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End Of Chapter Exercises

End of chapter exercises

Exercise 3.4

Find the first five terms of the quadratic sequence defined by:

\({T}_{n}={n}^{2}+2n+1\)
\(-4; 9; 16; 25; 36\)

Determine whether each of the following sequences is:

  • a linear sequence,
  • a quadratic sequence,
  • or neither.

\(6;9;14;21;30;...\)

\begin{align*} \text{First differences: } &= 3; 5; 7; 9; \\ \text{Second difference: } &= 2 \end{align*}

Quadratic sequence

\(1;7;17;31;49;...\)

\begin{align*} \text{First differences: } &= 6; 10; 14; 18 \\ \text{Second difference: } &= 4 \end{align*}

Quadratic sequence

\(8;17;32;53;80;...\)

\begin{align*} \text{First differences: } &= 9; 15; 21; 27 \\ \text{Second difference: } &= 6 \end{align*}

Quadratic sequence

\(9;26;51;84;125;...\)

\begin{align*} \text{First differences: } &= 17; 25; 33; 41 \\ \text{Second difference: } &= 8 \end{align*}

Quadratic sequence

\(2;20;50;92;146;...\)

\begin{align*} \text{First differences: } &= 18; 30; 42; 54 \\ \text{Second difference: } &= 12 \end{align*}

Quadratic sequence

\(5;19;41;71;109;...\)

\begin{align*} \text{First differences: } &= 14; 22; 30; 38 \\ \text{Second difference: } &= 8 \end{align*}

Quadratic sequence

\(2;6;10;14;18;...\)

\[\text{First difference: } = 4\]

Linear sequence

\(3;9;15;21;27;...\)

\[\text{First difference: } = 6\]

Linear sequence

\(1;\text{2,5};5;\text{8,5};13;...\)

\begin{align*} \text{First differences: } &= \text{1,5};~ \text{2,5};~ \text{3,5};~ \text{4,5} \\ \text{Second difference: } &= 1 \end{align*}

Quadratic sequence

\(10;24;44;70;102;...\)

\begin{align*} \text{First differences: } &= 14; 20; 26; 32 \\ \text{Second difference: } &= 16 \end{align*}

Quadratic sequence

\(2\frac{1}{2}; 6; 10\frac{1}{2}; 16; 22\frac{1}{2}; \ldots\)

\begin{align*} \text{First differences: } &= \text{3,5};~ \text{4,5};~ \text{5,5};~ \text{6,5} \\ \text{Second difference: } &= 1 \end{align*}

Quadratic sequence

\(3p^2; 6p^2; 9p^2; 12p^2; 15p^2; \ldots\)

\[\text{First difference: } = 3p^2\]

Linear sequence

\(2k; 8k; 18k; 32k; 50k; \ldots\)

\begin{align*} \text{First differences: } &= 6k; 10k; 14k; 18k \\ \text{Second difference: } &= 4k \end{align*}

Quadratic sequence

Given the pattern: \(16; x; 46; \ldots\), determine the value of \(x\) if the pattern is linear.

\begin{align*} x - 16 &= 46 - x \\ 2x &= 62 \\ \therefore x &= 31 \end{align*}

Given \(T_n = 2n^2\), for which value of \(n\) does \(T_n = 242\)?

\begin{align*} 2n^2 &= 242 \\ n^2 &= 121 \\ \therefore n &= 11 \end{align*}

Given \({T}_{n}=3{n}^{2}\), find \({T}_{11}\).

\begin{align*} T_n &= 3n^2 \\ \therefore T_11 &= 3(11)^2 \\ &= 363 \end{align*}

Given \(T_n = n^2+4\), for which value of \(n\) does \(T_n = 85\)?

\begin{align*} n^2 + 4 &= 85 \\ n^2 &= 81 \\ \therefore n &= 9 \end{align*}

Given \({T}_{n}=4{n}^{2}+3n-1\), find \({T}_{5}\).

\begin{align*} T_n &= 4n^2 + 3n - 1 \\ \therefore T_5 &= 4(5)^2 + 3(5) - 1\\ &= 100 + 15 - 1 \\ &= 114 \end{align*}

Given \(T_n = \dfrac{3}{2}n^2\), for which value of \(n\) does \(T_n = 96\)?

\begin{align*} \dfrac{3}{2}n^2 &= 96 \\ n^2 &= 96 \times \frac{2}{3} \\ &= 64 \\ \therefore n &= 8 \end{align*}

For each of the following patterns, determine:

  • the next term in the pattern,
  • and the general term,
  • the tenth term in the pattern.

