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Sequences & Series

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(BLUE = Paper 1, RED = Paper 2)

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

Sequences and Shapes

Arithmetic Sequences

Arithmetic Series

Quadratic Sequences

2022 Paper 1 Question 6(a)

The \(n\)th term of an arithmetic sequence is given by the following expression, for \(n\in\mathbb{N}\):

\begin{align}T_n=-254+(n-1)(4)\end{align}

(a)

(i) Find the value of \(T_1\) , the first term of this sequence.

(ii) Find the value of the common difference for this sequence (that is, \(T_2-T_1\)).

Answer

(i) \(-254\)

(ii) \(4\)

Solution

(i)

\begin{align}T_1&=-254+(1-1)(4)\\&=-254\end{align}

(ii)

\begin{align}T_2&=-254+(2-1)(4)\\&=-250\end{align}

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

\begin{align}d&=T_2-T_1\\&=-250-(-254)\\&=4\end{align}

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2021 Paper 1 Question 6

(a) The first three terms of an arithmetic sequence are \(-5,k,1\).

(i) Find \(k\) and hence or otherwise show that the common difference is \(3\).

(ii) Find \(T_{10}\), the \(10\)th term in the sequence.

(iii) Find which term in the sequence has a value of \(247\). 

Answer

(i) \(k=-2\)

(ii) \(T_{10}=22\)

(iii) \(n=85\)

Solution

(i)

\begin{align}T_2-T_1=T_3-T_2\end{align}

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

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

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

\begin{align}2k=-4\end{align}

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

\begin{align}k=-2\end{align}

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

\begin{align}T_2-T_1&=k-(-5)\\&=-2+5\\&=3\end{align}

as required.

(ii)

\begin{align}a=-5&&d=3\end{align}

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

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

\begin{align}T_{10}&=-5+(10-1)(3)\\&=22\end{align}

(iii)

\begin{align}T_n=247\end{align}

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

\begin{align}-5+(n-1)(3)=247\end{align}

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

\begin{align}-5+3n-3=247\end{align}

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

\begin{align}3n=255\end{align}

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

\begin{align}n=85\end{align}

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(b) The first three terms of a different arithmetic sequence are \(4,9,14\).
Find \(S_{50}\), the sum of the first \(50\) terms of the sequence.

Answer

\(S_{50}=6{,}325\)

Solution

\begin{align}a=4&&d=5\end{align}

\begin{align}S_n=\frac{n}{2}[2a+(n-1)d]\end{align}

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

\begin{align}S_{50}&=\frac{50}{2}[2(4)+(50-1)(5)]\\&=6{,}325\end{align}

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2021 Paper 1 Question 9

(a) The first three patterns in a sequence of patterns containing dots and crosses are shown below. 

pattern 1pattern 2pattern 3

(i) Draw the fourth pattern in the sequence into the box below.

(ii) Find a formula, in \(n\), for the number of dots in pattern \(n\) of the sequence \(T_n\)).

(iii) Find the total number of dots in the first 20 patterns of the sequence.

(iv) The table shows the number of crosses for the first two patterns. Complete the table, and hence find a formula for the number of crosses in pattern \(n\) of the sequence.

Pattern \(1\) \(2\) \(3\) \(4\) \(5\) \(6\)

Number of crosses

\(1\)

\(4\)

(v) Find the number of crosses in pattern \(20\) of the sequence.

(vi) Find the number of shapes (dots and crosses) in pattern \(10\) of the sequence.

Answer

(i)

pattern 4

(ii) \(T_n=n\)

(iii) \(210\)

(iv)

Pattern \(1\) \(2\) \(3\) \(4\) \(5\) \(6\)

Number of crosses

\(1\)

\(4\)

\(9\)

\(16\)

\(25\)

\(36\)

\begin{align}T_n=n^2\end{align}

(v) \(144\)

(vi) \(110\)

Solution

(i)

pattern 4

(ii)

\begin{align}a=1&&d=1\end{align}

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

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

(iii)

\begin{align}a=1&&d=1\end{align}

\begin{align}S_n=\frac{n}{2}[2a+(n-1)d]\end{align}

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

\begin{align}S_{20}&=\frac{20}{2}[2(1)+(20-1)(1)]\\&=210\end{align}

(iv)

