$ .

a.

Candidates probably had the most success with this question with many good solutions which were written with the working clearly shown. Many used the alternate approach of $${u_n} = 3n – 1$$ .

b.

## Question

Consider the infinite geometric sequence $$3{\text{, }}3(0.9){\text{, }}3{(0.9)^2}{\text{, }}3{(0.9)^3}{\text{, }} \ldots$$ .

Write down the 10th term of the sequence. Do not simplify your answer.

[1]
a.

Consider the infinite geometric sequence $$3{\text{, }}3(0.9){\text{, }}3{(0.9)^2}{\text{, }}3{(0.9)^3}{\text{, }} \ldots$$ .

Find the sum of the infinite sequence.

[4]
b.

## Markscheme

$${u_{10}} = 3{(0.9)^9}$$Â  Â  A1Â Â Â Â  N1

[1 mark]

a.

recognizing $$r = 0.9$$Â Â Â  Â (A1)

correct substitutionÂ Â Â Â  A1

e.g.Â  $$S = \frac{3}{{1 – 0.9}}$$

$$S = \frac{3}{{0.1}}$$Â  Â  (A1)

$$S = 30$$Â Â Â  A1Â Â Â Â  N3

[4 marks]

b.

## Examiners report

This question was well done by most candidates. There were a surprising number of candidates who lost a mark for not simplifying $$\frac{3}{{0.1}}$$Â to 30Â , and there were a few candidates who used the formula for the finite sum unsuccessfully.

a.

This question was well done by most candidates. There were a surprising number of candidates who lost a mark for not simplifying $$\frac{3}{{0.1}}$$Â to 30, and there were a few candidates who used the formula for the finite sum unsuccessfully.

b.

## Question

The first three terms of an infinite geometric sequence are 32, 16 and 8.

Write down the value of r .

[1]
a.

Find $${u_6}$$ .

[2]
b.

Find the sum to infinity of this sequence.

[2]
c.

## Markscheme

$$r = \frac{{16}}{{32}}\left( { = \frac{1}{2}} \right)$$Â Â Â Â  A1 Â  Â  N1

[1 mark]

a.

correct calculation or listing termsÂ Â Â Â  (A1)

e.g. $$32 \times {\left( {\frac{1}{2}} \right)^{6 – 1}}$$ , $$8 \times {\left( {\frac{1}{2}} \right)^3}$$ , 32, $$\ldots$$ 4, 2, 1

$${u_6} = 1$$Â Â Â Â Â A1Â Â Â Â  N2

[2 marks]

b.

evidence of correct substitution in $${S_\infty }$$Â Â Â Â Â Â A1

e.g. $$\frac{{32}}{{1 – \frac{1}{2}}}$$ , $$\frac{{32}}{{\frac{1}{2}}}$$

$${S_\infty } = 64$$Â Â Â Â Â A1Â Â Â Â  N1

[2 marks]

c.

## Examiners report

This question was very well done by the majority of candidates. There were some who used a value of r greater than one, with the most common error being $$r = 2$$ .

a.

This question was very well done by the majority of candidates. There were some who used a value of r greater than one, with the most common error being $$r = 2$$ .Â

b.

A handful of candidates struggled with the basic computation involved in part (c).

c.

## Question

In an arithmetic sequence, $${u_1} = 2$$ and $${u_3} = 8$$ .

Find dÂ .

[2]
a.

Find $${u_{20}}$$ .

[2]
b.

Find $${S_{20}}$$ .

[2]
c.

## Markscheme

attempt to find dÂ Â Â Â  (M1)

e.g. $$\frac{{{u_3} – {u_1}}}{2}$$ , $$8 = 2 + 2d$$

$$d = 3$$Â Â Â Â Â A1 Â  Â  N2

[2 marks]

a.

correct substitutionÂ Â Â Â  (A1)

e.g. $${u_{20}} = 2 + (20 – 1)3$$ , $${u_{20}} = 3 \times 20 – 1$$

$${u_{20}} = 59$$Â Â Â Â Â A1Â Â Â Â  N2

[2 marks]

b.

correct substitutionÂ Â Â Â  (A1)

e.g. $${S_{20}} = \frac{{20}}{2}(2 + 59)$$ , $${S_{20}} = \frac{{20}}{2}(2 \times 2 + 19 \times 3)$$

$${S_{20}} = 610$$Â Â Â  A1 Â  Â  N2

[2 marks]

c.

## Examiners report

This question was answered correctly by the large majority of candidates. The few mistakes seen were due to either incorrect substitution into the formula or simple arithmetic errors. Even where candidates made mistakes, they were usually able to earn full follow-through marks in the subsequent parts of the question.

a.

This question was answered correctly by the large majority of candidates. The few mistakes seen were due to either incorrect substitution into the formula or simple arithmetic errors. Even where candidates made mistakes, they were usually able to earn full follow-through marks in the subsequent parts of the question.

b.

This question was answered correctly by the large majority of candidates. The few mistakes seen were due to either incorrect substitution into the formula or simple arithmetic errors. Even where candidates made mistakes, they were usually able to earn full follow-through marks in the subsequent parts of the question.

c.

## Question

The first three terms of a infinite geometric sequence are $$m – 1,{\text{ 6, }}m + 4$$, where $$m \in \mathbb{Z}$$.

Write down an expression for the common ratio, $$r$$.

[2]
a(i).

Hence, show that $$m$$ satisfies the equation $${m^2} + 3m – 40 = 0$$.

[2]
a(ii).

Find the two possible values of $$m$$.

[3]
b(i).

Find the possible values of $$r$$.

[3]
b(ii).

The sequence has a finite sum.

State which value of $$r$$ leads to this sum and justify your answer.

[3]
c(i).

The sequence has a finite sum.

Calculate the sum of the sequence.

[3]
c(ii).

