L.C. MATHS

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Past Papers

## Sequences & Series

NOTE: Clicking an entry in the right column will take you directly to that question!

(BLUE = Paper 1, RED = Paper 2)

Since 2015, the following subtopic... Has explicitly appeared in...

Arithmetic Sequences

Arithmetic Series

Geometric Sequences

Geometric Series

Other Sequences

## 2022 Paper 1 Question 4

(a) A sequence $$u_1,u_2,u_3…$$ is defined as follows, for $$n\in\mathbb{N}$$:

\begin{align}u_1=2, && u_2=64, && u_{n+1}=\sqrt{\frac{u_n}{u_{n-1}}} \end{align}

Write $$u_3$$ in the form $$2^p$$, where $$p\in\mathbb{R}$$.

$$2^{5/2}$$

Solution

\begin{align}u_3&=\sqrt{\frac{u_2}{u_1}}\\&=\sqrt{\frac{64}{2}}\\&=\sqrt{32}\\&=\sqrt{2^5}\\&=2^{5/2}\end{align}

Video Walkthrough
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(b) The first three terms in an arithmetic sequence are as follows, where $$k\in\mathbb{R}$$:

\begin{align}5e^{-k},&&13,&&5e^k\end{align}

(i) By letting $$y=e^k$$ in this arithmetic sequence, show that:

\begin{align}5y^2-26y+5=0\end{align}

(ii) Use the equation in $$y$$ in part (b)(i) to find the two possible values of $$k$$.
Give each value in the form $$\ln p$$ or $$-\ln p$$, where $$p\in\mathbb{N}$$.

(ii) $$k=-\ln5$$ or $$k=\ln5$$

Solution

(i)

\begin{align}13-5e^{-k}=5e^k-13\end{align}

\begin{align}\downarrow\end{align}

\begin{align}13-\frac{5}{y}=5y-13\end{align}

\begin{align}\downarrow\end{align}

\begin{align}13y-5=5y^2-13y\end{align}

\begin{align}\downarrow\end{align}

\begin{align}5y^2-26y+5=0\end{align}

as required.

(ii)

\begin{align}5y^2-26y+5=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(5y-1)(y-5)=0\end{align}

\begin{align}\downarrow\end{align}

$$y=\dfrac{1}{5}$$ or $$y=5$$

\begin{align}\downarrow\end{align}

$$e^k=\dfrac{1}{5}$$ or $$e^k=5$$

\begin{align}\downarrow\end{align}

$$k=\ln\left(\dfrac{1}{5}\right)$$ or $$k=\ln5$$

\begin{align}\downarrow\end{align}

$$k=-\ln5$$ or $$k=\ln5$$

Video Walkthrough

## 2022 Paper 1 Question 9(c)-(f)

Alex gets injections of a medicinal drug. Each injection has $$15\mbox{ mg}$$ of the drug.
Each day, the amount of the drug left in Alex’s body from an injection decreases by $$40\%$$.
So, the amount of the drug (in mg) left in Alex’s body $$t$$ days after a single injection is given by:

\begin{align}15(0.6)^t\end{align}

where $$t\in\mathbb{R}$$.

Alex is given a $$15\mbox{ mg}$$ injection of the drug at the same time every day for a long period of time.

(c) Explain why the total amount of the drug, in mg, in Alex’s body immediately after the $$4$$th injection is given by:

\begin{align}15+15(0.6)+15(0.6)^2+15(0.6)^3\end{align}

Solution

On the fourth day:

• $$15(0.6^3)$$ is the amount left from the first injection.
• $$15(0.6^2)$$ is the amount left from the second injection.
• $$15(0.6)$$ is the amount left from the third injection.
• $$15$$ is the amount in the fourth injection that was just given.

The total amount on the fourth day is then the sum of the above.

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(d) Find the total amount of the drug in Alex’s body immediately after the $$10$$th injection.
Give your answer in $$\mbox{mg}$$, correct to $$2$$ decimal places.

$$37.27\mbox{ mg}$$

Solution

\begin{align}S_n&=\frac{a(1-r^n)}{1-r}\\&=\frac{15(1-0.6^{10})}{1-0.6}\\&\approx37.27\mbox{ mg}\end{align}

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(e) Use the formula for the sum to infinity of a geometric series to estimate the amount of the drug (in $$\mbox{mg}$$) in Alex’s body, after a long period of time during which he gets daily injections.

$$37.5\mbox{ mg}$$

Solution

\begin{align}S_\infty&=\frac{a}{1-r}\\&=\frac{15}{1-0.6}\\&=37.5\mbox{ mg}\end{align}

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(f) Jessica also gets daily injections of a medicinal drug at the same time every day.
She gets $$d\mbox{ mg}$$ of the drug in each injection, where $$d\in\mathbb{R}$$.
Each day, the amount of the drug left in Jessica’s body from an injection decreases by $$15\%$$.

