DP (Grade 11 & 12) · Maths AA SL
Number & Algebra
Sequences & Series
QUESTION 1 [6 marks]
Easy
The table shows the first four terms of three sequences: $u_n$, $v_n$, and $w_n$.
| n | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| $u_n$ | 15 | 25 | 35 | 45 |
| $v_n$ | 15 | 30 | 60 | 120 |
| $w_n$ | 15 | 25 | 37 | 55 |
a.
State which sequence is (i) arithmetic; (ii) geometric.
[2] Show Solution
$u_n$ is arithmetic (constant difference of $10$). $v_n$ is geometric (constant ratio of $2$).
b.
Find the sum of the first 20 terms of the arithmetic sequence.
[2] Show Solution
$$S_{20} = \frac{n}{2}(2u_1+(n-1)d) = 2200$$
c.
Find the exact value of the 10th term of the geometric sequence.
[2] Show Solution
$$v_{10} = u_1 r^{n-1} = 15 \times 2^{9} = 7680$$
QUESTION 2 [6 marks]
Medium
Consider the sum $$S = \sum_{k=3}^{l} (5k+1), \text{ where } l \text{ is a positive integer greater than } 3.$$
a.
Write down the first three terms of the series.
[2] Show Solution
$$16, 21, 26, \ldots$$
b.
Write down the number of terms in the series, in terms of $l$.
[1] Show Solution
$$\text{number of terms} = l - 3 + 1$$
c.
Given that $S = 1270$, find the value of $l$.
[3] Show Solution
Using $S_n = \frac{n}{2}(u_1+u_n)$ with $u_1=16$, and solving for $l$ gives $$l = 22$$
QUESTION 3 [6 marks]
Easy
A 3D printer builds a set of 49 scale-model towers of decreasing size. The height of the largest tower is 80 cm. The height of each successive smaller tower is 95% of the preceding tower's height.
a.
Find the height of the smallest tower in this set.
[3] Show Solution
$$h_{49} = 80 \times (0.95)^{48} = 6.821\text{ cm}$$
b.
Find the total height if all 49 towers were placed one on top of another.
[3] Show Solution
$$S_{49} = \frac{80(1-0.95^{49})}{1-0.95} = 1470\text{ cm}$$
QUESTION 4 [4 marks]
Medium
The first term of an arithmetic sequence is 8 and the common difference is 4. The first and fourth terms of this arithmetic sequence are the first two terms of a geometric sequence.
a.
Find $u_{60}$, the 60th term of the arithmetic sequence.
[2] Show Solution
$$u_{60} = u_1 + 59d = 8 + 59(4) = 244$$
b.
Calculate the sixth term of the geometric sequence.
[2] Show Solution
The geometric sequence starts $8, 20, \ldots$, so $r = \frac{20}{8} = 2.5$. $$v_6 = v_1 r^5 = 8 \times (2.5)^5 = 781.2$$
QUESTION 5 [6 marks]
Medium
Consider the following sequence of figures made of connected line segments, where each new figure adds a fixed number of segments to the previous one. Figure 1 contains 7 line segments, and each subsequent figure adds 5 more segments than the previous.
a.
Given that Figure $n$ contains 122 line segments, show that $n = 24$.
[3] Show Solution
Segments in Figure $n$: $u_n = 7 + (n-1)(5)$. Setting $u_n = 122$: $$7+(n-1)(5)=122 \Rightarrow n = 24$$
b.
Find the total number of line segments in the first 20 figures.
[3] Show Solution
$$S_{20} = \frac{20}{2}(2(7)+(20-1)(5)) = 1090$$
QUESTION 6 [6 marks]
Medium
Maya walks to school each morning. During the first minute, she travels 150 metres. In each subsequent minute, she travels 6 metres less than the distance travelled during the previous minute. The distance from home to school is 800 metres. Maya leaves home at 8:05 am and must be at school by 8:15 am. Will Maya arrive on time? Justify your answer.
Show Solution
This is an arithmetic series, first term 150, common difference -6. Over 10 minutes: $$S_{10} = \frac{10}{2}(2(150)+(10-1)(-6)) = 1230 \text{ m}$$ Since 1230 ? 800 m, Maya will arrive on time.
QUESTION 7 [6 marks]
Medium
The sum of the first three terms of a geometric sequence is 414.5, and the sum of the infinite sequence is 571.4. Find the common ratio.
Show Solution
Let the ratio be $r$. $$u_1(1+r+r^2)=414.5 \quad \text{and} \quad \frac{u_1}{1-r}=571.4$$ Solving simultaneously gives $r=0.65$.
QUESTION 8 [6 marks]
Easy
A geometric sequence has $u_1 = 20$, $u_2 = 40$, $u_3 = 80$.
a.
Find the common ratio, $r$.
[2] Show Solution
$$r = \frac{u_2}{u_1} = \frac{40}{20} = 2$$
b.
Find $u_{6}$.
[2] Show Solution
$$u_{6} = u_1 r^{n-1} = 20 \times \left(2\right)^{5} = 640$$
c.
Find $S_{7}$.