\(3; 7; 11; 15; \ldots\)

\begin{align*} d &= 7 - 3 \\ &= 4 \\ T_5 &= 15 + 4 \\ &= 19 \\ T_n &= 4n - 1 \\ T_{10} &= 4(10) - 1 \\ &= 39 \end{align*}

\(17; 12; 7; 2; \ldots\)

\begin{align*} d &= 12 - 17 \\ &= -5 \\ T_5 &= 2 - 5 \\ &= -3 \\ T_n &= 22 - 5n \\ T_{10} &= 22 - 5(10) \\ &= -28 \end{align*}

\(\frac{1}{2}; 1; 1\frac{1}{2}; 2; \ldots\)

\begin{align*} d &= 1 - \frac{1}{2} \\ &= \frac{1}{2} \\ T_5 &= 2 + \frac{1}{2} \\ &= 2 \frac{1}{2} \\ T_n &= \frac{1}{2}n \\ T_{10} &= \frac{1}{2}(10) \\ &= 5 \end{align*}

\(a; a+b; a+2b; a+3b; \ldots\)

\begin{align*} d &= a + b - a \\ &= b \\ T_5 &= a + 3b + b \\ &= a + 4b \\ T_n &= a - b + bn \\ T_{10} &= a - b + b(10) \\ &= a + 9b \end{align*}

\(1; -1; -3; -5; \ldots\)

\begin{align*} d &= -1 - 1 \\ &= -2 \\ T_5 &= -5 -2 \\ &= -7 \\ T_n &= 3 - 2n \\ T_{10} &= 3 - 2(10) \\ &= -17 \end{align*}

For each of the following sequences, find the equation for the general term and then use the equation to find \({T}_{100}\).

\(4;7;12;19;28;...\)

9fdaa83be730b67b978cc74ef7b3bf4d.png

\(2;8;14;20;26;...\)

\begin{align*} d &= 8 - 2 \\ &= 6 \\ \therefore T_n &= 6n - 4 \end{align*} \begin{align*} T_n &= 6n - 4 \\ \therefore T_{100} &= 6(100) - 4 \\ &= 596 \end{align*}

\(7;13;23;37;55;...\)

3052bf142e789e4b56d73485661f1a6e.png

\(5;14;29;50;77;...\)

b0136aaccb251d0c2a9156e802c332d7.png

Given: \(T_n = 3n-1\)

Write down the first five terms of the sequence.

\(2; 5; 8; 11; 14\)

What do you notice about the difference between any two consecutive terms?

Constant difference, \(d=3\)

Will this always be the case for a linear sequence?

Yes

Given the following sequence: \(-15; -11; -7; \ldots ; 173\)

Determine the equation for the general term.

\begin{align*} d &= -11 - (-15) \\ &= 4 \\ T_n &= 4n - 19 \end{align*}

Calculate how many terms there are in the sequence.

\begin{align*} T_n &= 4n - 19 \\ 173 &= 4n - 19 \\ 192 &= 4n \\ \therefore 48 &= n \end{align*}

Given \(3; 7; 13; 21; 31; \ldots\)

Thabang determines that the general term is \(T_n = 4n - 1\). Is he correct? Explain.

\begin{align*} \text{First differences: } &= 4; 6; 8; 10 \\ \text{Second difference: } &= 2 \end{align*}

Thabang's general term \(T_n = 4n - 1\) describes a linear sequence and this is a quadratic sequence.

Cristina determines that the general term is \(T_n = n^2 + n + 1\). Is she correct? Explain.

\begin{align*} T_n &= an^2 + bn + c \\ \text{Second difference: } 2a &= 2 \\ \therefore a &= 1 \\ 3a + b &= 4 \\ b &= 4 - 3(1) \\ \therefore b &= 1 \\ a + b + c &= 3 \\ \therefore c &= 3 - a - b \\ &= 3 - 1 - 1 \\ \therefore c &= 1 \\ \therefore T_n &= n^2 + n + 1 \end{align*}

Given the following pattern of blocks:

80b29662534c658fdb3c6af28c0680db.png

Draw pattern \(\text{5}\).