(i) Use the sum of a geometric series to show that the total amount of the drug (in $$\mbox{mg}$$) in Jessica’s body immediately after the $$n$$th injection, where $$n\in\mathbb{N}$$, is:

\begin{align}\frac{20d(1-0.85^n)}{3}\end{align}

(ii) Immediately after the $$7$$th injection, there are $$50\mbox{ mg}$$ of the drug in Jessica’s body.

Find the amount of the drug in one of Jessica’s daily injections.
Give your answer correct to the nearest $$\mbox{mg}$$.

(ii) $$11\mbox{ mg}$$

Solution

(i)

\begin{align}S_n&=\frac{a(1-r^n)}{1-r}\\&=\frac{d(1-0.85^n)}{1-0.85}\\&\frac{20d(1-0.85^n)}{3}\end{align}

as required

(ii)

\begin{align}\frac{20d(1-0.85^7)}{3}=50\end{align}

\begin{align}\downarrow\end{align}

\begin{align}d&=\frac{(3)(50)}{20(1-0.85^7)}\\&\approx11\mbox{ mg}\end{align}

Video Walkthrough

## 2021 Paper 1 Question 4(b)

(b) $$p,p+7,p+14,p+21,…$$ is an arithmetic sequence, where $$p\in\mathbb{N}$$.

(i) Find the $$n^{\mbox{th}}$$ term, $$T_n$$, in terms of $$n$$ and $$p$$, where $$n\in\mathbb{N}$$.

(ii) Find the smallest value of $$p$$ for which $$2021$$ is a term in the sequence.

(i) $$T_n=p+7n-7$$

(ii) $$p=5$$

Solution

(i)

\begin{align}T_n=a+(n-1)d\end{align}

\begin{align}a=p && d=7\end{align}

\begin{align}\downarrow\end{align}

\begin{align}T_n&=p+(n-1)(7)\\&=p+7n-7\end{align}

(ii)

\begin{align}p+7n-7=2021\end{align}

\begin{align}\downarrow\end{align}

\begin{align}p=2028-7n\end{align}

Since $$p\in\mathbb{N}$$, the smallest value of $$p$$ occurs for the largest value of $$n\in\mathbb{N}$$ that satisfies $$2028-7n>0$$, i.e. $$n=289$$, and therefore

\begin{align}p&=2028-7(289)\\&=5\end{align}

Video Walkthrough

## 2021 Paper 1 Question 7

The tip of the pendulum of a grandfather clock swings initially through an arc length of $$45\mbox{ cm}$$.
On each successive swing the length of the arc is $$90\%$$ of the previous length.

(a)

(i) Complete the table below by filling in the missing lengths.

Swing $$1$$ $$2$$ $$3$$ $$4$$ $$5$$

Length of Arc (cm)

$$45$$

$$\dfrac{729}{20}$$

(ii) $$T_n=45(0.9)^{n-1}$$ is the arc length of swing $$n$$.
Find the arc length of swing $$25$$, correct to $$1$$ decimal place.

(iii) Find the total distance travelled by the tip of the pendulum when it has completed swing $$40$$. Give your answer, in $$\mbox{cm}$$, correct to the nearest whole number.

(iv) Swing $$p$$ is the first swing which has an arc length of less than $$2\mbox{ cm}$$. Find the value of $$p$$.

(i)

Swing $$1$$ $$2$$ $$3$$ $$4$$ $$5$$

Length of Arc (cm)

$$45$$

$$\dfrac{81}{2}$$

$$\dfrac{729}{20}$$

$$\dfrac{6{,}561}{200}$$

$$\dfrac{59{,}049}{2{,}000}$$

(ii) $$3.6\mbox{ cm}$$

(iii) $$443\mbox{ cm}$$

(iv) $$p=31$$

Solution

(i)

Swing $$1$$ $$2$$ $$3$$ $$4$$ $$5$$

Length of Arc (cm)

$$45$$

$$\dfrac{81}{2}$$

$$\dfrac{729}{20}$$

$$\dfrac{6{,}561}{200}$$

$$\dfrac{59{,}049}{2{,}000}$$

(ii)

\begin{align}T_n=45(0.9)^{n-1}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}T_{25}&=45(0.9)^{25-1}\\&\approx3.6\mbox{ cm}\end{align}

(iii)

\begin{align}S_n=\frac{a(1-r^n)}{1-r}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}S_{40}&=\frac{45(1-0.9^{40})}{1-0.9}\\&\approx443\mbox{ cm}\end{align}

(iv)

\begin{align}45(0.9)^{n-1}=2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(0.9)^{n-1}=\frac{2}{45}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}n-1=\log_{0.9}\left(\frac{2}{45}\right)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}n&=\log_{0.9}\left(\frac{2}{45}\right)+1\\&=30.551…\end{align}

\begin{align}\downarrow\end{align}

\begin{align}p=31\end{align}

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(b)

(i) If the length of the pendulum is $$1\mbox{ m}$$, show that the angle, $$\theta$$, of swing $$1$$ of the pendulum is $$26^{\circ}$$, correct to the nearest degree.