[2] Show Solution
$$S_{7} = \frac{u_1(r^n-1)}{r-1} = 2540$$
QUESTION 9 [6 marks]
Easy
In an arithmetic sequence, $u_{3} = -8$, $u_{4} = -13$.
a.
Find the common difference.
[2] Show Solution
$$d = \frac{u_{4} - u_{3}}{4-3} = \frac{-13 - (-8)}{1} = -5$$
b.
Find the first term.
[2] Show Solution
$$u_1 = u_{3} - (3-1)d = -8 - (2)(-5) = 2$$
c.
Find the sum of the first 100 terms.
[2] Show Solution
$$S_{100} = \frac{n}{2}(2u_1+(n-1)d) = -24550$$
QUESTION 10 [4 marks]
Medium
The first term of an arithmetic sequence is 10 and the common difference is 4. The first and fourth terms of this arithmetic sequence are the first two terms of a geometric sequence.
a.
Find $u_{60}$, the 60th term of the arithmetic sequence.
[2] Show Solution
$$u_{60} = u_1 + 59d = 10 + 59(4) = 246$$
b.
Calculate the sixth term of the geometric sequence.
[2] Show Solution
The geometric sequence starts $10, 22, \ldots$, so $r = \frac{22}{10} = 2.2$. $$v_6 = v_1 r^5 = 10 \times (2.2)^5 = 515.4$$
QUESTION 11 [6 marks]
Medium
The sum of the first three terms of a geometric sequence is 277.5, and the sum of the infinite sequence is 480. Find the common ratio.
Show Solution
Let the common ratio be $r$. $$u_1(1+r+r^2) = 277.5 \quad \text{and} \quad \frac{u_1}{1-r} = 480$$ Solving simultaneously gives $r = 0.75$.
QUESTION 12 [5 marks]
Hard
The first two terms of an infinite geometric sequence, in order, are $$3\log_5 x, \quad 2\log_5 x, \quad \text{where } x > 0.$$
a.
Find the common ratio, $r$.
[2] Show Solution
$$r = \frac{2\log_5 x}{3\log_5 x} = \frac{2}{3}$$
b.
Show that the sum of the infinite sequence is $9\log_{5} x$.
[3] Show Solution
$$S_\infty = \frac{u_1}{1-r} = \frac{3\log_5 x}{1-\frac{2}{3}} = \frac{3\log_5 x}{\frac{1}{3}} = 9\log_5 x$$
QUESTION 13 [6 marks]
Easy
A geometric sequence has $u_1 = 7$, $u_2 = 4.667$, $u_3 = 3.111$.
a.
Find the common ratio, $r$.
[2] Show Solution
$$r = \frac{4.667}{7} = \frac{2}{3}$$
b.
Find $u_{6}$.
[2] Show Solution
$$u_{6} = 7\times\left(\frac{2}{3}\right)^{5} = 0.9218$$
c.
Find $S_\infty$.
[2] Show Solution
$$S_\infty = \frac{7}{1-(\frac{2}{3})} = 21$$
QUESTION 14 [6 marks]
Easy
Only one of the following four sequences is arithmetic and only one of them is geometric: $$d_n = 11, 23, 24, 27, \ldots,\quad c_n = 2, 1, 0.5, 0.25, \ldots,\quad b_n = 1, 4, 9, 16, \ldots,\quad a_n = 3, 5, 7, 9, \ldots$$
a.
State which sequence is arithmetic and find the common difference.
[2] Show Solution
Sequence $a_n$ is arithmetic, with common difference $d = 2$.
b.
State which sequence is geometric and find the common ratio.
[2] Show Solution
Sequence $c_n$ is geometric, with common ratio $r = \frac{1}{2}$.
c.
For the geometric sequence, find the exact value of the 6th term. Give your answer as a fraction.
[2] Show Solution
$$c_{6} = 2 \times \left(\frac{1}{2}\right)^{5} = \frac{1}{16}$$
QUESTION 15 [5 marks]
Medium
An arithmetic sequence has first term 40 and common difference -1.
a.
Given that the $k$th term of the sequence is zero, find the value of $k$.
[2] Show Solution
$$0 = u_1+(k-1)d \Rightarrow k = 1 - \frac{u_1}{d} = 41$$
b.
Let $S_n$ denote the sum of the first $n$ terms. Find the largest value of $n$ such that $S_n > 100$.
[3] Show Solution
$$S_n = \frac{n}{2}(2u_1+(n-1)d) = -0.5n^2 + 40.5n$$ Setting up the inequality $$-0.5n^2 + 40.5n > 100$$ and solving (using the quadratic formula, then checking nearby integers since $n$ must be a positive integer), the largest value of $n$ satisfying this is $$n = 78$$
QUESTION 16 [4 marks]
Medium
The first term of an arithmetic sequence is 10 and the common difference is 2. The first and fourth terms of this arithmetic sequence are the first two terms of a geometric sequence.
a.
Find $u_{60}$, the 60th term of the arithmetic sequence.
[2] Show Solution
$$u_{60} = u_1 + 59d = 10 + 59(2) = 128$$
b.
Calculate the sixth term of the geometric sequence.
