$bold{Concept:}$ A sequence is arithmetic if there is a constant difference in each consecutive terms. A sequence is geometric if there is a constant ratio in each consecutive terms.

a) $t_n=5times 3^{n+1}$

$t_1=5(3^2)=45$

$t_2=5(3^3)=135$

$t_3=5(3^4)=405$

$t_4=5(3^5)=1215$

$t_5=5(3^6)=3645$

The general term is consistent with $t_n=t_1r^{n-1}$, thus it is a geometric sequence.

b) $t_n=dfrac{3n+2}{2n+1}$

$t_1=dfrac{3+2}{2+1}=dfrac{5}{3}$

$t_2=dfrac{3(2)+2}{2(2)+1}=dfrac{8}{5}$

$t_3=dfrac{3(3)+2}{2(3)+1}=dfrac{11}{7}$

$t_4=dfrac{3(4)+2}{2(4)+1}=dfrac{14}{9}$

$t_5=dfrac{3(5)+2}{2(5)+1}=dfrac{17}{11}$

$t_2-t_1neq t_4-t_3$ and $dfrac{t_2}{t_1}neq dfrac{t_3}{t_2}$

Thus, it is neither arithmetic nor geometric.

c) $t_n=5n$

$t_1=5$

$t_2=10$

$t_3=15$

$t_4=20$

$t_5=25$

A constant difference of 5 exists between consecutive terms, thus, it is an arithmetic sequence.

d) $t_1=5$ , $t_n=7t_{n-1}$ , where $n>1$

$t_1=5$

$t_2=7(5)=35$

$t_3=7(35)=245$

$t_4=7(245)=1715$

$t_5=7(1715)=12005$

A constant ratio of $7$ exists between consecutive terms, thus, it is a geometric sequence.

e) $t_1=19$ , $t_n=1-t-t_{n-1}$ where $n>1$

$t_1=19$

$t_2=1-19=-18$

$t_3=1-(-18)=19$

$t_4=1-19=-18$

$t_5=1-(-18)=19$

$t_2-t_1=-37$ while $t_3-t_2=37$

$dfrac{t_2}{t_1}=-dfrac{18}{19}$ while $dfrac{t_3}{t_2}=-dfrac{19}{18}$

Therefore, it is neither arithmetic nor geometric sequence.

$t_1=7$ , $t_2=13$, $t_n=2t_{n-1}-t_{n-2}$, where $n>2$

$t_1=7$

$t_2=13$

$t_3=2(13)-7=19$

$t_4=2(19)-13=25$

$t_5=2(25)-19=31$

A common difference of 6 exists between consecutive terms, thus it is arithmetic.

begin{table}[]
defarraystretch{1.6}%
begin{tabular}{l|c|c|}
cline{2-3}
& $bold{i)}$ & $bold{ii})$ \ cline{2-3}
a) & 45, 135, 405, 1215, 3645 & geometric \ cline{2-3}
b) & $frac{5}{3},;frac{8}{5},;frac{11}{7},;frac{14}{9},;frac{17}{11}$ & neither \ cline{2-3}
c) & 5, 10 , 15, 20, 25 & arithmetic \ cline{2-3}
d) & 5, 35, 245, 1715, 12 005 & geometric \ cline{2-3}
e) & 19, -18, 19, -18, 19 & neither \ cline{2-3}
f) & 7, 13, 19, 25, 31 & arithmetic \ cline{2-3}
end{tabular}
end{table}

a) The general term $t_n$ of geometric sequence with first term $a=t_1$ and common ratio $r$ is

$$
begin{equation*}t_n=acdot r^{n-1}end{equation*}
$$

While the recursive formula is

$$
begin{equation*} t_n=rcdot t_{n-1}; end{equation*}
$$

In this case, $a=-9$ and $r=-11$

General term: $t_n=(-9)(-11)^{n-1}$

Recursive formula: $t_1=-9,;;t_n=-11cdot t_{n-1}$ where $n>1$

b) The general term $t_n$ of an arithmetic sequence with first term $t_1$ and common difference $d$ is