48b4ed778ef8296ccb7362aad9f8e43f.png

Complete the table below:

pattern number \((n)\)\(\text{2}\)\(\text{3}\)\(\text{4}\)\(\text{5}\)\(\text{10}\)\(\text{250}\)\(n\)
number of white blocks \((w)\)\(\text{4}\)\(\text{8}\)
pattern number \((n)\)\(\text{2}\)\(\text{3}\)\(\text{4}\)\(\text{5}\)\(\text{10}\)\(\text{250}\)\(n\)
number of white blocks \((w)\)\(\text{4}\)\(\text{8}\)\(\text{12}\)\(\text{16}\)\(\text{36}\)\(\text{996}\)\(4n - 4\)

Is this a linear or a quadratic sequence?

Linear

Cubes of volume \(\text{1}\) \(\text{cm$^{3}$}\) are stacked on top of each other to form a tower:

75098e556f43cc4bc7216d6f7a259c7e.png

Complete the table for the height of the tower:

tower number \((n)\)\(\text{1}\)\(\text{2}\)\(\text{3}\)\(\text{4}\)\(\text{10}\)\(n\)
height of tower \((h)\)\(\text{2}\)
tower number \((n)\)\(\text{1}\)\(\text{2}\)\(\text{3}\)\(\text{4}\)\(\text{10}\)\(n\)
height of tower \((h)\)\(\text{2}\)\(\text{3}\)\(\text{4}\)\(\text{5}\)\(\text{11}\)\(n+1\)

What type of sequence is this?

Linear

Now consider the number of cubes in each tower and complete the table below:

tower number \((n)\)\(\text{1}\)\(\text{2}\)\(\text{3}\)\(\text{4}\)
number of cubes \((c)\)\(\text{3}\)
tower number \((n)\)\(\text{1}\)\(\text{2}\)\(\text{3}\)\(\text{4}\)
number of cubes \((c)\)\(\text{3}\)\(\text{6}\)\(\text{10}\)\(\text{15}\)

What type of sequence is this?

Quadratic

Determine the general term for this sequence.

\begin{align*} T_n &= an^2 + bn + c \\ \text{Second difference: } 2a &= 1 \\ \therefore a &= \frac{1}{2} \\ 3a + b &= 3 \\ b &= 3 - 3(\frac{1}{2}) \\ \therefore b &= \frac{3}{2} \\ a + b + c &= 3 \\ \therefore c &= 3 - a - b \\ &= 3 - \frac{1}{2} - \frac{3}{2} \\ \therefore c &= 1 \\ \therefore T_n &= \frac{1}{2}n^2 + \frac{3}{2}n + 1 \end{align*}

How many cubes are needed for tower number \(\text{21}\)?

\begin{align*} T_n &= \frac{1}{2}n^2 + \frac{3}{2}n + 1 \\ \therefore T_{21} &= \frac{1}{2}(21)^{2}+ \frac{3}{2}(21) + 1 \\ &= \frac{441}{2} + \frac{63}{2} + \frac{2}{2} \\ &= 253 \end{align*}

How high will a tower of \(\text{496}\) cubes be?

\begin{align*} T_n &= \frac{1}{2}n^2 + \frac{3}{2}n + 1\\ 496 &= \frac{1}{2}n^2 + \frac{3}{2}n + 1 \\ 0 &= \frac{1}{2}n^2 + \frac{3}{2}n - 495 \\ &= n^2 + 3n - 990 \\ &= (n - 30)(n + 33) \\ \therefore n = 30 &\text{ or } n = -33 \\ \therefore n &= 30 \\ \text{And } h &= n + 1 \\ &= 30 + 1 \\ &= \text{31}\text{ cm} \end{align*}

A quadratic sequence has a second term equal to \(\text{1}\), a third term equal to \(-\text{6}\) and a fourth term equal to \(-\text{14}\).

Determine the second difference for this sequence.

\begin{align*} T_3 - T_2 &= -6 -(1) \\ &= -7 \\ T_4 - T_3 &= -14 - (-6) \\ &= -8 \\ \therefore \text{Second difference} &= -1 \end{align*}

Hence, or otherwise, calculate the first term of the pattern.

\begin{align*} T_{1} &= 1 + 6 \\ &= 7 \end{align*}

There are \(\text{15}\) schools competing in the U16 girls hockey championship and every team must play two matches — one home match and one away match.