[2] Show Solution
The geometric sequence starts $10, 16, \ldots$, so $r = \frac{16}{10} = 1.6$. $$v_6 = v_1 r^5 = 10 \times (1.6)^5 = 104.9$$
QUESTION 17 [6 marks]
Easy
A geometric sequence has $u_1 = 12$, $u_2 = -6$, $u_3 = 3$.
a.
Find the common ratio, $r$.
[2] Show Solution
$$r = \frac{u_2}{u_1} = \frac{-6}{12} = \frac{-1}{2}$$
b.
Find $u_{7}$.
[2] Show Solution
$$u_{7} = u_1 r^{n-1} = 12 \times \left(\frac{-1}{2}\right)^{6} = 0.1875$$
c.
Find the sum to infinity, $S_\infty$.
[2] Show Solution
$$S_\infty = \frac{u_1}{1-r} = \frac{12}{1-(\frac{-1}{2})} = 8$$
QUESTION 18 [6 marks]
Easy
In an arithmetic sequence, $u_{4} = -11$, $u_{5} = -4$.
a.
Find the common difference.
[2] Show Solution
$$d = \frac{u_{5} - u_{4}}{5-4} = \frac{-4 - (-11)}{1} = 7$$
b.
Find the first term.
[2] Show Solution
$$u_1 = u_{4} - (4-1)d = -11 - (3)(7) = -32$$
c.
Find the sum of the first 25 terms.
[2] Show Solution
$$S_{25} = \frac{n}{2}(2u_1+(n-1)d) = 1300$$
QUESTION 19 [6 marks]
Easy
An arithmetic sequence has $u_1 = 24$, $u_2 = 17$, $u_3 = 10$.
a.
Find the common difference, $d$.
[2] Show Solution
$$d = u_2 - u_1 = 17 - (24) = -7$$
b.
Find $u_{9}$.
[2] Show Solution
$$u_{9} = u_1 + (n-1)d = 24 + (9-1)(-7) = -32$$
c.
Find $S_{8}$.
[2] Show Solution
$$S_{8} = \frac{n}{2}(2u_1+(n-1)d) = \frac{8}{2}(2(24)+(8-1)(-7)) = -4$$
QUESTION 20 [6 marks]
Easy
In an arithmetic sequence, $u_{5} = -14$, $u_{11} = 4$.
a.
Find the common difference.
[2] Show Solution
$$d = \frac{u_{11} - u_{5}}{11-5} = \frac{4 - (-14)}{6} = 3$$
b.
Find the first term.
[2] Show Solution
$$u_1 = u_{5} - (5-1)d = -14 - (4)(3) = -26$$
c.
Find the sum of the first 15 terms.
[2] Show Solution
$$S_{15} = \frac{n}{2}(2u_1+(n-1)d) = -75$$
QUESTION 21 [6 marks]
Easy
An arithmetic sequence has $u_1 = -5$, $u_2 = -14$, $u_3 = -23$.
a.
Find the common difference, $d$.
[2] Show Solution
$$d = u_2-u_1 = -14-(-5) = -9$$
b.
Find $u_{11}$.
[2] Show Solution
$$u_{11} = -5+(11-1)(-9) = -95$$
c.
Find $S_{11}$.
[2] Show Solution
$$S_{11} = \frac{11}{2}(2(-5)+(11-1)(-9)) = -550$$
QUESTION 22 [5 marks]
Medium
An arithmetic sequence has first term 20 and common difference -1.5.
a.
Given that the $k$th term of the sequence is zero, find the value of $k$.
[2] Show Solution
$$0 = u_1+(k-1)d \Rightarrow k = 1 - \frac{u_1}{d} = 14.33$$
b.
Let $S_n$ denote the sum of the first $n$ terms. Find the largest value of $n$ such that $S_n > 100$.
[3] Show Solution
$$S_n = \frac{n}{2}(2u_1+(n-1)d) = -0.75n^2 + 20.75n$$ Setting up the inequality $$-0.75n^2 + 20.75n > 100$$ and solving (using the quadratic formula, then checking nearby integers since $n$ must be a positive integer), the largest value of $n$ satisfying this is $$n = 21$$
QUESTION 23 [8 marks]
Hard
The sum, $S_n$, of the first $n$ terms of a geometric sequence, whose $n$th term is $u_n$, is given by $$S_n = \frac{5^n - a^n}{5^n}, \text{ where } a > 0.$$
a.
Find an expression for $u_n$.
[3] Show Solution
$$u_n = S_n - S_{n-1} = \left(1-\left(\frac{a}{5}\right)^n\right) - \left(1-\left(\frac{a}{5}\right)^{n-1}\right) = \left(\frac{a}{5}\right)^{n-1}\left(1-\frac{a}{5}\right)$$
b.
Find the first term and common ratio of the sequence.
[2] Show Solution
First term: $u_1 = 1-\frac{a}{5}$. Common ratio: $r = \frac{a}{5}$.
c.
(i) Determine the values of $a$ such that the sum to infinity exists. (ii) Find the sum to infinity when it exists.