## Markscheme

correct expression for $$r$$ Â  Â  A1 Â  Â  N1

eg Â  $$r = \frac{6}{{m – 1}},{\text{ }}\frac{{m + 4}}{6}$$

[2 marks]

a(i).

correct equation Â  Â  A1

eg Â  Â  $$\frac{6}{{m – 1}} = \frac{{m + 4}}{6},{\text{ }}\frac{6}{{m + 4}} = \frac{{m – 1}}{6}$$

correct working Â  Â  (A1)

eg Â  Â  $$(m + 4)(m – 1) = 36$$

correct working Â  Â  A1

eg Â  Â  $${m^2} – m + 4m – 4 = 36,{\text{ }}{m^2} + 3m – 4 = 36$$

$${m^2} + 3m – 40 = 0$$ Â  Â  AG Â  Â  N0

[2 marks]Â

a(ii).

valid attempt to solve Â  Â  (M1)

eg Â  Â  $$(m + 8)(m – 5) = 0,{\text{ }}m = \frac{{ – 3 \pm \sqrt {9 + 4 \times 40} }}{2}$$

$$m =Â – 8,{\text{ }}m = 5$$ Â  Â  A1A1 Â  Â  N3

[3 marks]

b(i).

attempt to substitute any value of $$m$$ to find $$r$$ Â  Â  (M1)

eg Â  Â  $$\frac{6}{{ – 8 – 1}},{\text{ }}\frac{{5 + 4}}{6}$$

$$r = \frac{3}{2},{\text{ }}r =Â – \frac{2}{3}$$ Â  Â  A1A1 Â  Â  N3

[3 marks]

b(ii).

$$r =Â – \frac{2}{3}$$ Â  (may be seen in justification) Â  Â  A1

valid reason Â  Â  R1 Â  Â  N0

eg Â  Â  $$\left| r \right| < 1,{\text{ }} – 1 < \frac{{ – 2}}{3} < 1$$

Â

Notes: AwardÂ R1Â for $$\left| r \right| < 1$$ only ifÂ A1Â awarded.

Â

[2 marks]

c(i).

finding the first term of the sequence which has $$\left| r \right| < 1$$ Â  Â Â (A1)

egÂ Â  Â  $$– 8 – 1,{\text{ }}6 \div \frac{{ – 2}}{3}$$

$${u_1} =Â – 9$$ Â  (may be seen in formula) Â  Â Â (A1)

correct substitution of $${u_1}$$ andÂ their $$r$$ into $$\frac{{{u_1}}}{{1 – r}}$$, as long as $$\left| r \right| < 1$$ Â  Â Â A1

egÂ Â  Â  $${S_\infty } = \frac{{ – 9}}{{1 – \left( { – \frac{2}{3}} \right)}},{\text{ }}\frac{{ – 9}}{{\frac{5}{3}}}$$

$${S_\infty } =Â – \frac{{27}}{5}{\text{ }}( =Â – 5.4)$$ Â  Â Â A1 Â  Â  N3

[4 marks]Â

c(ii).

[N/A]

a(i).

[N/A]

a(ii).

[N/A]

b(i).

[N/A]

b(ii).

[N/A]

c(i).

[N/A]

c(ii).

## Question

In an arithmetic sequence, the third term is 10 and the fifth term is 16.

Find the common difference.

[2]
a.

Find the first term.

[2]
b.

Find the sum of the first 20 terms of the sequence.

[3]
c.

## Markscheme

attempt to find $$d$$ Â  Â Â (M1)

eg Â  Â Â $$\frac{{16 – 10}}{2}{\text{, }}10 – 2d = 16 – 4d{\text{, }}2d = 6{\text{, }}d = 6$$

$$d = 3$$ Â  Â Â A1 Â  Â  N2

[2 marks]

a.

correct approach Â  Â Â (A1)

egÂ  Â  Â $$10 = {u_1} + 2 \times 3{\text{, }}10 – 3 – 3$$

$${u_1} = 4$$Â  Â  Â A1 Â  Â  N2

[2 marks]

b.

correct substitution into sum or term formula Â  Â Â (A1)

egÂ  Â  Â $$\frac{{20}}{2}(2 \times 4 + 19 \times 3){\text{, }}{u_{20}} = 4 + 19 \times 3$$

correct simplification Â  Â Â (A1)

egÂ  Â  Â $$8 + 57{\text{, }}4 + 61$$

$${S_{20}} = 650$$ Â  Â  A1 Â  Â  N2

[3 marks]

c.

[N/A]

a.

[N/A]

b.

[N/A]

c.

## Question

The sides of a square are 16 cm in length. The midpoints of the sides of this square are joinedÂ to form a new square and four triangles (diagram 1). The process is repeated twice, as shownÂ in diagrams 2 and 3.

Let $${x_n}$$Â denote the length of one of the equal sides of each new triangle.

Let $${A_n}$$Â denote the area of each new triangle.

Â

The following table gives the values of $${x_n}$$Â and $${A_n}$$, for $$1 \leqslant n \leqslant 3$$. Copy and completeÂ the table. (Do not write on this page.)

Â

 $$n$$ 1 2 3 $${x_n}$$ 8 Â 4 $${A_n}$$ 32 16 Â
[4]
a.

The process described above is repeated. Find $${A_6}$$.

[4]
b.

Consider an initial square of side lengthÂ $$k {\text{ cm}}$$. The process described above is repeatedÂ indefinitely. The total area of the shaded regions isÂ $$k {\text{ c}}{{\text{m}}^2}$$. Find the value of $$k$$.

[7]
c.

## Markscheme

valid method for finding side length Â  Â  (M1)

eg Â  $${8^2} + {8^2} = {c^2},{\text{ }}45 – 45 – 90{\text{ side ratios, }}8\sqrt 2 ,{\text{ }}\frac{1}{2}{s^2} = 16,{\text{ }}{x^2} + {x^2} = {8^2}$$

correct working for area Â  Â Â (A1)

eg Â  $$\frac{1}{2} \times 4 \times 4$$

 $$n$$ 1 2 3 $${x_n}$$ 8 $$\sqrt {32}$$ 4 $${A_n}$$ 32 16 8

Â  Â  Â A1A1 Â  Â  N2N2

[4 marks]

a.

METHOD 1

recognize geometric progression for $${A_n}$$ Â  Â  (R1)

eg Â  $${u_n} = {u_1}{r^{n – 1}}$$

$$r = \frac{1}{2}$$ Â  Â Â (A1)

correct working Â  Â  (A1)

egÂ  Â $$32{\left( {\frac{1}{2}} \right)^5};{\text{ 4, 2, 1, }}\frac{1}{2},{\text{ }}\frac{1}{4},{\text{ }} \ldots$$

$${A_6} = 1$$ Â  Â Â A1 Â  Â  N3

Â

METHOD 2

attempt to find $${x_6}$$ Â  Â  (M1)

egÂ  Â $$8{\left( {\frac{1}{{\sqrt 2 }}} \right)^5},{\text{ }}2\sqrt 2 ,{\text{ 2, }}\sqrt 2 ,{\text{ 1, }} \ldots$$

$${x_6} = \sqrt 2$$ Â  Â Â (A1)

correct working Â  Â  (A1)

egÂ  Â $$\frac{1}{2}{\left( {\sqrt 2 } \right)^2}$$

$${A_6} = 1$$ Â  Â  A1 Â  Â  N3

[4 marks]

b.

