$t_1,;t_2,;t_3;t_4…$ is a geometric sequence if

$dfrac{t_2}{t_1}=dfrac{t_3}{t_2}=dfrac{t_4}{t_3}=r$

In this case, $r$ is called the $bold{common;ratio}$

$therefore$ not geometric sequence

$therefore$ geometric sequence

$therefore$ geometric sequence

$therefore$ geometric sequence

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

$bold{Recursive;Formula;of;Geometric;Sequence}$ The recursive formula of geometric sequence is

$t_1 ;;, ;;t_n=t_{n-1}cdot r$ where $n>1$

General Term: $t_n=9cdot 4^{n-1}$

Recursive Formula: $t_1=9$ , $t_n=4cdot t_{n_1}$ where $n>1$

General Term: $t_n=625cdot 2^{n-1}$

Recursive Formula: $t_1=625$ , $t_n=2cdot t_{n-1}$ where $n>1$

General Term: $t_n=10125cdot left( dfrac{2}{3}right)^{n-1}$

Recursive Formula: $t_1=10125$ , $t_n=dfrac{2}{3}cdot t_{n-1}$ where $n>1$

b.) $t_n=625cdot 2^{n-1}$

c.) $t_n=10125cdot left(dfrac{2}{3}right)^{n-1}$

Thus, $r=dfrac{t_{32}}{t_{31}}=dfrac{1107}{123}=9$

To obtain $t_{33}$

$t_n=rcdot t_{n-1}$

$$
t_{33}=rcdot t_{32}=9cdot 1107=9963
$$

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

$dfrac{t_2}{t_1}=dfrac{t_3}{t_2}=dfrac{t_4}{t_3}=r$

$bold{General;Term;of;Geometric;Sequence:}$ The general term of a geometric sequence with first term $t_1$ and common ratio $r$ is

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

$bold{Recursive;Formula;of;Geometric;Sequence:}$ The recursive formula of geometric sequence is

$t_1 ;;, ;;t_n=t_{n-1}cdot r$ where $n>1$

General Term: $t_n=12cdot 2^{n-1}$

Recursive Formula: $t_1=12$ , $t_n=2cdot t_{n-1}$ where $n>1$

General Term: $t_n=5cdot(-3)^{n-1}$

Recursive Formula: $t_1=5$ , $t_n=(-3)cdot t_{n-1}$ where $n>1$

General Term: $t_n=125cdotleft(dfrac{2}{5}right)^{n-1}$

Recursive Formula: $t_1=125$ , $t_n=dfrac{2}{5}cdot t_{n-1}$ where $n>1$

b.) not geometric sequence

c.) not geometric sequence

d.) $t_n=5cdot (-3)^{n-1}$

e.) not geometric sequence

f.) $t_n=125cdot left(dfrac{2}{5}right)^{n-1}$

$dfrac{t_2}{t_1}=dfrac{t_3}{t_2}=dfrac{t_4}{t_3}=r$

$bold{General;Term;of;Geometric;Sequence:}$ The general term of a geometric sequence with first term $t_1$ and common ratio $r$ is

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

$bold{Recursive;Formula;of;Geometric;Sequence:}$ The recursive formula of geometric sequence is