[3] Show Solution
(i) $S_\infty$ exists when $|r|<1$, i.e. $\left|\frac{a}{5}\right|<1$. Since $a>0$, this means $0
QUESTION 24 [4 marks]
Medium
The first term of an arithmetic sequence is 9 and the common difference is 5. The first and fourth terms of this arithmetic sequence are the first two terms of a geometric sequence.
a.
Find $u_{60}$, the 60th term of the arithmetic sequence.
[2] Show Solution
$$u_{60} = 9+59(5) = 304$$
b.
Calculate the sixth term of the geometric sequence.
[2] Show Solution
$r = \frac{24}{9} = 2.667$. $$v_6 = 9\times(2.667)^5 = 1214$$
QUESTION 25 [5 marks]
Hard
The largest rectangle has dimensions 6 cm by 3 cm. A smaller rectangle is formed inside it with half the area, and this process is repeated indefinitely, with every other ring of the pattern shaded (similar to the pattern shown).
a.
Find the total area of the first 3 shaded regions.
[3] Show Solution
Each shaded ring's area follows a geometric sequence with first term $\frac{18}{4} = 4.5$ cm$^2$ and ratio $\frac{1}{2}$. $$S_{3} = \frac{4.5(1-0.5^{3})}{0.5} = 7.875\text{ cm}^2$$
b.
If the process is repeated indefinitely, find the total area of the shaded regions.
[2] Show Solution
$$S_\infty = \frac{4.5}{0.5} = 9\text{ cm}^2$$
QUESTION 26 [5 marks]
Medium
A ball is dropped from a height of 1.5 m onto a concrete floor. After hitting the floor, the ball rebounds back up to 85% of its previous height, and this pattern continues indefinitely.
a.
Show that the total distance travelled by the ball can be expressed as $$1.5 + 2(1.5)(0.85) + 2(1.5)(0.85)^2 + \cdots$$
[2] Show Solution
After falling 1.5 m, the ball rebounds to 1.5×0.85 m and falls that same distance again (up and down), giving $2(1.5)(0.85)$ for that bounce, then $2(1.5)(0.85)^2$ for the next, and so on.
b.
Find the total distance travelled by the ball.
[3] Show Solution
Total $$= 1.5 + \frac{2(1.5)(0.85)}{1-0.85} = 18.5\text{ m}$$
QUESTION 27 [7 marks]
Hard
The sides of a square are 10 cm long. A new square is formed by joining the midpoints of the sides of the original square, and two of the four resulting corner triangles are shaded. This process is repeated 5 more times (each time joining midpoints of the previous inner square), forming a spiral pattern of shaded triangles.
a.
Show that the area shaded at each stage forms a geometric sequence with common ratio $\frac{1}{2}$, and find the first term.
[3] Show Solution
Joining the midpoints of a square with area $A$ creates an inner square of area $\frac{A}{2}$, leaving 4 equal corner triangles totalling area $\frac{A}{2}$ (so shading 2 of them shades $\frac{A}{4}$). Since the area being subdivided halves at every stage, the shaded amount at each stage also halves — a geometric sequence with ratio $\frac{1}{2}$. First term (from the original square, area $100$): $$u_1 = \frac{100}{4} = 25\text{ cm}^2$$
b.
Find the total area of the shaded region after all 6 stages (the original subdivision plus 5 repeats).
[2] Show Solution
$$S_{6} = \frac{u_1(1-r^{6})}{1-r} = \frac{25(1-0.5^{6})}{0.5} = 49.22\text{ cm}^2$$
c.
Find the total area of the shaded region if the process is repeated indefinitely.
[2] Show Solution
$$S_\infty = \frac{u_1}{1-r} = \frac{25}{0.5} = 50\text{ cm}^2$$
QUESTION 28 [6 marks]
Easy
An arithmetic sequence has $u_1 = -28$, $u_2 = -21$, $u_3 = -14$.
a.
Find the common difference, $d$.
[2] Show Solution
$$d = u_2 - u_1 = -21 - (-28) = 7$$
b.
Find $u_{9}$.
[2] Show Solution
$$u_{9} = u_1 + (n-1)d = -28 + (9-1)(7) = 28$$
c.
Find $S_{8}$.
[2] Show Solution
$$S_{8} = \frac{n}{2}(2u_1+(n-1)d) = \frac{8}{2}(2(-28)+(8-1)(7)) = -28$$
QUESTION 29 [6 marks]
Easy
A geometric sequence has $u_1 = 20$, $u_2 = 5$, $u_3 = 1.25$.
a.
Find the common ratio, $r$.
[2] Show Solution
$$r = \frac{u_2}{u_1} = \frac{5}{20} = \frac{1}{4}$$
b.
Find $u_{9}$.
[2] Show Solution
$$u_{9} = u_1 r^{n-1} = 20 \times \left(\frac{1}{4}\right)^{8} = 0.0003052$$
c.
Find the sum to infinity, $S_\infty$.
[2] Show Solution
$$S_\infty = \frac{u_1}{1-r} = \frac{20}{1-(\frac{1}{4})} = 26.67$$
QUESTION 30 [6 marks]
Easy
An arithmetic sequence has $u_1 = -40$, $u_2 = -32$, $u_3 = -24$.
a.
