Math AA SL

(a)

d = 5 - 2

{\boldsymbol{\color{Purple} d = 3}}

(b)

Using the nth term of an arithmetic sequence formula, we get that:

u_{11} = 2 + (11-1)(3) = 2 + 30

\boldsymbol{{\color{Purple} u_{11} = 32}}

(c)

Using the sum of n terms of an arithmetic sequence formula (the second version in the Formula Booklet), we get that:

S_{11} = \frac{11}{2}\left ( 2 + 32 \right ) = 11 \times 17

\boldsymbol{{\color{Purple} S_{11} = 187}}

.

(a) (i)

\boldsymbol{{\color{Purple} \textbf{Sequence 4}}}

(a) (ii)

\boldsymbol{{\color{Purple} \textbf{Sequence 1}}}

(b) (i)

d = 2 - 6

\boldsymbol{{\color{Purple} d = -4}}

(b) (ii)

Using the nthe term of an arithmetic sequence formula:

u_{15} = 6 + (15-1)(-4)

\boldsymbol{{\color{Purple} u_{15} = -50}}

(c) (i)

\boldsymbol{{\color{Purple} r = \frac{1}{3}}}

(c) (ii)

Using the sum of nterms of a finite geometric sequence formula:

S_{7} = \frac{9\left ( 1 - \left (  \frac{1}{3}\right )^{7} \right )}{1 - \frac{1}{3}}

\boldsymbol{{\color{Purple} S_{7} = \frac{1093}{81}}}

.

(a)

r = \frac{112}{-280}

\boldsymbol{{\color{Purple} r = -0.4}}

(b)

Using the nthe term of a geometric sequence formula:

u_{12} = -280 \times \left ( -0.4 \right )^{12-1}

\boldsymbol{{\color{Purple} u_{12} \approx 0.0117}}

(c)

Using the sum of an infinite geometric sequence formula:

S_{\infty } = \frac{-280}{1-(-0.4)}

\boldsymbol{{\color{Purple} S_{\infty } = -200}}

.

(a)

15 = \frac{6}{2} (u_{1} + (-5))

15 = 3u_{1} -15

\boldsymbol{{\color{Purple} u_{1} = 10}}

(b)

Finding the common difference:

-5 = 10 + (6-1)d

d = -3

Finding the tenth term:

u_{10} = 10 + (10 - 1)(-3)

\boldsymbol{{\color{Purple} u_{10} = -17}}

.

.(a) (i)

d = \frac{u_{5} - u_{3}}{2} = \frac{14-8}{2}

\boldsymbol{{\color{Purple} d = 3}}

(a) (ii)

u_{1} = u_{3} - 2d = 8 - 2(3)

\boldsymbol{{\color{Purple} u_{1} = 2}}

(b)

Using the nth term of an arithmetic sequence formula:

35 = 2 + (n - 1)(3)

35 = 3n - 1

\boldsymbol{{\color{Purple} n = 12}}

.

(a)

Using the compound interest formula (you can also use the Finance app on your calculator):

6000\left ( 1 + \frac{8.5}{100 \times 2} \right )^{2 \times 7}\approx

\boldsymbol{{\color{Purple} \approx 10745}}

(b)

Using the compound interest formula (you can also use the Finance app on your calculator):

13000 = 6000\left ( 1 + \frac{8.5}{100 \times 2} \right )^{2 \times n}

Using the calculator to solve for n:

n = 9.288, therefore

\boldsymbol{{\color{Purple} n = 10}}

(c)

Using the compound interest formula (you can also use the Finance app on your calculator):

14000 = 6000\left ( 1 + \frac{r}{100 \times 4} \right )^{4 \times 11}

Using the calculator to solve for r we get that:

\boldsymbol{{\color{Purple} r \approx 7.78}}

(d)

Adding 70\: \% of what was added the previous year means that the amounts added each year form a geometric sequence where r = 0.7.

Using the sum of an infinite geometric sequence formula:

S_{\infty} = \frac{4500}{1 - 0.7} = 15000

\boldsymbol{{\color{Purple} \textbf{Since}\:  15000 < 17500, \textbf{Alma will not reach her goal.}}}

(e)

Using the sum of n terms of a geometric sequence formula:

17500 = \frac{4500(1 - r^{20})}{1 - r}

Using the calculator to solve for r:

r \approx 0.743543, so:

\boldsymbol{{\color{Purple} x \approx 74.4}}

.