Pattern \(1\) \(2\) \(3\) \(4\) \(5\) \(6\)

Number of crosses

\(1\)

\(4\)

\(9\)

\(16\)

\(25\)

\(36\)

\begin{align}T_n=n^2\end{align}

(v)

\begin{align}T_{12}&=12^2\\&=144\end{align}

(vi)

\begin{align}10^2+10=110\end{align}

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(b) The first two patterns in a sequence of patterns of crosses are shown below.

pattern 1pattern 2

The number of crosses in pattern \(n\) is \(T_n\).
The general term describing \(T_n\) can be written in the form:

\(T_n=\dfrac{n^2}{2}+bn+c\), where \(b,c\in\mathbb{R}\).

(i) Use substitution for \(n\) to write \(T_1\) and \(T_2\) in terms of \(b\), \(c\) and a number.

(ii) Hence, or otherwise, find the value of \(b\) and the value of \(c\). 

Answer

(i) \(T_1=b+c+\dfrac{1}{2}\) and \(T_2=2b+c+2\)

(ii) \(b=\dfrac{1}{2}\) and \(c=0\)

Solution

(i)

\begin{align}T_1&=\frac{1^2}{2}+b(1)+c\\&=b+c+\frac{1}{2}\end{align}

and

\begin{align}T_2&=\frac{2^2}{2}+b(2)+c\\&=2b+c+2\end{align}

(ii)

\begin{align}T_1=1\end{align}

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

\begin{align}b+c+\frac{1}{2}=1\end{align}

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

\begin{align}b+c=\frac{1}{2}\end{align}

and

\begin{align}T_2=3\end{align}

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

\begin{align}2b+c+2=3\end{align}

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

\begin{align}2b+c=1\end{align}

We therefore have the following two equations with two unknowns:

\begin{align}b+c=\frac{1}{2}\end{align}

\begin{align}2b+c=1\end{align}

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

\begin{align}b=\frac{1}{2}\end{align}

and

\begin{align}c&=\frac{1}{2}-b\\&=\frac{1}{2}-\frac{1}{2}=0\end{align}

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2020 Paper 1 Question 9

The following sequence of patterns is created using matchsticks to form equilateral triangles.

Pattern 1Pattern 2Pattern 3

(a) Complete the table below to show the number of matchsticks required to make each of the first six patterns of the above sequence. 

Pattern Number \(1\) \(2\) \(3\) \(4\) \(5\) \(6\)

Number of matchsticks

\(3\)

\(7\)

Answer
Pattern Number \(1\) \(2\) \(3\) \(4\) \(5\) \(6\)

Number of matchsticks

\(3\)

\(7\)

\(11\)

\(15\)

\(19\)

\(23\)

Solution
Pattern Number \(1\) \(2\) \(3\) \(4\) \(5\) \(6\)

Number of matchsticks

\(3\)

\(7\)

\(11\)

\(15\)

\(19\)

\(23\)

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(b)

(i) How many matchsticks are required to make pattern \(10\) of the sequence?

(ii) Find a formula for \(T_n\), the number of matchsticks required to make pattern \(n\) of the sequence.

(iii) Pattern \(k\) has \(147\) matchsticks, where \(k\in\mathbb{N}\). Find the value of \(k\).

Answer

(i) \(39\)

(ii) \(T_n=4n-1\)

(iii) \(37\)

Solution

(i)

\begin{align}3+(10-1)4=39\end{align}

(ii)

\begin{align}T_n&=3+(n-1)4\\&=3+4n-4\\&=4n-1\end{align}

(iii)

\begin{align}T_k=147\end{align}

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

\begin{align}4k-1=147\end{align}

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

\begin{align}k&=\frac{147+1}{4}\\&=37\end{align}

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(c)

(i) Find a formula for \(S_n\), the total number of matchsticks required to make the first \(n\) patterns.

(ii) Find the total number of complete patterns in the sequence that can be made using \(820\) matchsticks.