Find the common difference, $d$.
[2] Show Solution
$$d = u_2 - u_1 = -32 - (-40) = 8$$
b.
Find $u_{12}$.
[2] Show Solution
$$u_{12} = u_1 + (n-1)d = -40 + (12-1)(8) = 48$$
c.
Find $S_{12}$.
[2] Show Solution
$$S_{12} = \frac{n}{2}(2u_1+(n-1)d) = \frac{12}{2}(2(-40)+(12-1)(8)) = 48$$
QUESTION 31 [6 marks]
Medium
The first term and the common ratio of a geometric series are denoted, respectively, by $u_1$ and $r$, where $u_1, r \in \mathbb{Q}$. Given that the fourth term is 40 and the sum to infinity is 481.7, find the value of $u_1$ and the value of $r$.
Show Solution
From $u_1 r^3 = 40$ and $\frac{u_1}{1-r} = 481.7$: solving simultaneously gives $r \approx 0.5848$ and $u_1 = 200$ (verify: $u_1(1-r) = 481.7$, checks out).
QUESTION 32 [6 marks]
Easy
In an arithmetic sequence, $u_{9} = 10$, $u_{11} = 16$.
a.
Find the common difference.
[2] Show Solution
$$d = \frac{u_{11} - u_{9}}{11-9} = \frac{16 - (10)}{2} = 3$$
b.
Find the first term.
[2] Show Solution
$$u_1 = u_{9} - (9-1)d = 10 - (8)(3) = -14$$
c.
Find the sum of the first 25 terms.
[2] Show Solution
$$S_{25} = \frac{n}{2}(2u_1+(n-1)d) = 550$$
QUESTION 33 [6 marks]
Easy
The table shows the first four terms of three sequences: $u_n$, $v_n$, and $w_n$.
| n | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| $u_n$ | 24 | 36 | 48 | 60 |
| $v_n$ | 24 | 48 | 96 | 192 |
| $w_n$ | 24 | 36 | 50 | 66 |
a.
State which sequence is (i) arithmetic; (ii) geometric.
[2] Show Solution
$u_n$ is arithmetic (constant difference of $12$). $v_n$ is geometric (constant ratio of $2$).
b.
Find the sum of the first 20 terms of the arithmetic sequence.
[2] Show Solution
$$S_{20} = \frac{n}{2}(2u_1+(n-1)d) = 2760$$
c.
Find the exact value of the 12th term of the geometric sequence.
[2] Show Solution
$$v_{12} = u_1 r^{n-1} = 24 \times 2^{11} = 49152$$
QUESTION 34 [6 marks]
Easy
An arithmetic sequence has $u_1 = -28$, $u_2 = -20$, $u_3 = -12$.
a.
Find the common difference, $d$.
[2] Show Solution
$$d = u_2 - u_1 = -20 - (-28) = 8$$
b.
Find $u_{12}$.
[2] Show Solution
$$u_{12} = u_1 + (n-1)d = -28 + (12-1)(8) = 60$$
c.
Find $S_{12}$.
[2] Show Solution
$$S_{12} = \frac{n}{2}(2u_1+(n-1)d) = \frac{12}{2}(2(-28)+(12-1)(8)) = 192$$
QUESTION 35 [6 marks]
Easy
A geometric sequence has $u_1 = 2$, $u_2 = 4$, $u_3 = 8$.
a.
Find the common ratio, $r$.
[2] Show Solution
$$r = \frac{u_2}{u_1} = \frac{4}{2} = 2$$
b.
Find $u_{8}$.
[2] Show Solution
$$u_{8} = u_1 r^{n-1} = 2 \times \left(2\right)^{7} = 256$$
c.
Find $S_{7}$.
[2] Show Solution
$$S_{7} = \frac{u_1(r^n-1)}{r-1} = 254$$
QUESTION 36 [6 marks]
Medium
Consider the sum $$S = \sum_{k=5}^{l} (3k-1), \text{ where } l \text{ is a positive integer greater than } 5.$$
a.
Write down the first three terms of the series.
[2] Show Solution
$$14, 17, 20, \ldots$$
b.
Write down the number of terms in the series, in terms of $l$.
[1] Show Solution
$$\text{number of terms} = l - 5 + 1$$
c.
Given that $S = 711$, find the value of $l$.
[3] Show Solution
Using $S_n = \frac{n}{2}(u_1+u_n)$ with $u_1=14$, and solving for $l$ gives $$l = 22$$
QUESTION 37 [6 marks]
Medium
The first term and the common ratio of a geometric series are denoted, respectively, by $u_1$ and $r$, where $u_1, r \in \mathbb{Q}$. Given that the fourth term is 30 and the sum to infinity is 426.7, find the value of $u_1$ and the value of $r$.
Show Solution
From $u_1 r^3 = 30$ and $\frac{u_1}{1-r} = 426.7$: solving simultaneously gives $r \approx 0.5313$ and $u_1 = 200$ (verify: $u_1(1-r) = 426.7$, checks out).