$t_1 ;;, ;;t_n=t_{n-1}cdot r$ where $n>1$

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

Recursive Formula: $t_1=4$ , $t_n=5cdot t_{n-1}$ where $n>1$

Sixth term: $t_6=4cdot(5)^{6-1}=12;500$

General Term: $t_n=(-11)cdot 2^{n-1}$

Recursive Formula: $t_1=-11$ , $t_n=2cdot t_{n-1}$ where $n>1$

Sixth term: $t_6=(-11)cdot 2^{6-1}=-352$

General Term: $t_n=(15)cdot (-4)^{n-1}$

Recursive Formula: $t_1=15$ , $t_n=(-4)cdot t_{n-1}$ where $n>1$

Sixth term: $t_6=(15)cdot (-4)^{6-1}=-15;360$

General Term: $t_n=(896)cdot left(dfrac{1}{2}right)^{n-1}$

Recursive Formula: $t_1=896$ , $t_n=dfrac{1}{2}cdot t_{n-1}$ where $n>1$

Sixth term: $t_6=(896)cdot left(dfrac{1}{2}right)^{6-1}=28$

General Term: $t_n=(6)cdot left(dfrac{1}{3}right)^{n-1}$

Recursive Formula: $t_1=6$ , $t_n=dfrac{1}{3}cdot t_{n-1}$ where $n>1$

Sixth term: $t_6=(6)cdot left(dfrac{1}{3}right)^{6-1}=dfrac{2}{81}$

General Term: $t_n=(1)cdot (0.2)^{n-1}$

Recursive Formula: $t_1=1$ , $t_n=0.2cdot t_{n-1}$ where $n>1$

Sixth term: $t_6=(1)cdot left(0.2right)^{6-1}=dfrac{32}{100;000}$

b.) -352

c.) -15 360

d.) 28

e.) $dfrac{2}{81}$

f.) $dfrac{32}{100;000}$

For a given sequence $t_1,;t_2,;t_3,;t_4…$
if

$dfrac{t_2}{t_1}=dfrac{t_3}{t_2}=dfrac{t_4}{t_3}=r implies$ geometric sequence

$t_2-t_1=t_3-t_2=t_4-t_3=dimplies$ arithmetic sequence

$bold{General;Term}$

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

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

$bold{Recursive;Formula}$

geometric: $t_1 ;;, ;;t_n=t_{n-1}cdot r$ where $n>1$

arithmetic: $t_1 ;,; t_n=t_{n-1}+d$ where $n>1$

$t_1=9$ , $d=4$

General Term: $t_n=9+(n-1)d=4n+5$

$t_1=7$ , $r=-3$

General Term: $t_n=7cdot (-3)^{n-1}$

$t_1=18$ , $r=-1$

General Term: $t_n=18cdot (-1)^{n-1}$

$t_1=29$ , $d=-10$

General Term: $t_n=29+(n-1)(-10)=39-10n$

$t_1=128$ , $r=dfrac{3}{4}$

General Term: $t_n=128cdot left( dfrac{3}{4}right)^{n-1}$

b.) -352

c.) -15 360

d.) 28

e.) $dfrac{2}{81}$

f.) $dfrac{32}{100;000}$

$dfrac{t_2}{t_1}=dfrac{t_3}{t_2}=dfrac{t_4}{t_3}=r$

$bold{General;Term;of;Geometric;Sequence:}$ The general term of a geometric sequence with first term $t_1$ and common ratio $r$ is

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

$bold{Recursive;Formula;of;Geometric;Sequence:}$ The recursive formula of geometric sequence is

$t_1 ;;, ;;t_n=t_{n-1}cdot r$ where $n>1$

Recursive fomula: $t_1=19$ , $t_n=5cdot t_{n-1}$ where $n>1$

General term: $t_n=10cdot 5^{n-1}$

Recursive formula: $t_1=-9$ , $t_n=-4t_{n-1}$ where $n>1$

General term: $t_n=(-9)cdot (-4)^{n-1}$

Recursive formula: $t_1=144$ , $t_n=dfrac{t_{n-1}}{4}$ where $n>1$

General term: $t_n=144cdot left(dfrac{1}{4}right)^{n-1}$

Recursive formula: $t_1=900$ , $t_n=dfrac{1}{6}cdot t_{n-1}$ where $n>1$

General term: $t_n=900cdot left(dfrac{1}{6}right)^{n-1}$

b.) $t_n=(-9)cdot(-4)^{n-1}$

c.) $t_n=144cdot left(dfrac{1}{4}right)^{n-1}$

d.) $t_n=900cdot left(dfrac{1}{6}right)^{n-1}$

$t_1 ;;, ;;t_n=t_{n-1}cdot r$ where $n>1$

This means that the ratio between terms is NOT constant and varies as the value of $n$ increases. Thus, this is not a geometric sequence.

$t_1=-8$

$t_2=-3(-8)=24$

$t_3=-3(24)=-72$

$t_4=-3(-72)=216$

$$
t_5=-3(216)=-648
$$

$t_1=123$

$t_2=dfrac{1}{3}cdot(123)=41$

$t_3=dfrac{1}{3}cdot(41)=dfrac{41}{3}$

$t_4=dfrac{1}{3}cdotdfrac{41}{3}=dfrac{41}{9}$

$$
t_5=dfrac{1}{3}cdotdfrac{41}{9}=dfrac{41}{27}
$$

$t_1=10$

$t_2=20$

$t_3=4cdot t_1=40$

$t_4=4cdot t_2=4(2)=80$

$t_5=4cdot t_3=4(40)=160$

Notice the a common ratio of $r=2$ exists between consecutive terms, thus it must be geometric sequence.