METHOD 1

recognize infinite geometric series Â  Â  (R1)

eg Â  $${S_n} = \frac{a}{{1 – r}},{\text{ }}\left| r \right| < 1$$

area of first triangle in terms of $$k$$Â  Â  Â (A1)

egÂ  Â $$\frac{1}{2}{\left( {\frac{k}{2}} \right)^2}$$

attempt to substitute into sum of infinite geometric series (must have $$k$$) Â  Â  (M1)

eg Â  $$\frac{{\frac{1}{2}{{\left( {\frac{k}{2}} \right)}^2}}}{{1 – \frac{1}{2}}},{\text{ }}\frac{k}{{1 – \frac{1}{2}}}$$

correct equation Â  Â  A1

eg Â  $$\frac{{\frac{1}{2}{{\left( {\frac{k}{2}} \right)}^2}}}{{1 – \frac{1}{2}}} = k,{\text{ }}k = \frac{{\frac{{{k^2}}}{8}}}{{\frac{1}{2}}}$$

correct working Â  Â  (A1)

eg Â  $${k^2} = 4k$$

valid attempt to solve their quadratic Â  Â  (M1)

eg Â  $$k(k – 4),{\text{ }}k = 4{\text{ or }}k = 0$$

$$k = 4$$ Â  Â Â A1 Â  Â  N2

METHOD 2

recognizing that there are four sets of infinitely shaded regions with equal area Â  Â  R1

area of original square is $${k^2}$$ Â  Â  (A1)

so total shaded area is $$\frac{{{k^2}}}{4}$$ Â  Â Â (A1)

correct equation $$\frac{{{k^2}}}{4} = k$$ Â  Â Â A1

$${k^2} = 4k$$ Â  Â (A1)

valid attempt to solve their quadratic Â  Â  (M1)

eg Â  $$k(k – 4),{\text{ }}k = 4{\text{ or }}k = 0$$

$$k = 4$$ Â  Â Â A1 Â  Â  N2

[7 marks]

c.

[N/A]

a.

[N/A]

b.

[N/A]

c.

## Question

The sums of the terms of a sequence follow the pattern

$${S_1} = 1 + k,{\text{ }}{S_2} = 5 + 3k,{\text{ }}{S_3} = 12 + 7k,{\text{ }}{S_4} = 22 + 15k,{\text{ }} \ldots ,{\text{ where }}k \in \mathbb{Z}.$$

Given that $${u_1} = 1 + k$$, find $${u_2},{\text{ }}{u_3}$$ and $${u_4}$$.

[4]
a.

Find a general expression for $${u_n}$$.

[4]
b.

## Markscheme

valid method Â  Â  (M1)

eg Â  Â  $${u_2} = {S_2} – {S_1},{\text{ }}1 + k + {u_2} = 5 + 3k$$

$${u_2} = 4 + 2k,{\text{ }}{u_3} = 7 + 4k,{\text{ }}{u_4} = 10 + 8k$$ Â  Â  A1A1A1 Â  Â  N4

[4 marks]

a.

correct AP or GP Â  Â Â (A1)

eg Â  Â  finding common difference is $$3$$, common ratio is $$2$$

valid approach using arithmetic and geometric formulas Â  Â  (M1)

eg Â  Â  $$1 + 3(n – 1)$$Â  and $${r^{n – 1}}k$$

$${u_n} = 3n – 2 + {2^{n – 1}}k$$ Â  Â  A1A1 Â  Â  N4

Note: Award A1 for $$3n – 2$$, A1 for $${2^{n – 1}}k$$.

[4 marks]

b.
b.

## Question

In an arithmetic sequence, the first term isÂ $$2$$ and the second term is $$5$$.

Find the common difference.

[2]
a.

Find the eighth term.

[2]
b.

Find the sum of the first eight terms of the sequence.

[2]
c.

## Markscheme

correct approach Â  Â  (A1)

eg$$\;\;\;d = {u_2} – {u_1},{\text{ }}5 – 2$$

$$d = 3$$ Â  Â  A1 Â  Â  N2

[2 marks]

a.

correct approach Â  Â Â (A1)

eg$$\;\;\;{u_8} = 2 + 7 \times 3$$, listing terms

$${u_8} = 23$$ Â  Â  A1 Â  Â  N2

[2 marks]

b.

correct approachÂ Â Â Â  (A1)

eg$$\;\;\;{S_8} = \frac{8}{2}(2 + 23)$$, listing terms, $$\frac{8}{2}\left( {2(2) + 7(3)} \right)$$

$${S_8} = 100$$ Â  Â  A1 Â  Â  N2

[2 marks]

Total [6 marks]

c.
c.

## Question

Ann and Bob play a game where they each have an eight-sided die. Annâ€™s die has three green faces and five red faces; Bobâ€™s die has four green faces and four red faces. They take turns rolling their own die and note what colour faces up. The first player to roll green wins. Ann rolls first. Part of a tree diagram of the game is shown below.

Find the probability that Ann wins on her first roll.

[2]
a.

(i) Â  Â  The probability that Ann wins on her third roll is $$\frac{5}{8} \times \frac{4}{8} \times p \times q\ \times \frac{3}{8}$$.

Write down the value of $$p$$ and of $$q$$.

(ii) Â  Â  The probability that Ann wins on her tenth roll is $$\frac{3}{8}{r^k}$$ where $$r \in \mathbb{Q},{\text{ }}k \in \mathbb{Z}$$.

Find the value of $$r$$ and of $$k$$.

[6]
b.

Find the probability that Ann wins the game.

[7]
c.

## Markscheme

recognizing Ann rolls green Â  Â  (M1)

eg$$\;\;\;{\text{P(G)}}$$

$$\frac{3}{8}$$ Â  Â  A1 Â  Â  N2

[2 marks]

a.

(i) Â  Â  $$p = \frac{4}{8},{\text{ }}q = \frac{5}{8}$$ or $$q = \frac{4}{8},{\text{ }}p = \frac{5}{8}$$ Â  Â  A1A1 Â  Â  N2

(ii) Â  Â  recognizes Ann and Bob loseÂ $$9$$ times Â  Â  (M1)

eg$$\;\;\;\(\overbrace {{A_L}{B_\(\overbrace {{A_L}{B_ \ldots \(\overbrace {{A_L}{B_{\text{ 9 times, }}\underbrace {\left( {\frac{5}{8} \times \frac{4}{8}} \right) \timesÂ \ldotsÂ \times \left( {\frac{5}{8} \times \frac{4}{8}} \right)}_{{\text{9 times}}}$$

$$k = 9\;\;\;$$(seen anywhere) Â  Â  A1 Â  Â  N2

correct working Â  Â  (A1)

eg$$\;\;\;{\left( {\frac{5}{8} \times \frac{4}{8}} \right)^9} \times \frac{3}{8},{\text{ }}\left( {\frac{5}{8} \times \frac{4}{8}} \right) \timesÂ \ldotsÂ \times \left( {\frac{5}{8} \times \frac{4}{8}} \right) \times \frac{3}{8}$$

$$r = \frac{{20}}{{64}}\;\;\;\left( { = \frac{5}{{16}}} \right)$$ Â  Â  A1 Â  Â  N2