QUESTION 38 [6 marks]
Easy
An arithmetic sequence has $u_1 = 40$, $u_2 = 47$, $u_3 = 54$.
a.
Find the common difference, $d$.
[2] Show Solution
$$d = u_2 - u_1 = 47 - (40) = 7$$
b.
Find $u_{8}$.
[2] Show Solution
$$u_{8} = u_1 + (n-1)d = 40 + (8-1)(7) = 89$$
c.
Find $S_{15}$.
[2] Show Solution
$$S_{15} = \frac{n}{2}(2u_1+(n-1)d) = \frac{15}{2}(2(40)+(15-1)(7)) = 1335$$
QUESTION 39 [6 marks]
Easy
An arithmetic sequence has $u_1 = 40$, $u_2 = 44$, $u_3 = 48$.
a.
Find the common difference, $d$.
[2] Show Solution
$$d = u_2 - u_1 = 44 - (40) = 4$$
b.
Find $u_{8}$.
[2] Show Solution
$$u_{8} = u_1 + (n-1)d = 40 + (8-1)(4) = 68$$
c.
Find $S_{15}$.
[2] Show Solution
$$S_{15} = \frac{n}{2}(2u_1+(n-1)d) = \frac{15}{2}(2(40)+(15-1)(4)) = 1020$$
QUESTION 40 [6 marks]
Medium
Maya walks to school each morning. During the first minute, she travels 160 metres. In each subsequent minute, she travels 6 metres less than the distance travelled during the previous minute. The distance from home to school is 1000 metres. Maya leaves home at 8:05 am and must be at school by 8:15 am. Will Maya arrive on time? Justify your answer.
Show Solution
This is an arithmetic series, first term 160, common difference -6. Over 10 minutes: $$S_{10} = \frac{10}{2}(2(160)+(10-1)(-6)) = 1330 \text{ m}$$ Since 1330 ? 1000 m, Maya will arrive on time.
QUESTION 41 [6 marks]
Easy
In an arithmetic sequence, $u_{3} = -14$, $u_{9} = 4$.
a.
Find the common difference.
[2] Show Solution
$$d = \frac{u_{9} - u_{3}}{9-3} = \frac{4 - (-14)}{6} = 3$$
b.
Find the first term.
[2] Show Solution
$$u_1 = u_{3} - (3-1)d = -14 - (2)(3) = -20$$
c.
Find the sum of the first 15 terms.
[2] Show Solution
$$S_{15} = \frac{n}{2}(2u_1+(n-1)d) = 15$$
QUESTION 42 [6 marks]
Easy
Only one of the following four sequences is arithmetic and only one of them is geometric: $$c_n = 4, 12, 36, 108, \ldots,\quad a_n = 3, 7, 11, 15, \ldots,\quad d_n = 2, 5, 10, 14, \ldots,\quad b_n = 1, 4, 9, 16, \ldots$$
a.
State which sequence is arithmetic and find the common difference.
[2] Show Solution
Sequence $a_n$ is arithmetic, with common difference $d = 4$.
b.
State which sequence is geometric and find the common ratio.
[2] Show Solution
Sequence $c_n$ is geometric, with common ratio $r = 3$.
c.
For the geometric sequence, find the exact value of the 8th term. Give your answer as a fraction.
[2] Show Solution
$$c_{8} = 4 \times \left(3\right)^{7} = 8748$$
QUESTION 43 [9 marks]
Medium
Diego is planning to eat a tub of ice cream, one spoonful at a time. He believes the mass of ice cream remaining can be modelled by an arithmetic sequence. He puts the tub on a set of scales and records the following data.
| Number of spoonfuls | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Mass remaining (g) | 307 | 285 | 266 | 241 | 219 | 197 |
a.
Find an estimate of the common difference, by using an average of the differences.
[2] Show Solution
Average of consecutive differences $= -22.0$ g per spoonful.
b.
Find a model for the $n$th term $u_n$ (mass remaining after $n$ spoonfuls).
[2] Show Solution
$$u_n = 307 + (n-1)(-22.0)$$
c.
Use your model to estimate the original mass of ice cream in the tub (before any spoonfuls), and give a reason to support whether this is an over- or under-estimate.
[3] Show Solution
Extrapolating one step back ($n=0$): $$u_0 = 307 - (-22.0) = 329\text{ g}$$ This is a reasonably accurate estimate, since the first actual difference (-22 g) matches the average difference used in the model almost exactly, so there's no strong evidence of bias in either direction.
d.
Use your model to estimate the number of spoonfuls required to eat all the ice cream in the tub.
[2] Show Solution
Setting $u_n = 0$: $$0 = 307 + (n-1)(-22.0) \Rightarrow n \approx 14.95$$ so approximately 15 spoonfuls.
QUESTION 44 [6 marks]
Easy
A geometric sequence has $u_1 = 12$, $u_2 = 4$, $u_3 = 1.333$.
a.
Find the common ratio, $r$.
[2] Show Solution
$$r = \frac{u_2}{u_1} = \frac{4}{12} = \frac{1}{3}$$
b.
Find $u_{9}$.