$$
begin{equation*} t_n=t_1+(n-1)dend{equation*}
$$

While the recursive formula is

$$
begin{equation*} t_n=t_{n-1}+d end{equation*}
$$

In this case, $t_2=123$ and $t_3=-456$ which implies

$d=t_3-t_2=-456-123=-579$

$t_1=t_2-d=123-(-579)=702$

Therefore,

General term: $t_n=702+(n-1)(-579)implies t_n=1281-579n$

Recursive formula: $t_1=702$, $t_n=t_{n-1}-579$ where $n>1$

begin{table}[]
defarraystretch{1.5}%
begin{tabular}{l|c|c|}
cline{2-3}
& $bold{i)}$ General Term & $bold{ii})$ Recursive Formula \ cline{2-3}
a) & $t_n=(-9)(-11)^{n-1}$ & $t_1=-9$ , $t_n=-11cdot t_{n-1}$ , where $n>1$ \ cline{2-3}
b) & $t_n=1281-579n$ & $t_1=702$ , $t_n=t_{n-1}-579$ , where $n>1$ \ cline{2-3}
end{tabular}
end{table}

a) We can observe that $25-18=7$ and $32-25=7$, thus it is an arithmetic sequence with $d=7$

We can then use the general term for arithmetic sequence

$$
begin{align*} t_n&=t_1+(n-1)d\
193&=18+(n-1)(7)\
7(n-1)&=193-18\
n&=dfrac{193-18}{7}+1\
n&=26end{align*}
$$

Therefore, there are 26 terms in the sequence.

b) We can observe that $dfrac{-10}{2}=-5$ and $dfrac{50}{-10}=-5$, thus it is a geometric sequence with $r=-5$

We can then use the general term for geometric sequence

$$
begin{align*} t_n&=t_1cdot r^{n-1}\
-156;250 &=2cdot 5^{n-1}\
5^{n-1}&=dfrac{-156;250}{2}\
5^{n-1}&=78125\
5^{n-1}&=5^7\
n-1&=7\
n&=8end{align*}
$$

Therefore, there are 8 terms in the sequence.

a) $n=26$

b) $n=8$

a) Here, the exponent is $4$ so the coefficients based on Pascal’s triangle is 1,;4,;6,;4,;1.\\
begin{tabular}{|l|l|l|l|}
hline
Coefficients & $x$ & ($-$5) & Product \ hline
1 & $x^4$ & $(-5)^0$ & $1(-5)^0x^4=x^4$ \ hline
4 & $x^3$ & $(-5)^1$ & $4(-5)x^3=-20x^3$ \ hline
6 & $x^2$ & $(-5)^2$ & $6(-5)^2x^2=150x^2$ \ hline
4 & $x^1$ & $(-5)^3$ & $4(-5)^3x^1=-500x$ \ hline
1 & $x^0$ & $(-5)^4$ & $1(-5)^4x^0=625$ \ hline
end{tabular}
\\Therefore, \\
$(x-5)^4=5x^4-20x^3+150x^2-500x+625$

b) The exponent is 3 so the coefficients are 1, 3, 3, 1\\
begin{tabular}{|l|l|l|l|}
hline
Coefficients & $(2x)$ & $(3y)$ & Product \ hline
1 & $(2x)^3$ & $(3y)^0$ & $(1)(2x)^3(1)=8x^3$ \ hline
3 & $(2x)^2$ & $(3y)^1$ & $(3)(2x)^2(3y)=36x^2y$ \ hline
3 & $(2x)^1$ & $(3y)^2$ & $(3)(2x)^1(3y)^2=54xy^2$ \ hline
1 & $(2x)^0$ & $(3y)^3$ & $(1)(2x)^0(3y)^3=27y^3$ \ hline
end{tabular}
\\
Therefore\\
$(2x+3y)^3=8x^3+36x^3y+54xy^2+27y^3$

a) $(x-5)^4=5x^4-20x^3+150x^2-500x+625$

b) $(2x+3y)^3=8x^3+36x^3y+54xy^2+27y^3$

a) Observe that $33-19=14$ and $47-33=14$ so this is an arithmetic series with $t_1=19$ and $d=14$. Now, we need to know the number of terms using the general term of arithmetic sequence.