(ii) Hence, find the total accumulated angle that the pendulum swings through (i.e. the sum of all the angles it swings through until it stops swinging).

(iii) Hence, or otherwise, find the total distance travelled by the tip of the pendulum when it has moved through half of the total accumulated angle.
Give your answer, in $$\mbox{cm}$$, correct to the nearest integer.

(i) $$26^{\circ}$$

(ii) $$260^{\circ}$$

(iii) $$227\mbox{ cm}$$

Solution

(i)

\begin{align}l=2\pi r\left(\frac{\theta}{360^{\circ}}\right)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\theta&=\frac{(360^{\circ})(l)}{2\pi r}\\&=\frac{(360^{\circ})(45)}{2\pi(100)}\\&\approx26^{\circ}\end{align}

(ii)

\begin{align}S_{\infty}&=\frac{a}{1-r}\\&=\frac{26}{1-0.9}\\&=260^{\circ}\end{align}

(iii)

\begin{align}l&=2\pi r\left(\frac{\theta}{360^{\circ}}\right)\\&=2\pi(100\frac{130^{\circ}}{360^{\circ}}\\&\approx227\mbox{ cm}\end{align}

Video Walkthrough

## 2020 Paper 1 Question 7(a)-(b)

(a) A number of the form $$1+2+3+…+n$$ is sometimes called a triangular number because
it can be represented as an equilateral triangle.
The diagram below shows the first three terms in the sequence of triangular numbers.

(i) Complete the table below to list the next five triangular numbers.

Term $$T_1$$ $$T_2$$ $$T_3$$ $$T_4$$ $$T_5$$ $$T_6$$ $$T_7$$ $$T_8$$

Triangular Number

$$1$$

$$3$$

$$6$$

(ii) The $$n$$th triangular number can be found directly using the formula

\begin{align}T_n=\frac{n(n+1)}{2}\end{align}

Is $$1275$$ a triangular number? Give a reason for your answer.

(i)

Term $$T_1$$ $$T_2$$ $$T_3$$ $$T_4$$ $$T_5$$ $$T_6$$ $$T_7$$ $$T_8$$

Triangular Number

$$1$$

$$3$$

$$6$$

$$10$$

$$15$$

$$21$$

$$28$$

$$36$$

(ii) Yes

Solution

(i)

Term $$T_1$$ $$T_2$$ $$T_3$$ $$T_4$$ $$T_5$$ $$T_6$$ $$T_7$$ $$T_8$$

Triangular Number

$$1$$

$$3$$

$$6$$

$$10$$

$$15$$

$$21$$

$$28$$

$$36$$

(ii)

\begin{align}\frac{n(n+1)}{2}=1275\end{align}

\begin{align}\downarrow\end{align}

\begin{align}n^2+n=2550\end{align}

\begin{align}\downarrow\end{align}

\begin{align}n^2+n-2550=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(n-50)(n+51)=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}n=50\end{align}

Therefore, $$1275$$ is a triangular number.

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(b)

(i) The $$(n+1)$$th triangular number can be written as $$T_{n+1}=T_n+(n+1)$$, where $$n\in\mathbb{N}$$.

Write the expression $$\dfrac{n(n+1)}{2}+(n+1)$$ as a single fraction in its simplest form.

(ii) Prove that the sum of any two consecutive triangular numbers will always be a square number (a number in the form $$k^2$$, where $$k\in\mathbb{N}$$).

(iii) Two consecutive triangular numbers sum to $$12{,}544$$.
Find the smaller of these two numbers.

(i) $$\dfrac{(n+1)(n+2)}{2}$$

(iii) $$6{,}216$$

Solution

(i)

\begin{align}T_{n+1}&=T_n+(n+1)\\&=\frac{n(n+1)}{2}+(n+1)\\&=\frac{n(n+1)+2(n+1)}{2}\\&=\frac{(n+1)(n+2)}{2}\end{align}

(ii)

\begin{align}T_{n+1}+T_n&=\frac{(n+1)(n+2)}{2}+\frac{n(n+1)}{2}\\&=\frac{(n+1)(2n+2)}{2}\\&=2\frac{(n+1)(n+1)}{2}\\&=(n+1)^2\end{align}

(iii)

\begin{align}(n+1)^2=12{,}544\end{align}

\begin{align}\downarrow\end{align}

\begin{align}n+1=112\end{align}

\begin{align}\downarrow\end{align}

\begin{align}n=111\end{align}

\begin{align}\downarrow\end{align}

\begin{align}T_{111}&=\frac{(111)(111+1)}{2}\\&=6{,}216\end{align}

Video Walkthrough

## 2019 Paper 1 Question 7

The closed line segment $$[0,1]$$ is shown below. The first three steps in the construction of the
Cantor Set are also shown:

• Step $$1$$ removes the open middle third of the line segment $$[0,1]$$ leaving two closed line segments (i.e. the end points of the segments remain in the Cantor Set)
• Step $$2$$ removes the middle third of the two remaining segments leaving four closed line segments
• Step $$3$$ removes the middle third of the four remaining segments leaving eight closed line segments.