Answer

(i) \(S_n=n+2n^2\)

(ii) \(20\)

Solution

(i)

\begin{align}S_n&=\frac{n}{2}[a+(n-1)d]\\&=\frac{n}{2}[6+(n-1)(4)]\\&=\frac{n}{2}(6+4n-4)\\&=\frac{n}{2}(2+4n)\\&=n+2n^2\end{align}

(ii)

\begin{align}S_n=820\end{align}

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

\begin{align}n+2n^2=820\end{align}

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

\begin{align}2n^2+n-820=0\end{align}

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

\begin{align}(2n+41)(n-20)=0\end{align}

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

\begin{align}n=20\end{align}
(\(n\) is positive)

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(d)

(i) The table below shows the number of triangles formed by each pattern for the first
two patterns. Complete the table to show the number of triangles formed for patterns
three to six.

Pattern Number \(1\) \(2\) \(3\) \(4\) \(5\) \(6\)

Number of Triangles

\(1\)

\(3\)

Pattern 1Pattern 2Pattern 3

(ii) The area of each triangle is \(4\sqrt{3}\mbox{ cm}^2\).
Find, correct to the nearest \(\mbox{cm}^2\), the combined total area covered by the first \(15\) patterns in the sequence.

Answer

(i)

Pattern Number \(1\) \(2\) \(3\) \(4\) \(5\) \(6\)

Number of Triangles

\(1\)

\(3\)

\(5\)

\(7\)

\(9\)

\(11\)

(ii) \(1{,}559\mbox{ cm}^2\)

Solution

(i)

Pattern Number \(1\) \(2\) \(3\) \(4\) \(5\) \(6\)

Number of Triangles

\(1\)

\(3\)

\(5\)

\(7\)

\(9\)

\(11\)

(ii)

\begin{align}(4\sqrt{3})(S_{15})&=(4\sqrt{3})\left[\frac{15}{2}(2(1)+(15-1)(2))\right]\\&\approx1{,}559\mbox{ cm}^2\end{align}

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2019 Paper 1 Question 8(b)

(b) A company was set up in January \(2016\) to repair engines. In the first month of its existence the company made a loss of €\(4000\). This loss reduced by €\(250\) a month for each month that the company traded.

(i) Complete the table below to show the company’s loss/profit for each of the first six months of trading.

Month \(1\) \(2\) \(3\) \(4\) \(5\) \(6\)

\(\mathbf{h(t)}\)

\(-4000\)

\(-3750\)

\(-3250\)

(ii) Show that the profit the company makes in month \(n\) is given by the formula \(T_n=250n-4250\).

(iii) What profit does the company make in January 2018 (i.e. month 25)?

(iv) Find the month in which the company breaks even (i.e. €\(0\) profit).

(v) Find \(S_n\), the general term for the total profit of the company after \(n\) months.

(vi) Hence, or otherwise, find the total profit of the company at the end of January \(2019\) (i.e. month \(37\)).

Answer

(i)

Month \(1\) \(2\) \(3\) \(4\) \(5\) \(6\)

\(\mathbf{h(t)}\)

\(-4000\)

\(-3750\)

\(-3500\)

\(-3250\)

\(-3000\)

\(-2750\)

(ii) The answer is already in the question!

(iii) \(2{,}000\mbox{ euro}\)

(iv) The \(17\)th month

(v) \(S_n=-4125n+125n^2\)

(vi) \(18{,}500\mbox{ euro}\)

Solution

(i)

Month \(1\) \(2\) \(3\) \(4\) \(5\) \(6\)

\(\mathbf{h(t)}\)

\(-4000\)

\(-3750\)

\(-3500\)

\(-3250\)

\(-3000\)

\(-2750\)

(ii)

\begin{align}T_n&=a+(n-1)d\\&=-4000+(n-1)(250)\\&=-4000+250n-250\\&=250n-4250\end{align}

as required.