[6 marks]

b.

recognize the probability is an infinite sum Â  Â  (M1)

eg$$\;\;\;$$Ann wins on her $${{\text{1}}^{{\text{st}}}}$$ roll or $${{\text{2}}^{{\text{nd}}}}$$ roll or $${{\text{3}}^{{\text{rd}}}}$$ rollâ€¦, $${S_\infty }$$

recognizing GP Â  Â  (M1)

$${u_1} = \frac{3}{8}\;\;\;$$(seen anywhere) Â  Â  A1

$$r = \frac{{20}}{{64}}\;\;\;$$(seen anywhere) Â  Â  A1

correct substitution into infinite sum of GP Â  Â  A1

eg$$\;\;\;\frac{{\frac{3}{8}}}{{1 – \frac{5}{{16}}}},{\text{ }}\frac{3}{8}\left( {\frac{1}{{1 – \left( {\frac{5}{8} \times \frac{4}{8}} \right)}}} \right),{\text{ }}\frac{1}{{1 – \frac{5}{{16}}}}$$

correct working Â  Â  (A1)

eg$$\;\;\;\frac{{\frac{3}{8}}}{{\frac{{11}}{{16}}}},{\text{ }}\frac{3}{8} \times \frac{{16}}{{11}}$$

$${\text{P (Ann wins)}} = \frac{{48}}{{88}}\;\;\;\left( { = \frac{6}{{11}}} \right)$$ Â  Â  A1 Â  Â  N1

[7 marks]

Total [15 marks]

c.
c.

## Question

An arithmetic sequence has the first term $$\ln a$$ and a common difference $$\ln 3$$.

The 13th term in the sequence is $$8\ln 9$$. Find the value of $$a$$.

## Markscheme

Note: Â  Â  There are many approaches to this question, and the steps may be done in any order. There are 3 relationships they may need to apply at some stage, for the 3rd, 4th and 5th marks. These are

equating bases eg recognising 9 is $${{\text{3}}^2}$$

log rules: $$\ln b + \ln c = \ln (bc),{\text{ }}\ln b – \ln c = \ln \left( {\frac{b}{c}} \right)$$,

exponent rule: $$\ln {b^n} = n\ln b$$.

The exception to the FT rule applies here, so that if they demonstrate correct application of the 3 relationships, they may be awarded the A marks, even if they have made a previous error. However all applications of a relationship need to be correct. Once an error has been made, do not award A1FT for their final answer, even if it follows from their working.

Please check working and award marks in line with the markscheme.

correct substitution into $${u_{13}}$$ formula Â  Â  (A1)

eg$$\;\;\;\ln a + (13 – 1)\ln 3$$

set up equation for $${u_{13}}$$ in any form (seen anywhere) Â  Â  (M1)

eg$$\;\;\;\ln a + 12\ln 3 = 8\ln 9$$

correct application of relationships Â  Â  (A1)(A1)(A1)

$$a = 81$$ Â  Â  A1 Â  Â  N3

[6 marks]

Examples of application of relationships

Example 1

correct application of exponent rule for logs Â  Â  (A1)

eg$$\;\;\;\ln a + \ln {3^{12}} = \ln {9^8}$$

correct application of addition rule for logs Â  Â  (A1)

eg$$\;\;\;\ln (a{3^{12}}) = \ln {9^8}$$

substituting for 9 or 3 in ln expression in equation Â  Â  (A1)

eg$$\;\;\;\ln (a{3^{12}}) = \ln {3^{16}},{\text{ }}\ln (a{9^6}) = \ln {9^8}$$

Example 2

recognising $$9 = {3^2}$$Â  Â  Â  (A1)

eg$$\;\;\;\ln a + 12\ln 3 = 8\ln {3^2},{\text{ }}\ln a + 12\ln {9^{\frac{1}{2}}} = 8\ln 9$$

one correct application of exponent rule for logs relating $$\ln 9$$ to $$\ln 3$$ Â  Â  (A1)

eg$$\;\;\;\ln a + 12\ln 3 = 16\ln 3,{\text{ }}\ln a + 6\ln 9 = 8\ln 9$$

another correct application of exponent rule for logs Â  Â  (A1)

eg$$\;\;\;\ln a = \ln {3^4},{\text{ }}\ln a = \ln {9^2}$$

## Question

Consider the following sequence of figures.

Figure 1 contains 5 line segments.

Given that Figure $$n$$ contains 801 line segments, show that $$n = 200$$.

[3]
a.

Find the total number of line segments in the first 200 figures.

[3]
b.

## Markscheme

recognizing that it is an arithmetic sequence Â  Â  (M1)

eg$$\,\,\,\,\,$$$$5,{\text{ }}5 + 4,{\text{ }}5 + 4 + 4,{\text{ }} \ldots ,{\text{ }}d = 4,{\text{ }}{u_n} = {u_1} + (n – 1)d,{\text{ }}4n + 1$$

correct equation Â  Â  A1

eg$$\,\,\,\,\,$$$$5 + 4(n – 1) = 801$$

correct working (do not accept substituting $$n = 200$$) Â  Â  A1

eg$$\,\,\,\,\,$$$$4n – 4 = 796,{\text{ }}n – 1 = \frac{{796}}{4}$$

$$n = 200$$ Â  Â AG Â  Â  N0

[3 marks]

a.

recognition of sum Â  Â  (M1)

eg$$\,\,\,\,\,$$$${S_{200}},{\text{ }}{u_1} + {u_2} + Â \ldots Â + {u_{200}},{\text{ }}5 + 9 + 13 + Â \ldots Â + 801$$

correct working for AP Â  Â  (A1)

eg$$\,\,\,\,\,$$$$\frac{{200}}{2}(5 + 801),{\text{ }}\frac{{200}}{2}{\text{ }}\left( {2(5) + 199(4)} \right)$$

$$80\,600$$Â Â  Â  A1 Â  Â  N2

[3 marks]

b.

## Question

Three consecutive terms of a geometric sequence are $$x – 3$$, 6 and $$x + 2$$.

Find the possible values of $$x$$.