$$
begin{align*} t_n&=t_1+(n-1)d\
439 &= 19+(n-1)(14)\
n&=dfrac{439-19}{14}+1\
n&=31end{align*}
$$

The sum of an arithmetic series is given by

$$
begin{align*}S_n&=dfrac{n}{2}(t_1+t_{n-1})\
S_{31}&=dfrac{31}{2}(19+439)\
S_{31}&=boxed{7099}
end{align*}
$$

b) Observe that $dfrac{12;000}{10;000}=dfrac{14;400}{12;000}=1.2$. Therefore, this is a geometric series

with $r=1.2$= and $t_1=10;000$

The sum of a geometric series is given by

$$
begin{align*} S_n&=dfrac{t_1(r^n-1)}{r-1}\
S_{10}&=dfrac{10;000(1.2)^{10}-1}{1.2-1}\
S_{10}&=boxed{259;586.8211}end{align*}
$$

a) $7099$

b) $259;586.8211$

Let’s examine the first few terms:

$t_1=4$

$t_2=5$

$t_3=dfrac{5+1}{4}=1.5$

$t_4=dfrac{1.5+1}{5}=0.5$

$t_5=dfrac{0.5+1}{1.5}=1$

$t_6=dfrac{1+1}{0.5}=4$

$t_7=dfrac{4+1}{1}=5$

$t_8=dfrac{5+1}{4}=1.5$

Notice that the cycle repeats every 5 terms, implying that $t_n=1$ if $n$ is divisible by 5.

This means $t_{120}=1$
$t_{121}=4$
$t_{122}=5$
$t_{123}=1.5$

$$
t_{123}=1.5
$$

The initial term is $100$ (on your first birthday) and the amount they add to it increases by $75$ each year.

This corresponds to the following series:

$100+175+250+325…$

The amount of money in the account in your 21st birthday would be the sum of the first 21 terms of the arithmetic series with $t_1=100$ and $d=75$

$$
begin{align*}S_n&=dfrac{n}{2}[2t_1+(n-1)d]\
S_{21}&=dfrac{21}{2}[2(100)+(21-1)(75)]\
S_{21}&=17;850end{align*}
$$

Therefore, there will be $$17;850$ on your 21st birthday.

$$
$17;850
$$

a) Notice that

$t_3=1+7=8$

$t_4=7+8=15$

$t_5=8+15=23$

$t_6=15+23=38$

In other words, the next is term is the sum of the 2 previous terms. With this information, we can predict that

$t_7=23+38=61$

$t_8=38+61=99$

$$
t_9=61+99=160
$$

b) Notice that the exponent of $p$ increases by 1 in each consecutive term and the magnitude of coefficient of $q$ also increases by 1 but the sign is alternating. Therefore, we can predict that

$t_5=p^6+6q$

$t_6=p^7-7q$

$$
t_7=p^8+8q
$$

c) Notice that consecutive numerators have constant difference of $-6$ while the denominators have constant difference of 3. With this information, we can predict that

$t_6=dfrac{1-6}{15+3}=dfrac{-5}{18}$

$t_7=dfrac{-5-6}{18+3}=dfrac{-11}{21}$

$$
t_8=dfrac{-11-6}{21+3}=dfrac{-17}{24}
$$

a) $61$ , $99$ , $160$

b) $p^6+6p$ , $p^7+7p$, $p^8+8p$

c) $-dfrac{5}{18}$ , $-dfrac{11}{21}$ , $-dfrac{17}{24}$