(iii)

\begin{align}T_{25}&=250(25)-4250\\&=2{,}000\mbox{ euro}\end{align}

(iv)

\begin{align}T_n=0\end{align}

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

\begin{align}250n-4250=0\end{align}

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

\begin{align}n&=\frac{4250}{250}\\&=17\end{align}

(v)

\begin{align}S_n&=\frac{n}{2}[2a+(n-1)d]\\&=\frac{n}{2}[2(-4000)+(n-1)(250)]\\&=\frac{n}{2}(-8000+250n-250)\\&=-4125n+125n^2\end{align}

(vi)

\begin{align}S_{37}&=-4125(37)+125(37^2)\\&=18{,}500\mbox{ euro}\end{align}

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2018 Paper 1 Question 4

The first three stages in a pattern of grey and white tiles are shown in the diagram below.

Stage 1Stage 2Stage 4Stage 3

(a) Draw the next stage of tiles (Stage 4) onto the diagram above

Answer
Stage 1Stage 2Stage 4Stage 3
Solution
Stage 1Stage 2Stage 4Stage 3
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(b) Based on the pattern shown, complete the table below.

Stage (\(n\)) Number of Grey Tiles Number of White Tiles Total

\(1\)

\(4\)

\(6\)

\(2\)

\(3\)

\(4\)

\(5\)

Answer
Stage (\(n\)) Number of Grey Tiles Number of White Tiles Total

\(1\)

\(4\)

\(2\)

\(6\)

\(2\)

\(9\)

\(2\)

\(11\)

\(3\)

\(16\)

\(2\)

\(18\)

\(4\)

\(25\)

\(2\)

\(18\)

\(5\)

\(36\)

\(2\)

\(38\)

Solution
Stage (\(n\)) Number of Grey Tiles Number of White Tiles Total

\(1\)

\(4\)

\(2\)

\(6\)

\(2\)

\(9\)

\(2\)

\(11\)

\(3\)

\(16\)

\(2\)

\(18\)

\(4\)

\(25\)

\(2\)

\(18\)

\(5\)

\(36\)

\(2\)

\(38\)

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(c) Assuming the pattern continues, the total number of tiles in stage ݊\(n\) \(T_n\) is given by the formula \(T_n=n^2+bn+c\), where \(b\) and \(c\in\mathbb{N}\).
Find the value of \(b\) and the value of \(c\).

Answer

\(b=2\) and \(c=3\)

Solution

\begin{align}T_1=6\end{align}

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

\begin{align}1^2+b(1)+c=6\end{align}

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

\begin{align}b+c=5\end{align}

and

\begin{align}T_2=11\end{align}

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

\begin{align}2^2+b(2)+c=11\end{align}

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

\begin{align}2b+c=7\end{align}

We therefore have the following two equations with two unknowns:

\begin{align}b+c=5\end{align}

\begin{align}2b+c=7\end{align}

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

\begin{align}b=2\end{align}

and

\begin{align}c&=5-b\\&=5-2\\&=3\end{align}

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(d) Find the number of the stage which has \(443\) tiles in total.

Answer

\(n=20\)

Solution

\begin{align}T_n=443\end{align}

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

\begin{align}n^2+2n+3=443\end{align}

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

\begin{align}n^2+2n-440=0\end{align}

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

\begin{align}(n+22)(n-2)=0\end{align}

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

\begin{align}n=20\end{align}
(as \(n>0\))

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2018 Paper 1 Question 7(a) - (d)

Part of the seating arrangement in a theatre is shown in the diagram below. The seats are arranged in rows. Row \(1\) is nearest the stage and has \(28\) seats. Each subsequent row behind that contains one more seat. i.e. row \(2\) has \(29\) seats, row \(3\) has \(30\) seats, and so on.

Stage

(a) Find the number of seats in row \(10\).

Answer

\(37\)

Solution

\begin{align}28+(10-1)(1)=37\end{align}

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(b) There are \(50\) seats in the last row. How many rows of seats are there in the theatre? 

Answer

\(23\)

Solution

\begin{align}28+(n-1)(1)=50\end{align}

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

\begin{align}n&=50-28+1\\&=23\end{align}

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(c) Find the total number of seats in the theatre.