[2] Show Solution
$$u_{9} = u_1 r^{n-1} = 12 \times \left(\frac{1}{3}\right)^{8} = 0.001829$$
c.
Find the sum to infinity, $S_\infty$.
[2] Show Solution
$$S_\infty = \frac{u_1}{1-r} = \frac{12}{1-(\frac{1}{3})} = 18$$
QUESTION 45 [5 marks]
Medium
A ball is dropped from a height of 2 m onto a concrete floor. After hitting the floor, the ball rebounds back up to 75% of its previous height, and this pattern continues indefinitely.
a.
Show that the total distance travelled by the ball can be expressed as $$2 + 2(2)(0.75) + 2(2)(0.75)^2 + \cdots$$
[2] Show Solution
After falling 2 m, the ball rebounds to 2×0.75 m and falls that same distance again (up and down), giving $2(2)(0.75)$ for that bounce, then $2(2)(0.75)^2$ for the next, and so on.
b.
Find the total distance travelled by the ball.
[3] Show Solution
Total $$= 2 + \frac{2(2)(0.75)}{1-0.75} = 14\text{ m}$$
QUESTION 46 [6 marks]
Easy
The table shows the first four terms of three sequences: $u_n$, $v_n$, and $w_n$.
| n | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| $u_n$ | 12 | 22 | 32 | 42 |
| $v_n$ | 12 | 24 | 48 | 96 |
| $w_n$ | 12 | 22 | 34 | 52 |
a.
State which sequence is (i) arithmetic; (ii) geometric.
[2] Show Solution
$u_n$ is arithmetic (constant difference of $10$). $v_n$ is geometric (constant ratio of $2$).
b.
Find the sum of the first 30 terms of the arithmetic sequence.
[2] Show Solution
$$S_{30} = \frac{n}{2}(2u_1+(n-1)d) = 4710$$
c.
Find the exact value of the 12th term of the geometric sequence.
[2] Show Solution
$$v_{12} = u_1 r^{n-1} = 12 \times 2^{11} = 24576$$
QUESTION 47 [6 marks]
Medium
Consider the following sequence of figures made of connected line segments, where each new figure adds a fixed number of segments to the previous one. Figure 1 contains 6 line segments, and each subsequent figure adds 5 more segments than the previous.
a.
Given that Figure $n$ contains 81 line segments, show that $n = 16$.
[3] Show Solution
Segments in Figure $n$: $u_n = 6 + (n-1)(5)$. Setting $u_n = 81$: $$6+(n-1)(5)=81 \Rightarrow n = 16$$
b.
Find the total number of line segments in the first 20 figures.
[3] Show Solution
$$S_{20} = \frac{20}{2}(2(6)+(20-1)(5)) = 1070$$
QUESTION 48 [5 marks]
Medium
A rope of length 75 metres is cut into $n$ pieces of increasing lengths that form an arithmetic sequence with a common difference of $d$ metres. Given that the lengths of the shortest and longest pieces are 2 metres and 8 metres respectively, find the values of $n$ and $d$.
Show Solution
Sum of an arithmetic sequence: $S_n = \frac{n}{2}(u_1+u_n)$. Here $u_1=2$, $u_n=8$, and $S_n=75$: $$75 = \frac{n}{2}(2+8) \Rightarrow n = 15$$ Then using $u_n = u_1+(n-1)d$: $$8 = 2+(15-1)d \Rightarrow d = 0.4286\text{ m}$$
QUESTION 49 [6 marks]
Easy
Only one of the following four sequences is arithmetic and only one of them is geometric: $$c_n = 4, 2, 1, 0.5, \ldots,\quad b_n = 1, 4, 9, 16, \ldots,\quad d_n = 20, 25, 26, 28, \ldots,\quad a_n = 2, 7, 12, 17, \ldots$$
a.
State which sequence is arithmetic and find the common difference.
[2] Show Solution
Sequence $a_n$ is arithmetic, with common difference $d = 5$.
b.
State which sequence is geometric and find the common ratio.
[2] Show Solution
Sequence $c_n$ is geometric, with common ratio $r = \frac{1}{2}$.
c.
For the geometric sequence, find the exact value of the 8th term. Give your answer as a fraction.
[2] Show Solution
$$c_{8} = 4 \times \left(\frac{1}{2}\right)^{7} = \frac{1}{32}$$
QUESTION 50 [5 marks]
Medium
An arithmetic sequence has first term 45 and common difference -2.5.
a.
Given that the $k$th term of the sequence is zero, find the value of $k$.
[2] Show Solution
$$0 = u_1+(k-1)d \Rightarrow k = 1 - \frac{u_1}{d} = 19$$
b.
Let $S_n$ denote the sum of the first $n$ terms. Find the smallest value of $n$ such that $S_n < 0$.
[3] Show Solution
$$S_n = \frac{n}{2}(2u_1+(n-1)d) = -1.25n^2 + 46.25n$$ Setting up the inequality $$-1.25n^2 + 46.25n < 0$$ and solving (using the quadratic formula, then checking nearby integers since $n$ must be a positive integer), the smallest value of $n$ satisfying this is $$n = 38$$
QUESTION 51 [6 marks]
Easy
A geometric sequence has $u_1 = 9$, $u_2 = 2.25$, $u_3 = 0.5625$.
a.