(a)

Using the compound interest formula (you can also use the TVM Solver inside the Finance app), we get:

FV = 3500\left ( 1 + \frac{1.87}{100(12)} \right )^{(12)(8)} \approx

\boldsymbol{{\color{Purple} \approx \$4064.32}}

(b)

4064.32 - 3500 =

\boldsymbol{{\color{Purple} = \$ 564.32}}

(c)

Using the TVM Solver inside the Finance app (you can also use the compound interest formula), we get:

N = 8 \times 4 = 32

I\, \% = 2.05

FV = 4064.32

P/Y = 4

C/Y = 4

Solving for PV we get that:

\boldsymbol{{\color{Purple} P \approx 3451.00}}

.

(a)

Using the nth term of an arithmetic sequence formula we get:

u_{n} = -86 + (n-1) \times 4

Substituting zero for u_{n} and solving for n we get:

0 = -86 + (n-1) \times 4

n = 22.5

Since n must be a whole number, the final answer is:

\boldsymbol{{\color{Purple} n = 23}}

(b)

Using the sum of the first n terms of a arithmetic sequence formula we get:

S_{n} = \frac{n}{2}\left ( 2(-86) + (n-1) \times 4 \right )

Finding the minimum value of this expression, for example, by graphing on the calculator, we get that the minimum value of S_{n} is:

\boldsymbol{{\color{Purple} -968}}

.

(a)

\boldsymbol{{\color{Purple} r = 0.7}}

(b)

u_{1} = S_{1}= \sum_{k = 1}^{1}3 \times \left ( 0.7 \right )^{k} = 3 \times (0.7)^{1}

 \boldsymbol{{\color{Purple} u_{1} = 2.1}}

(c)

Using the sum of n terms of a finite geometric sequence formula:

\frac{2.1(1 - 0.7^{n})}{1-0.7}<6.999

Using the calculator to solve (graphing, Numeric Solver, or table):

S_{24} = 6.9986589

S_{25} = 6.9990613 

Therefore:

\boldsymbol{{\color{Purple} n = 24}}

.

Using the sum of an infinite geometric sequence formula and we can write that:

100 = \frac{u_{1}}{1 - r}

Using the sum of n terms of a finite geometric sequence we can write that:

67.232 = \frac{u_{1}(1 - r^{5})}{1 - r}

Rearranging the first equation we get that:

u_{1} = 100(1 - r)

Substituting 100(1 - r) for u_{1} in the second equation we can write that:

67.232 = \frac{(100(1 - r))(1 - r^{5})}{1 - r}

Simplifying gives us:

67.232 = 100(1 - r^{5})

Solving for r either by graphing or using the Solver function on the calculator, we get  that:

\boldsymbol{{\color{Purple} r = 0.8}}

.

Using the nth term of a geometric sequence formula we get that:

u_{2} = 5r and u_{3} = 5r^{2}

Using the nth term of an arithmetic sequence formula we get that:

v_{4} = 5 + (4-1)d = 5 + 3d and v_{16} = 5 + (16-1)d = 5 + 15d

Since u_{2} = v_{4} and u_{3} = v_{16}, we can write that:

5r = 5 + 3d and 5r^{2} = 5 + 15d

Multiplying the first equation by 5 and subtracting the result from the second equation gives us:

5r^{2} - 25r = -20

Rearranging for zero and dividing by 5 we get that:

r^{2} - 5r + 4 = 0

Solving for r we get that:

(r - 4)(r - 1) = 0, and since r \neq 1

\boldsymbol{{\color{Purple} r = 4}}

Substituting 4 for r into 5r = 5 + 3d we get:

5(4) = 5 + 3d and solving for d we get that:

\boldsymbol{{\color{Purple} d = 5}}

.

(a) (i)

d = u_{2} - u_{1} = u_{3} - u_{2}

k\textup{log}_{a}x - 2\textup{log}_{a}x =  \textup{log}_{a}x - k\textup{log}_{a}x

k - 2 = 1 - k, therefore:

\boldsymbol{{\color{Purple} k = \frac{3}{2}}}

(a) (ii)

d = u_{2} - u_{1} = \frac{3}{2}\textup{log}_{a}x - 2\textup{log}_{a}x = -\frac{1}{2}\textup{log}_{a}x

\boldsymbol{{\color{Purple} m = -\frac{1}{2}}}

(a) (iii)

Substituting into the sum of the first n terms of an arithmetic sequence formula:

4 = \frac{8}{2}\left ( 2 \times 2\textup{log}_{\sqrt{3}}\, x + (8-1)\left ( -\frac{1}{2}\textup{log}_{\sqrt{3}}\, x \right ) \right )

Simplifying and solving for x:

1 = 4\textup{log}_{\sqrt{3}}\, x - 3.5\textup{log}_{\sqrt{3}}\, x

\textup{log}_{\sqrt{3}}\, x = 2

x = (\sqrt{3})^{2}, so:

\boldsymbol{{\color{Purple} x = 3}}

(b) (i)

r = \frac{u_{2}}{u_{1}} = \frac{u_{3}}{u_{2}}

\frac{k\textup{log}_{a}x}{2\textup{log}_{a}x} = \frac{\textup{log}_{a}x}{k\textup{log}_{a}x}

k^{2} = 2 \Rightarrow k = \pm \sqrt{2}

\boldsymbol{{\color{Purple} k = -\sqrt{2}}}

(b) (ii)

r = \frac{\textup{log}_{a}x}{-\sqrt{2}\textup{log}_{a}x} = -\frac{1}{\sqrt{2}}

Using the sum of an infinite geometric sequence formula:

\frac{5 \sqrt{2}}{1 + \sqrt{2}} = \frac{2\textup{log}_{a}x}{1 - \left ( -\frac{1}{\sqrt{2}} \right )}

Simplifying and multiplying the fraction on the right-hand side by \frac{\sqrt{2}}{\sqrt{2}}:

\frac{5 \sqrt{2}}{1 + \sqrt{2}} = \frac{2 \sqrt{2}\textup{log}_{a}x}{1 + \sqrt{2}}

\textup{log}_{a}x = \frac{5}{2}\Rightarrow x = a^{\frac{5}{2}}

\boldsymbol{{\color{Purple} a = x^{\frac{2}{5}}}}

.

(a)

Using the concept of the common difference:

d = u_{2} - u_{1} = u_{3} - u_{2} = u_{4} - u_{3}

\textup{log}_{b}m - \textup{log}_{b}32 = \textup{log}_{b}n - \textup{log}_{b}m = \textup{log}_{b}108 - \textup{log}_{b}n

Using the logarithmic law \textup{log}_{a}\frac{x}{y} = \textup{log}_{a}x - \textup{log}_{a}y:

\textup{log}_{b}\frac{m}{32} = \textup{log}_{b}\frac{n}{m} = \textup{log}_{b}\frac{108}{n}

Therefore \frac{m}{32} = \frac{n}{m} = \frac{108}{n} = r so 32\, , \,m\, , \,n and 108 are four consecutive terms of a geometric sequence.

(b)

Using the nth term of a geometric sequence formula:

u_{4} = u_{1} \times r^{3}

108 = 32 \times r^{3}\Rightarrow r = \frac{3}{2}

Based on what we found in part (a):

\frac{m}{32} = \frac{3}{2}\Rightarrow \boldsymbol{{\color{Purple} m = 48}}

\frac{108}{n} = \frac{3}{2}\Rightarrow \boldsymbol{{\color{Purple} n = 72}}

.

(a)

We can calculate a 7\, \% decrease by multiplying the initial number of raspberry bushes by 0.93.

So for the number of raspberry bushes at the end of 2024 we get:

700 \times 0.93 + 80

For the number of raspberry bushes at the end of 2025 we get:

(700 \times 0.93 + 80) \times 0.93 + 80 = 700 \times 0.93^{2} + 80 \times 0.93 + 80 = 700 \times 0.93^{2} + 80(0.93 + 1)

For the number of raspberry bushes at the end of 2026 we get:

(700 \times 0.93^{2} + 80(0.93 + 1)) \times 0.93 + 80 = 700 \times 0.93^{3} + 80(0.93^{2} + 0.93 + 1)

Following this pattern, for the number of raspberry bushes at the end of 2030 we get:

700 \times 0.93^{7} + 80(0.93^{6} + 0.93^{5} + 0.93^{4} + 0.93^{3} + 0.93^{2} + 0.93 + 1) \approx

\boldsymbol{{\color{Purple} \approx 876 \: \textbf{bushes}}}

(b)

Let’s write an expression for the number of raspberry bushes n years after 2023:

700 \times 0.93^{n} + 80(0.93^{n-1} + 0.93^{n-2} + ... + 0.93 + 1)

The expression in the brackets is the sum of the first n terms of a geometric sequence with u_{1} = 1 and r = 0.93, so for the number of raspberry bushes n years after 2023 we get:

700 \times 0.93^{n} + 80\left ( \frac{1(1 - 0.93^{n})}{1 - 0.93} \right )

We are looking for the smallest value of n for which the number of raspberry bushes exceeds 1100, so simplifying and equating the expression to 1100 we get:

700 \times 0.93^{n} +  \frac{8000(1 - 0.93^{n})}{7} = 1100

Solving for n either by graphing or using the Solver function on the calculator, we get that n \approx 32.2, therefore the first year in which the number of raspberry bushes exceeds 1100 is 2023 + 33, so:

\boldsymbol{{\color{Purple} 2056}}

.