Answer

\(897\)

Solution

\begin{align}S_{23}&=\frac{23}{2}[2(28)+(23-1)(1)]\\&=897\end{align}

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(d) On a particular night \(600\) people are at a show in the theatre. Assume that people are seated only in the rows closest to the stage, i.e. they have filled the first ݊ rows and there are some people seated in the next row.
Find the value of \(n\) and find the number of people seated in the next row. 

Answer

\(n=16\) and the number of people in the next row is \(32\).

Solution

\begin{align}\frac{n}{2}[2(28)+(n-1)(1)]=600\end{align}

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

\begin{align}n^2+55n-1200=0\end{align}

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

\begin{align}n&=\frac{-55\pm\sqrt{55^2-4(1)(-1200)}}{2(1)}\end{align}

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

\begin{align}n\approx16.7\end{align}
(\(n>0\))

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

\begin{align}n=16\end{align}
(\(n\in\mathbb{N}\))

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

\begin{align}S_{16}&=\frac{16}{2}[2(28)+(16-1)(1)]\\&=568\end{align}

The number of people in the next row is therefore \(600-568=32\).

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2017 Paper 1 Question 7

The first three patterns in a sequence of patterns of tiles are shown in the diagram below.

Pattern 1Pattern 2Pattern 3

(a) Draw the next pattern of tiles.

Answer
Pattern 4
Solution
Pattern 4
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(b) Based on the patterns shown, complete the table below.

Pattern Number Number of Tiles

\(1\)

\(5\)

\(2\)

\(3\)

\(4\)

\(5\)

Answer
Pattern Number Number of Tiles

\(1\)

\(5\)

\(2\)

\(8\)

\(3\)

\(11\)

\(4\)

\(14\)

\(5\)

\(17\)

Solution
Pattern Number Number of Tiles

\(1\)

\(5\)

\(2\)

\(8\)

\(3\)

\(11\)

\(4\)

\(14\)

\(5\)

\(17\)

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(c)

(i) Assuming the pattern continues, the number of tiles in the \(n\)th pattern of the sequence is given by the formula \(T_n=pn+q\), where \(p\) and \(q\in\mathbb{N}\).
Find the value of \(p\) and the value of \(q\).

(ii) How many tiles are in the \(20\)th pattern?

(iii) Find which pattern has exactly \(290\) tiles.

Answer

(i) \(p=3\) and \(q=2\)

(ii) \(62\)

(iii) \(96\mbox{th}\)

Solution

(i)

\begin{align}a=5&&d=3\end{align}

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

\begin{align}T_n&=a+(n-1)d\\&=5+(n-1)3\\&=5+3n-3\\&=3n+2\end{align}

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

\(p=3\) and \(q=2\)

(ii)

\begin{align}T_{20}&=3(20)+2\\&=62\end{align}

(iii)

\begin{align}T_n=290\end{align}

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

\begin{align}3n+2=290\end{align}

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

\begin{align}n&=\frac{290-2}{3}\\&=96\end{align}

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(d)

(i) Show that \(S_n=\dfrac{3n^2+7n}{2}\) is a formula for the total number of tiles needed to build the first \(n\) patterns.

(ii) Find the total number of tiles needed to build the first \(30\) patterns.

Answer

(i) The answer is already in the question!

(ii) \(1{,}455\)

Solution

(i)

\begin{align}a=5&&d=3\end{align}

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

\begin{align}S_n&=\frac{n}{2}[2a+(n-1)d]\\&=\frac{n}{2}[2(5)+(n-1)3]\\&=\frac{n}{2}(10+3n-3)\\&=\frac{n}{2}(3n+7)\\&=\frac{3n^2+7n}{2}\end{align}

as required.

(ii)

\begin{align}S_{30}&=\frac{3(30^2)+7(30)}{2}\\&=1{,}455\end{align}

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2016 Paper 1 Question 5

The first five numbers in a pattern of numbers are given in the table below.

Term Number

\(U_1\)

\(13\)

\(U_2\)

\(15\)

\(U_3\)

\(19\)

\(U_4\)

\(25\)

\(U_5\)

\(33\)

\(U_6\)

\(U_7\)

\(U_8\)

(a)

(i) Follow the pattern in the table above to write the next three numbers into the table.