The general term can be written as $t_n=10cdot 2^{n-1}=5cdot 2^n$

b.) $-8$, $24$,$-72$, $216$, $-648$

c.) 123, 41, $dfrac{41}{3}$, $dfrac{41}{9}$, $dfrac{41}{27}$

d.) 10, 20, 40, 80, 160

$dfrac{t_2}{t_1}=dfrac{t_3}{t_2}=dfrac{t_4}{t_3}=r$

$bold{General;Term;of;Geometric;Sequence:}$ The general term of a geometric sequence with first term $t_1$ and common ratio $r$ is

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

$bold{Recursive;Formula;of;Geometric;Sequence:}$ The recursive formula of geometric sequence is

$t_1 ;;, ;;t_n=t_{n-1}cdot r$ where $n>1$

The first 5 terms are:

$t_1=4cdot 4^0=4$

$t_2=4cdot 4^1=16$

$t_3=4cdot 4^2=64$

$t_4=4cdot 4^3=256$

$t_5=4cdot 4^4=1024$

$t_n=7cdot(-5)^{n-4}$ as $t_n=7(-5)^{-3}cdot (-5)^{n-1}=-dfrac{7}{125}cdot (-5)^{n-1}$

which is the general term of geometric sequence with $t_1=dfrac{7}{125}$ and $r=-5$.

The first 5 terms are:

$t_1=-dfrac{7}{125}$

$t_2=-dfrac{7}{125}cdot (-5)^{2-1}=dfrac{7}{25}$

$t_3=-dfrac{7}{125}cdot (-5)^{3-1}=-dfrac{7}{5}$

$t_4=-dfrac{7}{125}cdot (-5)^{4-1}=7$

$t_5=-dfrac{7}{125}cdot (-5)^{5-1}=-35$

$t_n=dfrac{11}{13^n}$ as $t_n=dfrac{11}{13}cdot left(dfrac{1}{13}right)^{n-1}$

which is the general term of geometric sequence with $t_1=dfrac{11}{13}$ and $r=dfrac{1}{13}$.

The first 5 terms are:

$t_1=dfrac{11}{13}$

$t_2=dfrac{11}{13}cdot left( dfrac{1}{13}right)^{2-1}=dfrac{11}{169}$

$t_3=dfrac{11}{13}cdot left( dfrac{1}{13}right)^{3-1}=dfrac{11}{2197}$

$t_4=dfrac{11}{13}cdot left( dfrac{1}{13}right)^{4-1}=dfrac{11}{28;561}$

$t_5=dfrac{11}{13}cdot left( dfrac{1}{13}right)^{5-1}=dfrac{11}{371;293}$

b.) not geometric sequence

c.) not geometric sequence

d.) $-dfrac{7}{125}, dfrac{7}{25}, -dfrac{7}{5}, 7, -35$

e.) not geometric sequence

f.) $dfrac{11}{13},;dfrac{11}{169},;dfrac{11}{2197},;dfrac{11}{28;561},;dfrac{11}{371;293}$

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

Given that $t_8=360$ and $t_5=45$

$t_8=t_1cdot r^{8-1}$ and $t_5=t_1cdot r^{5-1}$

Take the ratio of the two equations.

$dfrac{t_8}{t_5}=dfrac{t_1cdot r^{8-1}}{t_1cdot r^{5-1}}$

$dfrac{360}{45}=8=r^3 implies r=2$

Now, the first term can be obtained as

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

$$
45=t_1cdot (2)^{5-1}implies t_1=dfrac{45}{2^4}=dfrac{45}{16}
$$

$$
t_{20}=dfrac{45}{16}cdot 2^{19}=1,474,560
$$

We shall determine how many bacteria would be there at 9th observation, that is, $t_9$

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

$t_{9}=5120cdot left(dfrac{3}{2}right)^{9-1}$

$$
boxed{t_9=131;220;;text{bacteria}}
$$

We shall determine how much would be there on the 10th year (i.e. after nine years), that is, $t_{10}$

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

$t_{10}=5;000cdot left(1.08right)^{10-1}$

$$
boxed{t_{10}=$9995.62}
$$

$$
t_4=90cdot (0.9)^{4-1}=65.61%
$$

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

$5=90cdot (0.9)^{n-1}$

Notice that at

$n=28implies 90(0.9)^{28-1}=5.23$

$n=29implies 90(0.9)^{29-1}=4.71$

Thus, 29 doses are needed to get the bacterial population to not more than $5%$.

b.) $29$ doses

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

If $j$ and $k$ are positive integers, we can find the relationship between $t_k$ and $t_j$ as

$t_j=t_1cdot r^{j-1}$

$t_k=t_1cdot r^{k-1}$

Taking the ratios of two equation

$dfrac{t_j}{t_k}=r^{j-k}$

$t_j=t_kcdot r^{j-k}$

Therefore, if we want to find $t_j$ given a value of $t_k$, just multiply it with $r^{j-k}$.

$t_8$ from $t_2$, multiply it with $r^{8-2}=r^6$.