The process continues indefinitely. The set of points in the line segment $$[0,1]$$ that are not removed during the process is the Cantor Set.

(a)

(i) Complete the table below to show the length of the line segment(s) removed at each step for the first $$5$$ steps. Give your answers as fractions.

Step Step $$1$$ Step $$2$$ Step $$3$$ Step $$4$$ Step $$5$$

Length Removed

$$\dfrac{1}{3}$$

$$\dfrac{2}{9}$$

(ii) Find the total length of all of the line segments removed from the initial line segment of length $$1$$ unit, after a finite number ($$n$$) of steps in the process.
Give your answer in terms of $$n$$.

(iii) Find the total length removed, from the initial line segment, after an infinite number of steps of the process.

(i)

Step Step $$1$$ Step $$2$$ Step $$3$$ Step $$4$$ Step $$5$$

Length Removed

$$\dfrac{1}{3}$$

$$\dfrac{2}{9}$$

$$\dfrac{4}{27}$$

$$\dfrac{8}{81}$$

$$\dfrac{16}{243}$$

(ii) $$1-\left(\dfrac{2}{3}\right)^n$$

(iii) $$1$$

Solution

(i)

Step Step $$1$$ Step $$2$$ Step $$3$$ Step $$4$$ Step $$5$$

Length Removed

$$\dfrac{1}{3}$$

$$\dfrac{2}{9}$$

$$\dfrac{4}{27}$$

$$\dfrac{8}{81}$$

$$\dfrac{16}{243}$$

(ii)

\begin{align}S_n&=\frac{a(1-r^n)}{1-r} \\&=\frac{\frac{1}{3}\left[1-\left(\frac{2}{3}\right)^n\right]}{1-\frac{2}{3}}\\&=1-\left(\frac{2}{3}\right)^n\end{align}

(iii)

\begin{align}S_{\infty}&=\frac{a}{1-r}\\&=\frac{\frac{1}{3}}{1-\frac{2}{3}}\\&=1\end{align}

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(b)

(i) Complete the table below to identify the end‐points labelled in the diagram.

Label $$A$$ $$B$$ $$C$$ $$D$$ $$E$$ $$F$$

End-point

(ii) Give a reason why $$\dfrac{1}{3}-\dfrac{1}{9}+\dfrac{1}{27}-\dfrac{1}{81}$$ is a point in the Cantor Set.

(iii) The limit of the series $$\dfrac{1}{3}-\dfrac{1}{9}+\dfrac{1}{27}-…$$ is a point in the Cantor Set. Find this point.

(i)

Label $$A$$ $$B$$ $$C$$ $$D$$ $$E$$ $$F$$

End-point

$$\dfrac{2}{3}$$

$$\dfrac{2}{9}$$

$$\dfrac{7}{9}$$

$$\dfrac{8}{9}$$

$$\dfrac{7}{27}$$

$$\dfrac{25}{27}$$

(ii) It is the end point of a segment.

(iii) $$\dfrac{1}{4}$$

Solution

(i)

Label $$A$$ $$B$$ $$C$$ $$D$$ $$E$$ $$F$$

End-point

$$\dfrac{2}{3}$$

$$\dfrac{2}{9}$$

$$\dfrac{7}{9}$$

$$\dfrac{8}{9}$$

$$\dfrac{7}{27}$$

$$\dfrac{25}{27}$$

(ii) It is the end point of a segment.

(iii)

\begin{align}S_{\infty}&=\frac{a}{1-r}\\&=\frac{\frac{1}{3}}{1-\left(-\frac{1}{3}\right)}\\&=\frac{1}{4}\end{align}

Video Walkthrough

## 2018 Paper 1 Question 2(a) & (c)

(a) The first three terms of a geometric series are $$x^2$$, $$5x-8$$, and $$x+8$$ where $$x\in\mathbb{R}$$.
Use the common ratio to show that $$x^3-17x^2+80x-64=0$$.

Solution

\begin{align}\frac{5x-8}{x^2}=\frac{x+8}{5x-8}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(5x-8)^2=x^2(x+8)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}25x^2-80x+64=x^3+8x^2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x^3-17x^2+80x-64=0\end{align}

as required.