Use the given information to complete the table:

no. of schoolsno. of matches
\(\text{1}\)\(\text{0}\)
\(\text{2}\)
\(\text{3}\)
\(\text{4}\)
\(\text{5}\)
no. of schoolsno. of matches
\(\text{1}\)\(\text{0}\)
\(\text{2}\)\(\text{2}\)
\(\text{3}\)\(\text{6}\)
\(\text{4}\)\(\text{12}\)
\(\text{5}\)\(\text{20}\)

Calculate the second difference.

\begin{align*} T_2 - T_1 &= 2 - 0 \\ &= 2 \\ T_3 - T_2 &= 6 - 2 \\ &= 4 \\ T_4 - T_3 &= 12 -6 \\ &= 6 \\ T_5 - T_4 &= 20 - 12 \\ &= 8 \\ \therefore \text{Second difference} &= 2 \end{align*}

Determine a general term for the sequence.

\begin{align*} T_n &= an^2 + bn + c \\ \text{Second difference: } 2a &= 2 \\ \therefore a &= 1 \\ 3a + b &= 2 \\ b &= 2 - 3(1) \\ \therefore b &= -1 \\ a + b + c &= 0 \\ \therefore c &= - a - b \\ &= -1 - (-1) \\ \therefore c &= 0 \\ \therefore T_n &= n^2 -n \end{align*}

How many matches will be played if there are \(\text{15}\) schools competing in the championship?

\begin{align*} T_n &= n^2 -n \\ T_{15} &= (15)^2 - 15 \\ &= 225 - 15 \\ &= 210 \end{align*}

If \(\text{600}\) matches must be played, how many schools are competing in the championship?

\begin{align*} T_n &= n^2 -n \\ 600 &= n^2 -n \\ 0 &= n^2 -n - 600\\ &= (n - 25)(n + 24)\\ \therefore n = 25 &\text{ or } n = -24 \\ \therefore n &= 25 \end{align*}

The first term of a quadratic sequence is \(\text{4}\), the third term is \(\text{34}\) and the common second difference is \(\text{10}\). Determine the first six terms in the sequence.

\begin{align*} \text{Let } T_2 &= x \\ \therefore T_2 - T_1 &= x - 4 \\ \text{And } T_3 - T_2 &= 34 - x \\ \text{Second difference } &= (T_3 - T_2) - (T_2 - T_1) \\ &= (34 - x) - (x - 4)\\ \therefore 10 &= 38 - 2x \\ 2x &= 28 \\ \therefore x &= 14 \end{align*}

\(4; 14; 34; 64; 104; 154\)

Challenge question:

Given that the general term for a quadratic sequences is \(T_n = an^2 + bn + c\), let \(d\) be the first difference and \(D\) be the second common difference.

Show that \(a = \dfrac{D}{2}\).

\begin{align*} T_n &= an^2 + bn +c \\ T_1 &= a(1)^2 + b(1) +c \\ &= a + b + c \\ T_2 &= a(2)^2 + b(2) +c \\ &= 4a + 2b + c\\ T_3 &= a(3)^2 + b(3) +c \\ &= 9a + 6b + c\\ \text{First difference } d &= T_2 - T_1 \\ \therefore d &= (4a + 2b + c) - (a + b + c ) \\ &= 3a + b \\ \therefore b &= d - 3a \\ \text{Second difference } D &= (T_3 - T_2) - (T_2 - T_1)\\ \therefore D &= (5a + b) - (3a + b) \\ &= 2a \\ \therefore a &= \dfrac{D}{2} \end{align*}

Show that \(b = d - \dfrac{3}{2}D\).

\begin{align*} a &= \dfrac{D}{2} \\ b &= d - 3a \\ \therefore b &= d - 3 \left( \dfrac{D}{2} \right) \\ &= d - \dfrac{3}{2} D \end{align*}

Show that \(c = T_1 - d + D\).

\begin{align*} T_1 &= a + b + c \\ \therefore c &= T_1 - a - b \\ &= T_1 - \left( \dfrac{D}{2} \right) - \left( d - \dfrac{3}{2}D \right) \\ &= T_1 - \dfrac{D}{2} - d + \dfrac{3}{2}D \\ &= T_1 - d + D \end{align*}

Hence, show that \(T_n = \dfrac{D}{2}n^2 + \left(d - \dfrac{3}{2}D\right)n + \left(T_1 -d + D\right)\).

\begin{align*} T_n &= an^2 + bn +c \\ \therefore T_n &= \dfrac{D}{2}n^2 + \left( d - \dfrac{3}{2}D \right)n + \left( T_1 - d + D \right) \end{align*}