(ii) Use the data in the table to show that the pattern is quadratic.

Answer

(i)

Term Number

\(U_1\)

\(13\)

\(U_2\)

\(15\)

\(U_3\)

\(19\)

\(U_4\)

\(25\)

\(U_5\)

\(33\)

\(U_6\)

\(43\)

\(U_7\)

\(55\)

\(U_8\)

\(69\)

(ii) The second difference is a constant value of \(2\).

Solution

(i)

Term Number

\(U_1\)

\(13\)

\(U_2\)

\(15\)

\(U_3\)

\(19\)

\(U_4\)

\(25\)

\(U_5\)

\(33\)

\(U_6\)

\(43\)

\(U_7\)

\(55\)

\(U_8\)

\(69\)

(ii)

First Difference

\(U_2-U_1\)

\(2\)

\(U_3-U_2\)

\(4\)

\(U_4-U_3\)

\(6\)

\(U_5-U_4\)

\(8\)

\(U_6-U_5\)

\(10\)

\(U_7-U_6\)

\(12\)

\(U_8-U_7\)

\(14\)

As there is therefore a constant second difference of \(2\), the pattern is quadratic.

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(b) \(U_n=n^2+6n+c\) where \(b,c\in\mathbb{Z}\). Find the value of \(b\) and the value of \(c\).

Answer

\(b=-1\) and \(c=13\)

Solution

\begin{align}1^2+b(1)+c=13\end{align}

\begin{align}2^2+b(2)+c=15\end{align}

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

\begin{align}b+c=12\end{align}

\begin{align}2b+c=11\end{align}

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

\begin{align}b=-1\end{align}

and

\begin{align}c&=12-b\\&=12-(-1)\\&=13\end{align}

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(c) The table below shows the first five terms of an arithmetic sequence.
Find an expression for \(T_n\), the \(n\)th term of the sequence.
Hence, or otherwise, find the value of \(T_{30}\), the \(30\)th term.

Term Number

\(T_1\)

\(12\)

\(T_2\)

\(14\)

\(T_3\)

\(16\)

\(T_4\)

\(18\)

\(T_5\)

\(20\)

Answer

\(T_n=2n+10\) and \(T_{30}=70\)

Solution

\begin{align}a=12&&d=2\end{align}

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

\begin{align}T_n&=a+(n-1)d\\&=12+(n-1)(2)\\&=12+2n-2\\&=2n+10\end{align}

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

\begin{align}T_{30}&=2(30)+10\\&=70\end{align}

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2015 Paper 1 Question 7

The first three patterns in a sequence of patterns are shown below.

123

(a) Draw the fourth pattern in the sequence.

Answer
4
Solution
4
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(b) Complete the table below.

\(t\) Number of Black Triangles Number of White Triangles Total Number of Small Triangles

Pattern 1

\(3\)

\(1\)

\(T_1=4\)

Pattern 2

\(T_2=...\)

Pattern 3

\(T_3=...\)

Pattern 4

\(T_4=...\)

Pattern 5

\(T_5=...\)

Answer
\(t\) Number of Black Triangles Number of White Triangles Total Number of Small Triangles

Pattern 1

\(3\)

\(1\)

\(T_1=4\)

Pattern 2

\(6\)

\(3\)

\(T_2=9\)

Pattern 3

\(10\)

\(6\)

\(T_3=16\)

Pattern 4

\(15\)

\(10\)

\(T_4=25\)

Pattern 5

\(21\)

\(15\)

\(T_5=36\)

Solution
\(t\) Number of Black Triangles Number of White Triangles Total Number of Small Triangles

Pattern 1

\(3\)

\(1\)

\(T_1=4\)

Pattern 2

\(6\)

\(3\)

\(T_2=9\)

Pattern 3

\(10\)

\(6\)

\(T_3=16\)

Pattern 4

\(15\)

\(10\)

\(T_4=25\)

Pattern 5

\(21\)

\(15\)

\(T_5=36\)

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(c) Show that the numbers of black triangles form a quadratic sequence.

Answer
111345636101521