## Markscheme

METHOD 1

valid approach Â  Â  (M1)

eg$$\,\,\,\,\,$$$$r = \frac{6}{{x – 3}},{\text{ }}(x – 3) \times r = 6,{\text{ }}(x – 3){r^2} = x + 2$$

correct equation in terms of $$x$$ only Â  Â  A1

eg$$\,\,\,\,\,$$$$\frac{6}{{x – 3}} = \frac{{x + 2}}{6},{\text{ }}(x – 3)(x + 2) = {6^2},{\text{ }}36 = {x^2} – x – 6$$

correct working Â  Â  (A1)

eg$$\,\,\,\,\,$$$${x^2} – x – 42,{\text{ }}{x^2} – x = 42$$

valid attempt to solve their quadratic equation Â  Â  (M1)

eg$$\,\,\,\,\,$$factorizing, formula, completing the square

evidence of correct working Â  Â  (A1)

eg$$\,\,\,\,\,$$$$(x – 7)(x + 6),{\text{ }}\frac{{1 \pm \sqrt {169} }}{2}$$

$$x = 7,{\text{ }}x = Â – 6$$ Â  Â Â A1 Â  Â  N4

METHOD 2 (finding r first)

valid approach Â  Â  (M1)

eg$$\,\,\,\,\,$$$$r = \frac{6}{{x – 3}},{\text{ }}6r = x + 2,{\text{ }}(x – 3){r^2} = x + 2$$

correct equation in terms of $$r$$ only Â  Â  A1

eg$$\,\,\,\,\,$$$$\frac{6}{r} + 3 = 6r – 2,{\text{ }}6 + 3r = 6{r^2} – 2r,{\text{ }}6{r^2} – 5r – 6 = 0$$

evidence of correct working Â  Â  (A1)

eg$$\,\,\,\,\,$$$$(3r + 2)(2r – 3),{\text{ }}\frac{{5 \pm \sqrt {25 + 144} }}{{12}}$$

$$r = Â – \frac{2}{3},{\text{ }}r = \frac{3}{2}$$ Â  Â A1

substituting their values of $$r$$ to find $$x$$ Â  Â  (M1)

eg$$\,\,\,\,\,$$$$(x – 3)\left( {\frac{2}{3}} \right) = 6,{\text{ }}x = 6\left( {\frac{3}{2}} \right) – 2$$

$$x = 7,{\text{ }}x = Â – 6$$ Â  Â A1 Â  Â  N4

[6 marks]

## Question

The first two terms of an infinite geometric sequence, in order, are

$$2{\log _2}x,{\text{ }}{\log _2}x$$, where $$x > 0$$.

The first three terms of an arithmetic sequence, in order, are

$${\log _2}x,{\text{ }}{\log _2}\left( {\frac{x}{2}} \right),{\text{ }}{\log _2}\left( {\frac{x}{4}} \right)$$, where $$x > 0$$.

Let $${S_{12}}$$ be the sum of the first 12 terms of the arithmetic sequence.

Find $$r$$.

[2]
a.

Show that the sum of the infinite sequence is $$4{\log _2}x$$.

[2]
b.

Find $$d$$, giving your answer as an integer.

[4]
c.

Show that $${S_{12}} = 12{\log _2}x – 66$$.

[2]
d.

Given that $${S_{12}}$$Â is equal to half the sum of the infinite geometric sequence, find $$x$$, giving your answer in the form $${2^p}$$, where $$p \in \mathbb{Q}$$.

[3]
e.

## Markscheme

evidence of dividing terms (in any order) Â  Â  (M1)

eg$$\,\,\,\,\,$$$$\frac{{{\mu _2}}}{{{\mu _1}}},{\text{ }}\frac{{2{{\log }_2}x}}{{{{\log }_2}x}}$$

$$r = \frac{1}{2}$$ Â  Â A1 Â  Â  N2

[2 marks]

a.

correct substitution Â  Â  (A1)

eg$$\,\,\,\,\,$$$$\frac{{2{{\log }_2}x}}{{1 – \frac{1}{2}}}$$

correct working Â  Â  A1

eg$$\,\,\,\,\,$$$$\frac{{2{{\log }_2}x}}{{\frac{1}{2}}}$$

$${S_\infty } = 4{\log _2}x$$ Â  Â Â AG Â  Â  N0

[2 marks]

b.

evidence of subtracting two terms (in any order) Â  Â  (M1)

eg$$\,\,\,\,\,$$$${u_3} – {u_2},{\text{ }}{\log _2}x – {\log _2}\frac{x}{2}$$

correct application of the properties of logs Â  Â  (A1)

eg$$\,\,\,\,\,$$$${\log _2}\left( {\frac{{\frac{x}{2}}}{x}} \right),{\text{ }}{\log _2}\left( {\frac{x}{2} \times \frac{1}{x}} \right),{\text{ }}({\log _2}x – {\log _2}2) – {\log _2}x$$

correct working Â  Â  (A1)

eg$$\,\,\,\,\,$$$${\log _2}\frac{1}{2},{\text{ }} – {\log _2}2$$

$$d = Â – 1$$ Â  Â A1 Â  Â  N3

[4 marks]

c.

correct substitution into the formula for the sum of an arithmetic sequence Â  Â  (A1)

eg$$\,\,\,\,\,$$$$\frac{{12}}{2}\left( {2{{\log }_2}x + (12 – 1)( – 1)} \right)$$

correct working Â  Â  A1

eg$$\,\,\,\,\,$$$$6(2{\log _2}x – 11),{\text{ }}\frac{{12}}{2}(2{\log _2}x – 11)$$

$$12{\log _2}x – 66$$ Â  Â AG Â  Â  N0

[2 marks]

d.

correct equation Â  Â  (A1)

eg$$\,\,\,\,\,$$$$12{\log _2}x – 66 = 2{\log _2}x$$

correct working Â  Â  (A1)

eg$$\,\,\,\,\,$$$$10{\log _2}x = 66,{\text{ }}{\log _2}x = 6.6,{\text{ }}{2^{66}} = {x^{10}},{\text{ }}{\log _2}\left( {\frac{{{x^{12}}}}{{{x^2}}}} \right) = 66$$

$$x = {2^{6.6}}$$Â (accept $$p = \frac{{66}}{{10}}$$) Â  Â  A1 Â  Â  N2

[3 marks]

e.
e.

## Question

The first three terms of a geometric sequence are $$\ln {x^{16}}$$, $$\ln {x^8}$$, $$\ln {x^4}$$, for $$x > 0$$.

Find the common ratio.

[3]
a.

Solve $$\sum\limits_{k = 1}^\infty Â {{2^{5 – k}}\ln x = 64}$$.

[5]
b.