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(c) For only one of $$x=1$$ or $$x=8$$, the terms in part (a) will generate a geometric series with a finite sum to infinity.
Which of these two values does this occur for? For this value, find the sum to infinity.

$$x=8$$ and $$S_{\infty}=128$$

Solution

For $$x=1$$, the series $$1^2,5(1)-8,1+8…=1,-3,9$$ doesn’t have a sum to infinity (as $$|r|>1$$).

For $$x=8$$, the series $$8^2,5(8)-8,8+8…=64,32,16$$ does have a sum to infinity (as $$|r|<1$$).

\begin{align}\downarrow\end{align}

\begin{align}S_{\infty}&=\frac{a}{1-r}\\&=\frac{64}{1-\frac{1}{2}}\\&=128\end{align}

Video Walkthrough

## 2018 Paper 1 Question 5

(a) The Sieve of Sundaram is an infinite table of arithmetic sequences.
The terms in the first $$4$$ rows and the first $$4$$ columns of the table are shown below.

(i) Find the difference between the sums of the first $$45$$ terms in the first two rows.

(ii) Find the number which is in the $$60$$th row and $$70$$th column of the table.

(i) $$2{,}115$$

(ii) $$8{,}530$$

Solution

(i)

\begin{align}S_{2,45}-S_{1,45}&=\frac{45}{2}[2(7)+(45-1)(5)]-\frac{45}{2}[2(4)+(45-1)(3)]\\&=2{,}115\end{align}

(ii)

\begin{align}T_{60,1}&=4+(60-1)(3)\\&=181\end{align}

and

\begin{align}T_{60,2}&=7+(60-1)(5)\\&=302\end{align}

\begin{align}\downarrow\end{align}

\begin{align}T_{60,70}&=181+(70-1)(302-181)\\&=8{,}530\end{align}

Video Walkthrough
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(b) The first two terms of a sequence are ܽ$$a_1=4$$ and ܽ$$a_2=2$$.
The general term is defined by ܽ$$a_n=a_{n-1}-a_{n-2}$$, when $$n\geq3$$.
Write out the next $$6$$ terms of the sequence and hence find the value of ܽ$$a_{2019}$$.

$$-2,-4,-2,2,4,2$$ and $$a_{2019}=-2$$

Solution

\begin{align}a_3&=a_2-a_1\\&=2-4\\&=-2\end{align}

and

\begin{align}a_4&=a_3-a_2\\&=-2-2\\&=-4\end{align}

and

\begin{align}a_5&=a_4-a_3\\&=-4-(-2)\\&=-2\end{align}

and

\begin{align}a_6&=a_5-a_4\\&=-2-(-4)\\&=2\end{align}

and

\begin{align}a_7&=a_6-a_5\\&=2-(-2)\\&=4\end{align}

and

\begin{align}a_8&=a_7-a_6\\&=4-2\\&=2\end{align}

The following pattern therefore repeats

\begin{align}4,2,-2,-4,-2,2,…\end{align}

As $$2019=2016+3$$, and as $$2016$$ is a multiple of $$6$$:

\begin{align}a_{2019}=-2\end{align}

Video Walkthrough

## 2018 Paper 1 Question 9

The diagram below shows the first $$4$$ steps of an infinite pattern which creates the Sierpinski Triangle. The sequence begins with a black equilateral triangle. Each step is formed by removing an equilateral triangle from the centre of each black triangle in the previous step, as shown. Each equilateral triangle that is removed is formed by joining the midpoints of the sides of a black triangle from the previous step.

(a) The table below shows the number of black triangles at each of the first $$4$$ steps and the fraction of the original triangle remaining at each step. Complete the table.

Step $$0$$ $$1$$ $$2$$ $$3$$

Number of black triangles

$$1$$

Fraction of the original triangle remaining

$$1$$

$$\dfrac{9}{16}$$

Step $$0$$ $$1$$ $$2$$ $$3$$

Number of black triangles

$$1$$

$$3$$

$$9$$

$$27$$

Fraction of the original triangle remaining

$$1$$

$$\dfrac{3}{4}$$

$$\dfrac{9}{16}$$

$$\dfrac{27}{64}$$

Solution
Step $$0$$ $$1$$ $$2$$ $$3$$

Number of black triangles

$$1$$

$$3$$

$$9$$

$$27$$

Fraction of the original triangle remaining

$$1$$

$$\dfrac{3}{4}$$

$$\dfrac{9}{16}$$

$$\dfrac{27}{64}$$

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(b)

(i) Write an expression in terms of $$n$$ for the number of black triangles in step $$n$$ of the pattern.

(ii) Step $$k$$ is the first step of the pattern in which the number of black triangles exceeds one thousand million (i.e. $$1\times10^9$$) for the first time. Find the value of $$k$$.