Find the common ratio, $r$.
[2] Show Solution
$$r = \frac{2.25}{9} = \frac{1}{4}$$
b.
Find $u_{7}$.
[2] Show Solution
$$u_{7} = 9\times\left(\frac{1}{4}\right)^{6} = 0.002197$$
c.
Find $S_\infty$.
[2] Show Solution
$$S_\infty = \frac{9}{1-(\frac{1}{4})} = 12$$
QUESTION 52 [6 marks]
Easy
An arithmetic sequence has $u_1 = 55$, $u_2 = 53$, $u_3 = 51$.
a.
Find the common difference, $d$.
[2] Show Solution
$$d = u_2-u_1 = 53-(55) = -2$$
b.
Find $u_{9}$.
[2] Show Solution
$$u_{9} = 55+(9-1)(-2) = 39$$
c.
Find $S_{14}$.
[2] Show Solution
$$S_{14} = \frac{14}{2}(2(55)+(14-1)(-2)) = 588$$
QUESTION 53 [6 marks]
Easy
A geometric sequence has $u_1 = 12$, $u_2 = 18$, $u_3 = 27$.
a.
Find the common ratio, $r$.
[2] Show Solution
$$r = \frac{u_2}{u_1} = \frac{18}{12} = \frac{3}{2}$$
b.
Find $u_{9}$.
[2] Show Solution
$$u_{9} = u_1 r^{n-1} = 12 \times \left(\frac{3}{2}\right)^{8} = 307.5$$
c.
Find $S_{7}$.
[2] Show Solution
$$S_{7} = \frac{u_1(r^n-1)}{r-1} = 386.1$$
QUESTION 54 [6 marks]
Medium
A ball is dropped from a height of 20 m. Every time it hits the ground it bounces back to 81\% of its previous height.
a.
Find the maximum height it reaches after the 10th bounce.
[3] Show Solution
$$h_{10} = 20 \times (0.81)^{10} = 2.432\text{ m}$$
b.
Find the total distance travelled by the ball until it comes to rest. (Assume no friction and no loss of elasticity.)
[3] Show Solution
$$\text{Total} = 20 + \frac{2(20)(0.81)}{1-0.81} = 190.5\text{ m}$$
QUESTION 55 [6 marks]
Medium
Leo cycles to the gym each morning. During the first minute, he travels 190 metres. In each subsequent minute, he travels 85% of the distance travelled during the previous minute. The distance from home to the gym is 1700 metres. Leo leaves home at 6:40 am and must arrive by 6:52 am. Will Leo arrive on time? Justify your answer.
Show Solution
This is a geometric series with first term 190 and ratio 0.85. After 12 minutes, total distance travelled $= \frac{190(1-0.85^{12})}{1-0.85} \approx 1086$ m. Since 1086 < 1700, Leo will NOT arrive on time.
QUESTION 56 [6 marks]
Medium
Consider the sum $$S = \sum_{k=4}^{l} (2k-1), \text{ where } l \text{ is a positive integer greater than } 4.$$
a.
Write down the first three terms of the series.
[2] Show Solution
$$7, 9, 11, \ldots$$
b.
Write down the number of terms in the series, in terms of $l$.
[1] Show Solution
$$\text{number of terms} = l - 4 + 1$$
c.
Given that $S = 315$, find the value of $l$.
[3] Show Solution
Using $S_n = \frac{n}{2}(u_1+u_n)$ with $u_1=7$, and solving for $l$ gives $$l = 18$$
QUESTION 57 [9 marks]
Medium
An experiment was undertaken to investigate the relationship between the length of a spring and the mass hanging from it. The table shows the extension of the spring (cm) for each mass (g).
It is believed the data can be modelled by an arithmetic sequence, according to Hooke's Law.
| Mass (g) | 12 | 24 | 36 | 48 | 60 | 72 | 84 | 96 |
|---|---|---|---|---|---|---|---|---|
| Extension (cm) | 3.5 | 6.91 | 10.39 | 13.87 | 17.45 | 20.83 | 24.44 | 27.92 |
a.
Find an estimate of the common difference, by using an average of the differences.
[3] Show Solution
Consecutive differences: $3.41, 3.48, 3.48, 3.58, 3.38, 3.61, 3.48$. Average $= 3.489$ cm.
b.
Find a model for the $n$th term $u_n$, where $n = \dfrac{\text{mass}}{12}$.
[3] Show Solution
$$u_n = 3.5 + (n-1)(3.489)$$
c.
(i) Use your model to estimate the extension for a mass of 72 g. (ii) Calculate the percentage error in this estimate.
[3] Show Solution
(i) $n = 72/12 = 6$. $$u_{6} = 3.5 + (6-1)(3.489) = 20.945\text{ cm}$$ (ii) Actual measured value at 72 g is 20.83 cm. Percentage error $= \left|\dfrac{20.945-20.83}{20.83}\right| \times 100 = 0.552\%$