To obtain $t_7$ from $t_5$ multiply it with $r^{7-5}=r^2$.

To obtain $t_{29}$ from $t_7$, multiply it by $r^{29-7}=r^{22}$

To obtain $t_{100}$ from $t_3$, multiply it by $r^{100-3}=r^{97}$

We can obtain the number of shaded triangles at the $6^{th}$ stage as

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

$$
t_6=1cdot 3^{6-1}=243
$$

We can obtain the shaded area at $20^{th}$ stage as

$t_{20}=80left(dfrac{3}{4}right)^{20-1}=0.338;$ cm$^2$

b.) 0.338 $cm^2$

(1) Both sequences can be expressed as a general term and a recursive formula.

(2) Both contain the expression $(n-1)$ in its standard general term.

(3) Both have a domain of discrete positive integers.

(1) Consecutive terms have constant difference for arithmetic sequence while consecutive terms have constant ratio for geometric sequence.

(2) In arithmetic sequence, the next term can be obtained by addition, while for geometric sequence, the next term is obtained by multiplication.

(3) The general term for arithmetic sequence is a linear function while that of geometric sequence is an exponential function.

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

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

$t_n=1cdot left(dfrac{1}{2}right)^{n-1}$

Thus, the first four terms are:

$t_1=left(dfrac{1}{2}right)^0=1$

$t_2=left(dfrac{1}{2}right)^{2-1}=dfrac{1}{2}$

$t_3=left(dfrac{1}{2}right)^{3-1}=dfrac{1}{4}$

$$
t_4=left(dfrac{1}{2}right)^{4-1}=dfrac{1}{8}
$$

$S_1=t_1=1$

$S_2=S_1+t_2=1+dfrac{1}{2}=dfrac{3}{2}=1.5$

$S_3=S_2+t_3=dfrac{3}{2}+dfrac{1}{4}=dfrac{7}{4}=1.75$

$S_4=S_3+t_4=dfrac{7}{4}+dfrac{1}{8}=dfrac{25}{8}=1.875$

Observe that the sum approaches 2 when more terms are added.

$3,;10,;28,;72,;176$

To see a pattern clearly, we shall express them in factored form.

$3=1cdot 3$

$10=2cdot 5$

$28=4cdot 7$

$72=8cdot 9$

$$
176=16cdot 11
$$

For the arithmetic sequence: 3, 5, 7, 9, 11

$a_1=3$ and $d=2$ $implies a_n=3+(n-1)(2)=2n+1$

For the geometric sequence: 1, 2, 4, 8, 16

$g_1=1$ and $r=2$ $implies g_n=1cdot 2^{n-1}$

Thus, the general term is

$t_n=a_ncdot g_n=(2n+1)cdot 2^{n-1}$

For $n=10$

$$
t_{10}=2^{10-1}[2(10)+1]=10;752
$$

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

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

If a sequence is both arithmetic and geometric,

$t_1+(n-1)d=t_1cdot r^{n-1}$

Notice that this equation is true if $d=0$ and $r=1$.

This leads to a sequences like, $3,;3,;3,;3$ or when all terms in the sequence are identical.

Thus, the answer is YES.

example 1: 2, 4, 6, 8, 10, 12, 14, 16… which contains $2,4,8,16$

example 2: 3, 6, 9, 12, 15, 18, 21, 24, 27 which contains 3, 9, 27

If the square in stage 1 has a side of $s_1$, the square on the next stage can be obtained from Pythagorean relation

$s_2^2=(0.5s_1)^2+(0.5s_1)^2implies s_2=sqrt{0.5s_1^2}=sqrt{0.5}s_1$

This is a geometric sequence with common ratio $r=sqrt{0.5}$,

Thus, the general term is

$s_{n}=s_1(sqrt{0.5})^{n-1}$

The first term $s_1$ is $0.5(12)=6$

$s_n=6(sqrt{0.5})^{n-1}$

The area of an isosceles triangle is $A=dfrac{1}{2}timestext{base}timestext{height}$. In this case, the base and height are equal. Thus, the additional area formed in each stage is

$A_n=dfrac{1}{2}(s_n)^2$

$A_n=dfrac{1}{2}cdot left[6(sqrt{0.5})^{n-1}right]^2$

$$
A_n=18cdot 0.5^{n-1}
$$

$A_1=18$

$A_2=18(0.5)=9$

$A_3=18(0.5)^2=4.5$

$A_4=18(0.5)^3=2.25$

$A_5=18(0.5)^4=1.125$

$A_6=18(0.5)^5=0.6125$

The total area is
$A_{text{total}}=18+9+4.5+2.25+1.125+0.5625=35.4375$ cm$^2$