(i) $$3^n$$

(ii) $$k=19$$

Solution

(i) $$3^n$$

(ii)

\begin{align}3^k>1\times10^9\end{align}

\begin{align}\downarrow\end{align}

\begin{align}k>\log_3(1\times10^9)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}k>18.863…\end{align}

\begin{align}\downarrow\end{align}

\begin{align}k=19\end{align}

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(c)

(i) Step $$h$$ is the first step of the pattern in which the fraction of the original triangle remaining is less than $$\dfrac{1}{100}$$ of the original triangle. Find the value of $$h$$.

(ii) What fraction of the original triangle remains after an infinite number of steps of the
pattern?

(i) $$h=17$$

(ii) None of the triangle remains.

Solution

(i)

\begin{align}\left(\frac{3}{4}\right)^h<\frac{1}{100}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\ln\left(\frac{3}{4}\right)^h<\ln\left(\frac{1}{100}\right)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}h\ln\left(\frac{3}{4}\right)<\ln\left(\frac{1}{100}\right)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}h>\frac{\ln\left(\frac{1}{100}\right)}{\ln\left(\frac{3}{4}\right)}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}h>16.007…\end{align}

\begin{align}\downarrow\end{align}

\begin{align}h=17\end{align}

(ii) $$\left(\dfrac{3}{4}\right)^n$$ approaches zero as $$n$$ approaches infinity.

Therefore, none of the triangle remains.

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(d)

(i) The side length of the triangle in Step $$0$$ is $$1$$ unit. The table below shows the total perimeter of all the black triangles in each of the first $$5$$ steps.
Complete the table below.

Step $$0$$ $$1$$ $$2$$ $$3$$ $$4$$

Perimeter

$$3$$

$$\dfrac{27}{4}$$

(ii) Find the total perimeter of the black triangles in step $$35$$ of the pattern.

(iii) Use your answers to part (c)(ii) and part (d)(ii) to comment on the total area and the total perimeter of the black triangles in step ݊$$n$$ of the pattern, as ݊$$n$$ tends to infinity.

(i)

Step $$0$$ $$1$$ $$2$$ $$3$$ $$4$$

Perimeter

$$3$$

$$\dfrac{9}{2}$$

$$\dfrac{27}{4}$$

$$\dfrac{81}{8}$$

$$\dfrac{243}{16}$$

(ii) $$4{,}368{,}329$$

(iii) The area tends to zero and the perimeter tends to infinity.

Solution

(i)

Step $$0$$ $$1$$ $$2$$ $$3$$ $$4$$

Perimeter

$$3$$

$$\dfrac{9}{2}$$

$$\dfrac{27}{4}$$

$$\dfrac{81}{8}$$

$$\dfrac{243}{16}$$

(ii)

\begin{align}P_n=\frac{3^{n+1}}{2^n}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}P_{35}&=\frac{3^{35+1}}{2^35}\\&=4{,}368{,}329\end{align}

(iii) The area tends to zero and the perimeter tends to infinity.

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## 2018 Paper 2 Question 7(a)

A section of a garden railing is shown below. This section consists of nine cylindrical bars, labelled $$A$$ to $$I$$, with a solid sphere attached to the centre of the top of each bar.
The volume of each sphere from $$B$$ to $$E$$ is $$1.75$$ times the volume of the previous sphere.

(a) The radius of sphere $$A$$ is $$3\mbox{ cm}$$. Find the sum of the volumes of the five spheres $$A$$, $$B$$, $$C$$, $$D$$ and $$E$$.
Give your answer correct to the nearest $$\mbox{cm}^3$$.

$$2324\mbox{ cm}^3$$

Solution

\begin{align}V_A&=\frac{4}{3}\pi r^3\\&=\frac{4}{3}\pi(3^3)\\&=36\pi\end{align}

\begin{align}\downarrow\end{align}

\begin{align}S_5&=\frac{36\pi(1-1.75^5)}{1-1.75}\\&\approx2324\mbox{ cm}^3\end{align}

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## 2017 Paper 1 Question 4

(a) The amount of a substance remaining in a solution reduces exponentially over time.
An experiment measures the percentage of the substance remaining in the solution.
The percentage is measured at the same time each day. The data collected over the first $$4$$ days are given in the table below. Based on the data in the table, estimate which is the first day on which the percentage of the substance in the solution will be less than $$0.01\%$$.

Day $$1$$ $$2$$ $$3$$ $$4$$

Percentage of substance (%)

$$95$$

$$42.75$$

$$19.2375$$

$$8.6569$$

The $$12$$th day.