## Markscheme

correct use $$\log {x^n} = n\log x$$ Â  Â  A1

eg$$\,\,\,\,\,$$$$16\ln x$$

valid approach to find $$r$$ Â  Â  (M1)

eg$$\,\,\,\,\,$$$$\frac{{{u_{n + 1}}}}{{{u_n}}},{\text{ }}\frac{{\ln {x^8}}}{{\ln {x^{16}}}},{\text{ }}\frac{{4\ln x}}{{8\ln x}},{\text{ }}\ln {x^4} = \ln {x^{16}} \times {r^2}$$

$$r = \frac{1}{2}$$ Â  Â  A1 Â  Â  N2

[3 marks]

a.

recognizing a sum (finite or infinite) Â  Â  (M1)

eg$$\,\,\,\,\,$$$${2^4}\ln x + {2^3}\ln x,{\text{ }}\frac{a}{{1 – r}},{\text{ }}{S_\infty },{\text{ }}16\ln x +Â \ldots$$

valid approach (seen anywhere) Â  Â  (M1)

eg$$\,\,\,\,\,$$recognizing GP is the same as part (a), using their $$r$$ value from part (a), $$r = \frac{1}{2}$$

correct substitution into infinite sum (only if $$\left| r \right|$$ is a constant and less than 1) Â  Â  A1

eg$$\,\,\,\,\,$$$$\frac{{{2^4}\ln x}}{{1 – \frac{1}{2}}},{\text{ }}\frac{{\ln {x^{16}}}}{{\frac{1}{2}}},{\text{ }}32\ln x$$

correct working Â  Â  (A1)

eg$$\,\,\,\,\,$$$$\ln x = 2$$

$$x = {{\text{e}}^2}$$ Â  Â  A1 Â  Â  N3

[5 marks]

b.

## Question

In an arithmetic sequence, the first term is 3 and the second term is 7.

Find the common difference.

[2]
a.

Find the tenth term.

[2]
b.

Find the sum of the first ten terms of the sequence.

[2]
c.

## Markscheme

attempt to subtract terms Â  Â  (M1)

eg$$\,\,\,\,\,$$$$d = {u_2} – {u_1},{\text{ }}7 – 3$$

$$d = 4$$ Â  Â  A1 Â  Â  N2

[2 marks]

a.

correct approach Â  Â  (A1)

eg$$\,\,\,\,\,$$$${u_{10}} = 3 + 9(4)$$

$${u_{10}} = 39$$ Â  Â  A1 Â  Â  N2

[2 marks]

b.

correct substitution into sum Â  Â  (A1)

eg$$\,\,\,\,\,$$$${S_{10}} = 5(3 + 39),{\text{ }}{S_{10}} = \frac{{10}}{2}(2 \times 3 + 9 \times 4)$$

$${S_{10}} = 210$$ Â  Â  A1 Â  Â  N2

[2 marks]

c

## Question

In an arithmetic sequence, the first term is 8 and the second term is 5.

Find the common difference.

[2]
a.

Find the tenth term.

[2]
b.

Find the sum of the first ten terms.

[2]
c.

## Markscheme

subtracting terms Â  Â  (M1)

eg$$\,\,\,\,\,$$$$5 – 8,{\text{ }}{u_2} – {u_1}$$

$$d =Â – 3$$ Â  Â  A1 Â  Â  N2

[2 marks]

a.

correct substitution into formula Â  Â  (A1)

eg$$\,\,\,\,\,$$$${u_{10}} = 8 + (10 – 1)( – 3),{\text{ }}8 – 27,{\text{ }} – 3(10) + 11$$

$${u_{10}} =Â – 19$$ Â  Â  A1 Â  Â  N2

[2 marks]

b.

correct substitution into formula for sum Â  Â  (A1)

eg$$\,\,\,\,\,$$$${S_{10}} = \frac{{10}}{2}(8 – 19),{\text{ 5}}\left( {2(8) + (10 – 1)( – 3)} \right)$$

$${S_{10}} =Â – 55$$ Â  Â  A1 Â  Â  N2

[2 marks]

c.

## Question

The following diagram shows [AB], with length 2 cm. The line is divided into an infinite number of line segments. The diagram shows the first three segments.

The length of the line segments are $$p{\text{ cm}},{\text{ }}{p^2}{\text{ cm}},{\text{ }}{p^3}{\text{ cm}},{\text{ }} \ldots$$, where $$0 < p < 1$$.

Show that $$p = \frac{2}{3}$$.

[5]
a.

The following diagram shows [CD], with length $$b{\text{ cm}}$$, where $$b > 1$$. Squares with side lengths $$k{\text{ cm}},{\text{ }}{k^2}{\text{ cm}},{\text{ }}{k^3}{\text{ cm}},{\text{ }} \ldots$$, where $$0 < k < 1$$, are drawn along [CD]. This process is carried on indefinitely. The diagram shows the first three squares.

The total sum of the areas of all the squares is $$\frac{9}{{16}}$$. Find the value of $$b$$.

[9]
b.

## Markscheme

infinite sum of segments is 2 (seen anywhere) Â  Â  (A1)

eg$$\,\,\,\,\,$$$$p + {p^2} + {p^3} +Â \ldotsÂ = 2,{\text{ }}\frac{{{u_1}}}{{1 – r}} = 2$$

recognizing GP Â  Â  (M1)

eg$$\,\,\,\,\,$$ratio is $$p,{\text{ }}\frac{{{u_1}}}{{1 – r}},{\text{ }}{u_n} = {u_1} \times {r^{n – 1}},{\text{ }}\frac{{{u_1}({r^n} – 1)}}{{r – 1}}$$

correct substitution into $${S_\infty }$$ formula (may be seen in equation) Â  Â  A1

eg$$\,\,\,\,\,$$$$\frac{p}{{1 – p}}$$

correct equation Â  Â  (A1)

eg$$\,\,\,\,\,$$$$\frac{p}{{1 – p}} = 2,{\text{ }}p = 2 – 2p$$

eg$$\,\,\,\,\,$$$$3p = 2,{\text{ }}2 – 3p = 0$$

$$p = \frac{2}{3}{\text{ (cm)}}$$ Â  Â  AG Â  Â  N0

[5 marks]

a.

