If a common difference $d$ is observed, the general term $t_n$ can be obtained as

$t_n=t_1+(n-1)d$

If a common ratio $r$ is observed, the general term is given by

$$
t_n=t_1cdot r^{n-1}
$$

(1) the first term

(2) the rule to get any term $t_n$ from the term before it $t_{n-1}$

If a common difference $d$ is observed, the recursive formula is

$t_n=t_{n-1}+d$ where $n>1$

If a common ratio $r$ is observed, the recursive formula is

$t_n=rcdot t_{n-1}$ where $n>1$

$t_4=19+4=23$

$t_5=23+4=27$

In this case, $t_1=7$, $d=4$

The general term is therefore

$t_n=7+(n-1)(4)$

$t_n=3+4n$

The recursive formula is

$t_1=7; ,; t_n=t_{n-1}+4$ where $n>1$

$t_4=-23-27=-50$

$t_5=-50-27=-77$

The general term is

$t_n=58+(n-1)(-27)$

$t_n=85-27n$

The recursive formula is

$t_1=58 , t_n=t_{n-1}-27$ where $n>1$

$t_4=320cdot 4 = 1280$

$t_5=1280cdot 4 = 5120$

The general term is

$t_n=5cdot 4^{n-1}$

The recursive formula is

$t_n=4cdot t_{n-1}$ where $n>1$

$t_4=250cdot (-0.5)=-125$

$t_5=-125cdot (-0.5)=62.5$

The general term is

$t_n=1000cdot(-0.5)^{n-1}$

The recursive formula is

$t_n=(-0.5)cdot t_{n-1}$ where $n>1$

b.) $t_n=85-27n$ ; $t_1=58 , t_n=t_{n-1}-27$ where $n>1$

c.) $t_n=5cdot 4^{n-1}$ ; $t_n=4cdot t_{n-1}$ where $n>1$

d.) $t_n=1000cdot(-0.5)^{n-1}$; $t_n=(-0.5)cdot t_{n-1}$ where $n>1$

If a common difference $d$ is observed (arithmetic sequence), the general term $t_n$ can be obtained as

$t_n=t_1+(n-1)d$

If a common ratio $r$ is observed (geometric sequence), the general term is given by

$$
t_n=t_1cdot r^{n-1}
$$

(1) the first term

(2) the rule to get any term $t_n$ from the term before it $t_{n-1}$

For arithmetic sequence, the recursive formula is

$t_n=t_{n-1}+d$ where $n>1$

For geometric sequence, the recursive formula is

$t_n=rcdot t_{n-1}$ where $n>1$

$t_n=t_1+(n-1)d$

$46=t_1+(4-1)cdot d$

$248=t_1+(6-1)cdot d$

Equate $t_1$ from each equation

$46-3d=248-5d$

$d=101$

Therefore,

$$
t_5=46+101=147
$$

Here $t_4=46$, $t_6=248$

Both $t_1$ and $d$ is unknown. Solve the system of equations with two unknowns.

$$
t_1=-257
$$

$t_n=-257+(n-1)(101)$

$$
t_n=-358+101n
$$

b.) $d$ = 101

c.) $t_1=-257$

d.) $t_{100}=9742$

If a common difference $d$ is observed (arithmetic sequence), the general term $t_n$ can be obtained as

$t_n=t_1+(n-1)d$

If a common ratio $r$ is observed (geometric sequence), the general term is given by

$$
t_n=t_1cdot r^{n-1}
$$

(1) the first term

(2) the rule to get any term $t_n$ from the term before it $t_{n-1}$

For arithmetic sequence, the recursive formula is

$t_n=t_{n-1}+d$ where $n>1$

For geometric sequence, the recursive formula is

$t_n=rcdot t_{n-1}$ where $n>1$

$r=dfrac{t_n}{t_{n-1}}$

$r=dfrac{9724.05}{9261}=1.05$

$r=dfrac{10210.2525}{9724.05}=1.05$

Thus, this is a geometric sequence.

$t_n=t_1cdot r^{n-1}$

$t_1=dfrac{9261}{1.05^{4-1}}=8000$

Thus, the recursive formula is

$t_1=8000; ;; t_n=1.05cdot t_{n-1}$ where $n>1$

$t_n=t_1cdot r^{n-1}$

$$
t_n=8000cdot 1.05^{n-1}
$$

b.) $t_1=8000; ;; t_n=1.05cdot t_{n-1}$ where $n>1$

c.) $t_n=8000cdot 1.05^{n-1}$

d.) $t_{10}=12410.62573$

$S_n=dfrac{n}{2}(t_1+t_n)$

$$
t_n=t_1+(n-1)cdot d
$$

$$
S_n=t_1cdot dfrac{1-r^{n}}{1-r}
$$

The tenth term is

$t_n=t_1+(n-1)cdot d$

$t_{10}=3+(10-1)cdot 2 = 21$

The sum of first ten terms is

$S_n=dfrac{n}{2}(t_1+t_n)$

$$
S_{10}=dfrac{10}{2}(3+21)=120
$$

If $d$ is given, it is arithmetic, if $r$, it is geometric.

The tenth term is

$t_n=t_1+(n-1)cdot d$

$t_{10}=-27+6(9)=27$

$S_n=dfrac{n}{2}(t_1+t_n)$

$$
S_{10}=dfrac{10}{2}(-27+27)=0
$$

If $d$ is given, it is arithmetic, if $r$, it is geometric.

The tenth term is

$t_{10}=48+9(-17)=-105$

$$
S_{10}=dfrac{10}{2}(48+(-105))=-285
$$

If $d$ is given, it is arithmetic, if $r$, it is geometric.

$S_{n}=t_1cdot dfrac{1-r^n}{1-r}$

$$
S_{10}=8192000cdot dfrac{1-(-0.5)^{10}}{1-(-0.5)^{10}}=5;456;000
$$

If $d$ is given, it is arithmetic, if $r$, it is geometric.

Thus, for the next three years, the populations are:

$t_2=200;000 cdot 1.05= 210;000$

$t_3=210;000 cdot 1.05=220;500$

$$
t_4=220;500 cdot 1.05=231;525
$$

$t_n=t_1cdot r^{n-1}$

Thus, in this case,

$t_n=200;000cdot 1.05^{n-1}$.

Since at present the population is $t_1$, after 10 years, the population is $t_{11}$

$$
t_{11}=200;000cdot 1.05^{11-1}=325;779
$$

b.) population after 10 years is $325, 779$

$2^x=4096$

$2^x=2^{12}$

Equate the exponents

$x=12$

We can confirm this graphically.

b) $x=3.477$

c) $x=11.264$

d) $8.723$

$y=4^{2x}$

$f(x)=5^{-0.5x}$

$$
g(t)=0.5t^{3x}
$$

Exponential function is function such that the independent variable is in the exponent.

This function is used to model quantities that grows or decays at a certain rate.

For example, a population that grows/decays at a rate of $r$ a year starting from initial population of $P_0$ can be modeled as

$P=P_0(1+r)^t$

where $t$ is the number of years.

If population doubles every $k$ time, the amount after time $t$ can be modeled by

$$
P=P_0times 2^{t/k}
$$