Solution

\begin{align}r&=\frac{42.75}{95}\\&=\frac{9}{20}\end{align}

\begin{align}T_n<0.01\end{align}

\begin{align}\downarrow\end{align}

\begin{align}ar^{n-1}<0.01\end{align}

\begin{align}\downarrow\end{align}

\begin{align}95\left(\frac{9}{20}\right)^{n-1}<0.01\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\left(\frac{9}{20}\right)^{n-1}<\frac{0.01}{95}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\ln\left(\frac{9}{20}\right)^{n-1}<\ln\left(\frac{0.01}{95}\right)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(n-1)\ln\left(\frac{9}{20}\right)<\ln\left(\frac{0.01}{95}\right)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}n>1+\frac{\ln\left(\frac{0.01}{95}\right)}{\ln\left(\frac{9}{20}\right)}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}n>12.47\end{align}

\begin{align}\downarrow\end{align}

$$12$$th day

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(b) A square has sides of length $$2\mbox{ cm}$$. The midpoints of the sides of this square are joined to form another square. This process is continued.
The first three squares in the process are shown below.
Find the sum of the perimeters of the squares if this process is continued indefinitely.
Give your answer in the form $$a+b\sqrt{c}\mbox{ cm}$$, where $$a$$, $$b$$ and $$c\in\mathbb{N}$$.

$$16+8\sqrt{2}\mbox{ cm}$$

Solution

\begin{align}S=4(2)+4\sqrt{2}+4+…\end{align}

\begin{align}a=8&&r=\frac{1}{\sqrt{2}}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}S_{\infty}&=\frac{a}{1-r}\\&=\frac{8}{1-\frac{1}{\sqrt{2}}}\\&=\frac{8}{1-\frac{1}{\sqrt{2}}}\times\frac{1+\frac{1}{\sqrt{2}}}{1+\frac{1}{\sqrt{2}}}\\&=\frac{8+\frac{8}{\sqrt{2}}}{1-\frac{1}{2}}\\&=16+\frac{16}{\sqrt{2}}\\&=16+8\sqrt{2}\mbox{ cm}\end{align}

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## 2017 Paper 1 Question 8(a)

(a) When a loan of €$$P$$ is repaid in equal repayments of amount €$$A$$ ,at the end of each of $$t$$ equal periods of time, where $$i$$ is the periodic compound interest rate (expressed as a decimal), the formula below can be used to find the amount of each repayment.

\begin{align}A=P\frac{i(1+i)^t}{((1+i)^t-1)}\end{align}

Show how this formula is derived. You may use the formula for the sum of a finite geometric series.

Solution

\begin{align}S_n=\frac{a(1-r^n)}{1-r}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}P=\frac{a(1-r^t)}{1-r}\end{align}

\begin{align}a=\frac{A}{1+i}&&r=\frac{1}{1+i}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}P&=\frac{\frac{A}{1+i}\left(1-\left(\frac{1}{1+i}\right)^t\right)}{1-\frac{1}{1+i}}\\&=\frac{A\left(1-\left(\frac{1}{1+i}\right)^t\right)}{1+i-1}\\&=\frac{A[(1+i)^t-1]}{i(1+i)^t}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}A=P\frac{i(1+i)^t}{(1+i)^t-1}\end{align}

as required.

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## 2016 Paper 1 Question 9

(a) At the first stage of a pattern, a point moves $$4$$ units from the origin in the positive direction along the $$x$$-axis. For the second stage, it turns left and moves $$2$$ units parallel to the $$y$$-axis. For the third stage, it turns left and moves $$1$$ unit parallel to the $$x$$-axis.
At each stage, after the first one, the point turns left and moves half the distance of the previous
stage, as shown

(i) How many stages has the point completed when the total distance it has travelled, along its path, is $$7.9375$$ units?

(ii) Find the maximum distance the point can move, along its path, if it continues in this pattern indefinitely.

(iii) Complete the second row of the table below showing the changes to the x co-ordinate, the first nine times the point moves to a new position. Hence, or otherwise, find the $$x$$ co-ordinate and the $$y$$ co-ordinate of the final position that the point is approaching, if it continues indefinitely in this pattern.

Stage 1st 2nd 3rd 4th 5th 6th 7th 8th 9th

Change in $$x$$

$$+4$$

$$0$$

$$-1$$

Change in $$y$$

(i) $$7$$

(ii) $$8\mbox{ units}$$

(iii)

Stage 1st 2nd 3rd 4th 5th 6th 7th 8th 9th

Change in $$x$$

$$+4$$

$$0$$

$$-1$$

$$0$$

$$\dfrac{1}{4}$$

$$0$$

$$-\dfrac{1}{16}$$

$$0$$

$$\dfrac{1}{64}$$

Change in $$y$$

$$0$$

$$2$$

$$0$$

$$-\dfrac{1}{2}$$

$$0$$

$$\dfrac{1}{8}$$

$$0$$

$$-\dfrac{1}{32}$$

$$0$$

$$\left(\dfrac{16}{5},\dfrac{8}{5}\right)$$

Solution

(i)

\begin{align}\frac{a(1-r^n)}{1-r}=7.9375\end{align}

\begin{align}a=4&&r=\frac{1}{2}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{4\left[1-\left(\frac{1}{2}\right)^n\right]}{1-\frac{1}{2}}=7.9375\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\end{align}

\begin{align}8-8\left(\frac{1}{2}\right)^n=7.9375\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\left(\frac{1}{2}\right)^n&=-\frac{7.9375-8}{8}\\&=\frac{1}{128}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}n=7\end{align}