recognizing infinite geometric series with squares Â  Â  (M1)

eg$$\,\,\,\,\,$$$${k^2} + {k^4} + {k^6} +Â \ldots ,{\text{ }}\frac{{{k^2}}}{{1 – {k^2}}}$$

correct substitution into $${S_\infty } = \frac{9}{{16}}$$ (must substitute into formula) Â  Â  (A2)

eg$$\,\,\,\,\,$$$$\frac{{{k^2}}}{{1 – {k^2}}} = \frac{9}{{16}}$$

correct working Â  Â  (A1)

eg$$\,\,\,\,\,$$$$16{k^2} = 9 – 9{k^2},{\text{ }}25{k^2} = 9,{\text{ }}{k^2} = \frac{9}{{25}}$$

$$k = \frac{3}{5}$$ (seen anywhere) Â  Â  A1

valid approach with segments and CD (may be seen earlier) Â  Â  (M1)

eg$$\,\,\,\,\,$$$$r = k,{\text{ }}{S_\infty } = b$$

correct expression for $$b$$ in terms of $$k$$ (may be seen earlier) Â  Â  (A1)

eg$$\,\,\,\,\,$$$$b = \frac{k}{{1 – k}},{\text{ }}b = \sum\limits_{n = 1}^\inftyÂ {{k^n},{\text{ }}b = k + {k^2} + {k^3} +Â \ldots }$$

substituting their value of $$k$$ into their formula for $$b$$ Â  Â  (M1)

eg$$\,\,\,\,\,$$$$\frac{{\frac{3}{5}}}{{1 – \frac{3}{5}}},{\text{ }}\frac{{\left( {\frac{3}{5}} \right)}}{{\left( {\frac{2}{5}} \right)}}$$

$$b = \frac{3}{2}$$ Â  Â  A1 Â  Â  N3

[9 marks]

b.

## Question

The first two terms of an infinite geometric sequence are u1 = 18 and u2Â = 12sin2 Î¸ ,Â where 0 < Î¸ < 2$$\pi$$ , and Î¸ â‰  $$\pi$$.

Find an expression for r in terms of Î¸.

[2]
a.i.

Find the possible values of r.

[3]
a.ii.

Show that the sum of the infinite sequence isÂ $$\frac{{54}}{{2 + {\text{cos}}\,\left( {2\theta } \right)}}$$.

[4]
b.

Find the values of Î¸ which give the greatest value of the sum.

[6]
c.

## Markscheme

valid approachÂ  Â  Â (M1)

egÂ  Â $$\frac{{{u_2}}}{{{u_1}}},\,\,\frac{{{u_1}}}{{{u_2}}}$$

$$r = \frac{{12\,{{\sin }^2}\,\theta }}{{18}}\left( { = \frac{{2\,{{\sin }^2}\,\theta }}{3}} \right)$$Â  Â  Â Â A1 N2

[2 marks]

a.i.

recognizing that sinÎ¸ is boundedÂ  Â  Â  (M1)

egÂ  Â  0 â‰¤ sin2Â Î¸Â â‰¤ 1,Â âˆ’1Â â‰¤Â sinÎ¸Â â‰¤ 1,Â âˆ’1Â <Â sinÎ¸Â <Â 1

0 < rÂ â‰¤Â $$\frac{2}{3}$$Â  Â  Â Â A2 N3

Note: If working shown, award M1A1 for correct values with incorrect inequality sign(s).
If no working shown, award N1 for correct values with incorrect inequality sign(s).

[3 marks]

a.ii.

correct substitution into formula for infinite sumÂ  Â  Â  Â A1

egÂ  $$\frac{{18}}{{1 – \frac{{2\,{\text{si}}{{\text{n}}^2}\,\theta }}{3}}}$$

evidence of choosing an appropriate rule for cos 2Î¸ (seen anywhere)Â  Â  Â  Â  Â (M1)

egÂ  Â cos 2Î¸ = 1Â âˆ’ 2 sin2Â Î¸

correct substitution of identity/working (seen anywhere)Â  Â  Â  (A1)

egÂ  Â $$\frac{{18}}{{1 – \frac{2}{3}\left( {\frac{{1 – {\text{cos}}\,2\theta }}{2}} \right)}},\,\,\frac{{54}}{{3 – 2\left( {\frac{{1 – {\text{cos}}\,2\theta }}{2}} \right)}},\,\,\frac{{18}}{{\frac{{3 – 2\,{\text{si}}{{\text{n}}^2}\,\theta }}{3}}}$$

correct working that clearly leads to the given answerÂ  Â  Â  Â A1

egÂ Â $$\frac{{18 \times 3}}{{2 + \left( {1 – 2\,{\text{si}}{{\text{n}}^2}\,\theta } \right)}},\,\,\frac{{54}}{{3 – \left( {1 – {\text{cos}}\,2\theta } \right)}}$$

$$\frac{{54}}{{2 + {\text{cos}}\left( {2\theta } \right)}}$$Â Â  Â AG N0

[4 marks]

b.

METHOD 1 (using differentiation)

recognizingÂ $$\frac{{{\text{d}}{S_\infty }}}{{{\text{d}}\theta }} = 0$$ (seen anywhere)Â  Â  Â  Â (M1)

finding any correct expression forÂ $$\frac{{{\text{d}}{S_\infty }}}{{{\text{d}}\theta }}$$Â  Â  Â  Â (A1)

egÂ Â $$\frac{{0 – 54 \times \left( { – 2\,{\text{sin}}\,2\,\theta } \right)}}{{{{\left( {2 + {\text{cos}}\,2\,\theta } \right)}^2}}},\,\, – 54{\left( {2 + {\text{cos}}\,2\,\theta } \right)^{ – 2}}\,\left( { – 2\,{\text{sin}}\,2\,\theta } \right)$$

correct workingÂ  Â  Â  Â (A1)

egÂ  sin 2Î¸ = 0

any correct value for sinâˆ’1(0) (seen anywhere)Â  Â  Â  Â (A1)

egÂ  0,Â $$\pi$$, â€¦ ,Â sketch of sine curve with x-intercept(s) marked both correct values for 2Î¸ (ignore additional values)Â  Â  Â  (A1)

2Î¸Â =Â $$\pi$$, 3$$\pi$$Â (accept values in degrees)

both correct answersÂ $$\thetaÂ = \frac{\pi }{2},\,\frac{{3\pi }}{2}$$Â  Â  Â Â A1 N4

Note: Award A0 if either or both correct answers are given in degrees.
Award A0Â if additional values are given.