(ii)

\begin{align}S_{\infty}&=\frac{a}{1-r}\\&=\frac{4}{1-\frac{1}{2}}\\&=8\mbox{ units}\end{align}

(iii)

Stage 1st 2nd 3rd 4th 5th 6th 7th 8th 9th

Change in $$x$$

$$+4$$

$$0$$

$$-1$$

$$0$$

$$\dfrac{1}{4}$$

$$0$$

$$-\dfrac{1}{16}$$

$$0$$

$$\dfrac{1}{64}$$

Change in $$y$$

$$0$$

$$2$$

$$0$$

$$-\dfrac{1}{2}$$

$$0$$

$$\dfrac{1}{8}$$

$$0$$

$$-\dfrac{1}{32}$$

$$0$$

\begin{align}S_{x,\infty}&=\frac{a_x}{1-r_x}\\&=\frac{4}{1-\left(-\frac{1}{4}\right)}\\&=\frac{16}{5}\mbox{ units}\end{align}

and

\begin{align}S_{y,\infty}&=\frac{a_y}{1-r_y}\\&=\frac{2}{1-\left(-\frac{1}{4}\right)}\\&=\frac{8}{5}\mbox{ units}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(x,y)=\left(\frac{16}{5},\frac{8}{5}\right)\end{align}

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(b) A male bee comes from an unfertilised egg, i.e. he has a female parent but he does not have a male parent. A female bee comes from a fertilised egg, i.e. she has a female parent and a male parent.

(i) The following diagram shows the ancestors of a certain male bee. We identify his generation as $$G_1$$ and our diagram goes back to $$G_4$$. Continue the diagram to $$G_5$$.

(ii) The number of ancestors of this bee in each generation can be calculated by the formula

\begin{align}G_{n+2}=G_{n+1}+G_n\end{align}

where $$G_1=1$$ and $$G_2=1$$, as in the diagram.
Use this formula to calculate the number of ancestors in $$G_6$$ and in $$G_7$$.

(iii) The number of ancestors in each generation can also be calculated by using the formula

\begin{align}G_n=\frac{(1+\sqrt{5})^n-(1-\sqrt{5})^n}{2^n\sqrt{5}}\end{align}

Use this formula to verify the number of ancestors in $$G_3$$.

(i)

(ii) $$G_6=8$$ and $$G_7=13$$

(iii) $$G_3=2$$

Solution

(i)

(ii)

\begin{align}G_6&=G_5+G_4\\&=5+3\\&=8\end{align}

and

\begin{align}G_7&=G_6+G_5\\&=8+5\\&=13\end{align}

(iii)

\begin{align}G_3&=\frac{(1+\sqrt{5})^3-(1-\sqrt{5})^3}{2^3\sqrt{5}}\\&=\frac{(1+\sqrt{5})(1+2\sqrt{5}+5)-[(1-\sqrt{5})(1-2\sqrt{5}+5)]}{8\sqrt{5}}\\&=\frac{1+2\sqrt{5}+5+\sqrt{5}+10+5\sqrt{5}-1+2\sqrt{5}-5+\sqrt{5}-10+5\sqrt{5}}{8\sqrt{5}}\\&=\frac{16\sqrt{5}}{8\sqrt{5}}\\&=2\end{align}

as required.

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## 2015 Paper 1 Question 1

Mary threw a ball onto level ground from a height of 2 m. Each time the ball hit the ground it bounced back up to $$\dfrac{3}{4}$$ of the height of the previous bounce, as shown.

(a) Complete the table below to show the maximum height, in fraction form, reached by the ball on each of the first four bounces.

Bounce $$1$$ $$2$$ $$3$$ $$4$$

Height (m)

$$\dfrac{2}{1}$$

Bounce $$1$$ $$2$$ $$3$$ $$4$$

Height (m)

$$\dfrac{2}{1}$$

$$\dfrac{3}{2}$$

$$\dfrac{9}{8}$$

$$\dfrac{27}{32}$$

Solution
Bounce $$1$$ $$2$$ $$3$$ $$4$$

Height (m)

$$\dfrac{2}{1}$$

$$\dfrac{3}{2}$$

$$\dfrac{9}{8}$$

$$\dfrac{27}{32}$$

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