METHOD 2 (using denominator)

recognizing when Sâˆž is greatestÂ  Â  Â  (M1)

eg 2 + cosÂ 2Î¸Â is a minimum, 1âˆ’r is smallest
correct workingÂ  Â  Â  (A1)

egÂ Â minimum value ofÂ 2 + cosÂ 2Î¸Â is 1, minimum r =Â $$\frac{2}{3}$$

correct workingÂ  Â  Â  (A1)

egÂ  $${\text{cos}}\,2\,\thetaÂ =Â – 1,\,\,\frac{2}{3}\,{\text{si}}{{\text{n}}^2}\,\thetaÂ = \frac{2}{3},\,\,{\text{si}}{{\text{n}}^2}\thetaÂ = 1$$

EITHER (using cos 2Î¸)

any correct value for cosâˆ’1(âˆ’1) (seen anywhere)Â  Â  Â  (A1)

egÂ Â $$\pi$$, 3$$\pi$$, â€¦ (accept values in degrees), sketch of cosine curve with x-intercept(s) marked

both correct values forÂ 2Î¸Â Â (ignore additional values)Â  Â  Â  (A1)

2Î¸Â =Â $$\pi$$, 3$$\pi$$Â (accept values in degrees)

OR (using sinÎ¸)

sinÎ¸ =Â Â±1Â  Â  Â (A1)

sinâˆ’1(1) =Â $$\frac{\pi }{2}$$Â (accept values in degrees) (seen anywhere)Â  Â  Â  A1

THEN

both correct answersÂ $$\thetaÂ = \frac{\pi }{2},\,\frac{{3\pi }}{2}$$Â Â  Â  Â Â A1 N4

Note:Â AwardÂ A0Â if either or both correct answers are given in degrees.
AwardÂ A0Â if additional values are given.

[6 marks]

c.

## Question

An arithmetic sequence hasÂ $${u_1} = {\text{lo}}{{\text{g}}_c}\left( p \right)$$ and $${u_2} = {\text{lo}}{{\text{g}}_c}\left( {pq} \right)$$, whereÂ $$c > 1$$ and $$p,\,\,q > 0$$.

Show thatÂ $$d = {\text{lo}}{{\text{g}}_c}\left( q \right)$$.

[2]
a.

LetÂ $$p = {c^2}$$ andÂ $$q = {c^3}$$. Find the value of $$\sum\limits_{n = 1}^{20} {{u_n}}$$.

[6]
b.

## Markscheme

valid approach involving addition or subtractionÂ  Â  Â  Â M1
egÂ Â $${u_2} = {\text{lo}}{{\text{g}}_c}\,p + d,\,\,{u_1} – {u_2}$$

correct application of log lawÂ  Â  Â  A1
egÂ Â $${\text{lo}}{{\text{g}}_c}\left( {pq} \right) = {\text{lo}}{{\text{g}}_c}\,p + {\text{lo}}{{\text{g}}_c}\,q,\,\,{\text{lo}}{{\text{g}}_c}\left( {\frac{{pq}}{p}} \right)$$

$$d = {\text{lo}}{{\text{g}}_c}\,q$$Â Â  Â AG N0

[2 marks]

a.

METHOD 1 (findingÂ $${u_1}$$ and d)

recognizingÂ $$\sum { = {S_{20}}}$$Â (seen anywhere)Â  Â  Â  (A1)

attempt to findÂ $${u_1}$$ or d usingÂ $${\text{lo}}{{\text{g}}_c}\,{c^k} = k$$Â  Â  Â (M1)
egÂ Â $${\text{lo}}{{\text{g}}_c}\,c$$,Â $${\text{3}}\,{\text{lo}}{{\text{g}}_c}\,c$$, correct value ofÂ $${u_1}$$ orÂ d

$${u_1}$$ = 2,Â d = 3Â (seen anywhere)Â  Â  Â  (A1)(A1)

correct workingÂ  Â  Â (A1)
egÂ Â $${S_{20}} = \frac{{20}}{2}\left( {2 \times 2 + 19 \times 3} \right),\,\,{S_{20}} = \frac{{20}}{2}\left( {2 + 59} \right),\,\,10\left( {61} \right)$$

$$\sum\limits_{n = 1}^{20} {{u_n}}$$ = 610Â  Â  Â A1 N2

METHOD 2 (expressing S in terms of c)

recognizingÂ $$\sum { = {S_{20}}}$$Â (seen anywhere)Â  Â  Â Â (A1)

correct expression for S in terms of cÂ  Â  Â  (A1)
egÂ Â $$10\left( {2\,{\text{lo}}{{\text{g}}_c}\,{c^2} + 19\,{\text{lo}}{{\text{g}}_c}\,{c^3}} \right)$$

$${\text{lo}}{{\text{g}}_c}\,{c^2} = 2,\,\,\,{\text{lo}}{{\text{g}}_c}\,{c^3} = 3$$Â Â (seen anywhere)Â  Â  Â (A1)(A1)

correct workingÂ  Â  Â  (A1)

egÂ Â $${S_{20}} = \frac{{20}}{2}\left( {2 \times 2 + 19 \times 3} \right),\,\,{S_{20}} = \frac{{20}}{2}\left( {2 + 59} \right),\,\,10\left( {61} \right)$$

$$\sum\limits_{n = 1}^{20} {{u_n}}$$ = 610Â  Â  Â A1 N2

METHOD 3Â (expressingÂ SÂ in terms ofÂ c)

recognizingÂ $$\sum { = {S_{20}}}$$Â (seen anywhere)Â  Â  Â Â (A1)

correct expression forÂ SÂ in terms ofÂ cÂ  Â  Â Â (A1)
egÂ Â $$10\left( {2\,{\text{lo}}{{\text{g}}_c}\,{c^2} + 19\,{\text{lo}}{{\text{g}}_c}\,{c^3}} \right)$$

correct application of log lawÂ  Â  Â (A1)
egÂ Â $$2\,{\text{lo}}{{\text{g}}_c}\,{c^2} = \,\,{\text{lo}}{{\text{g}}_c}\,{c^4},\,\,19\,{\text{lo}}{{\text{g}}_c}\,{c^3} = \,\,{\text{lo}}{{\text{g}}_c}\,{c^{57}},\,\,10\,\left( {{\text{lo}}{{\text{g}}_c}\,{{\left( {{c^2}} \right)}^2} + \,\,{\text{lo}}{{\text{g}}_c}\,{{\left( {{c^3}} \right)}^{19}}} \right),\,\,10\,\left( {{\text{lo}}{{\text{g}}_c}\,{c^4} + \,{\text{lo}}{{\text{g}}_c}\,{c^{57}}} \right),\,\,10\left( {{\text{lo}}{{\text{g}}_c}\,{c^{61}}} \right)$$

correct application of definition of logÂ  Â  Â  (A1)
egÂ Â $${\text{lo}}{{\text{g}}_c}\,{c^{61}} = 61,\,\,{\text{lo}}{{\text{g}}_c}\,{c^4} = 4,\,\,{\text{lo}}{{\text{g}}_c}\,{c^{57}} = 57$$

correct workingÂ  Â  Â (A1)
egÂ Â $${S_{20}} = \frac{{20}}{2}\left( {4 + 57} \right),\,\,10\left( {61} \right)$$

$$\sum\limits_{n = 1}^{20} {{u_n}}$$ = 610Â  Â  Â A1 N2

[6 marks]

b.