$bold{General;Term;of;a;Sequence}$

For an arithmetic sequence with first term $t_1$ and common difference $d$, the general term is

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

For a geometric sequence, with common ratio $r$, the general term is

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

$bold{Recursive;Formula}$. Recursive formula contains the first term and a rule on how to get the next term from the previous term.

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

For geometric sequence: $t_1;;, t_n=rcdot t_{n-1}$ where $n>1$

a.) The common difference is $d=21-29=-8$ and $t_1=29$

Recursive formula: $t_1=29$ , $t_n=t_{n-1}-8$ where $n>1$

General term: $t_n=29+(n-1)(-8)=37-8n$

Tenth term: $t_{10}=37-8(10)=-43$

b.) The common difference is $d=-16-(-8)=-8$ and $t_1=-8$

Recursive formula: $t_1=-8$ , $t_n=t_{n-1}-8$ where $n>1$

General term: $t_n=-8+(n-1)(-8)=-8n$

Tenth term: $t_{10}=-8(10)=-80$

c.) The common difference is $d=-9-(-17)=8$ and $t_1=-17$

Recursive formula: $t_1=-17$ , $t_n=t_{n-1}+8$ where $n>1$

General term: $t_n=-17+(n-1)(8)=-8n$

Tenth term: $t_{10}=8n-26=55$

d.) The common difference is $d=9.5-(3.25)=6.25$ and $t_1=3.25$

Recursive formula: $t_1=3.25$ , $t_n=t_{n-1}+6.25$ where $n>1$

General term: $t_n=3.25+(n-1)(6.25)=6.25n-3$

Tenth term: $t_{10}=6.25(10)-3=59.5$

e.) The common difference is $d=dfrac{2}{3}-dfrac{1}{2}=dfrac{1}{6}$ and $t_1=dfrac{1}{2}$

Recursive formula: $t_1=dfrac{1}{2}$ , $t_n=t_{n-1}+dfrac{1}{6}$ where $n>1$

General term: $t_n=dfrac{1}{2}+(n-1)left(dfrac{1}{6}right)=dfrac{1}{6}n+dfrac{1}{3}$

Tenth term: $t_{10}=dfrac{1}{6}(10)+dfrac{1}{3}=2$

f.) The common difference is $d=(3x+3y)-x=2x+3y$ and $t_1=x$

Recursive formula: $t_1=x$ , $t_n=t_{n-1}+2x+3y$ for $n>1$

General term:

$t_n=x+(n-1)(2x+3y)$

$t_n=x+2x(n-1)+3y(n-1)$

$t_n= x+2nx-2x+3ny-3$

$t_n=(-x+2nx)+3ny-3y$

$t_n=(2n-1)x+(3n-3)y$

Tenth term: $t_{10}=[2(10)-1](x)+[3(10)-3]y=19x+27y$

$bold{General;Term;of;a;Sequence}$

For an arithmetic sequence with first term $t_1$ and common difference $d$, the general term is

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

For a geometric sequence, with common ratio $r$, the general term is

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

$bold{Recursive;Formula}$. Recursive formula contains the first term and a rule on how to get the next term from the previous term.

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

For geometric sequence: $t_1;;, t_n=rcdot t_{n-1}$ where $n>1$

a.) The common difference is $d=11$ and $t_1=17$

Recursive formula: $t_1=17$ , $t_n=t_{n-1}-11$ where $n>1$

General term: $t_n=17+(n-1)(11)=11n+6$

b.) The common difference is $d=-7$ and $t_1=38$

Recursive formula: $t_1=38$ , $t_n=t_{n-1}-7$ where $n>1$

General term: $t_n=38+(n-1)(-7)=45-7n$

c.) The common difference is $d=18$ and $t_1=55$

Recursive formula: $t_1=55$ , $t_n=t_{n-1}+18$ where $n>1$

General term: $t_n=55+(n-1)(18)=18n+37$

d.) Remember that

$t_2=t_1+d$

$t_3=t_1+2d$

$t_4=t_1+3d$

.

.

$t_k=t_j+(k-j)d$

The common difference is $d=-38$ and $t_1=-34-2(-38)=42$

Recursive formula: $t_1=42$ , $t_n=t_{n-1}-38$ where $n>1$

General term: $t_n=42+(n-1)(-38)=80-38n$

d.) The common difference is

$d=dfrac{57-91}{2}=-17$ and $t_1=91-4(-17)=159$

Recursive formula: $t_1=159$ , $t_n=t_{n-1}-17$ where $n>1$

General term: $t_n=159+(n-1)left(-17right)=176-17n$

Since the number of seats between consecutive rows increases linearly (constant first difference), we shall model the number of seats as a function of row number as arithmetic sequence.

For arithmetic sequence with common difference $d$, the difference between the $j^{th}$ term and $k^{th}$ term is

$t_j-t_k=(j-k)d$

Given that $t_{13}=189$ and $t_{25}=225$

$t_{25}-t_{13}=(25-13)dimplies d=3$

Thus, we can obtain $t_{55}$ as

$t_{55}-t_{25}=(55-25)(d)$

$t_{55}=225+30(3)=315$

Thus, there are $boxed{bold{315;seats}}$ on the $55^{th}$ row.

$315$ seats

$bold{Identifying;Sequences}$ For a general sequence with terms

$t_1,;t_2,;t_3,;t_4,;…$

If $t_2=t_1=t_3-t_2=t_4-t_3=d implies$ $bold{arithmetic; sequence}$

If $dfrac{t_2}{t_1}=dfrac{t_3}{t_2}=dfrac{t_4}{t_3}=rimplies$ $bold{geometric; sequence}$

If neither of the conditions above are satisfied, the sequence is neither arithmetic nor geometric sequence.

$bold{General;Term;of;a;Sequence}$

For an arithmetic sequence with first term $t_1$ and common difference $d$, the general term is

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

For a geometric sequence, with common ratio $r$, the general term is

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

$bold{Recursive;Formula}$. Recursive formula contains the first term and a rule on how to get the next term from the previous term.

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

For geometric sequence: $t_1;;, t_n=rcdot t_{n-1}$ where $n>1$

a.) This is an arithmetic sequence with $d=30-15=15$ and $t_1=15$.

General term: $t_n=15+(n-1)(15)=15n$

Recursive formula: $t_1=15$ , $t_n=t_{n-1}+15$ where $n>1$

Sixth term: $t_6=15(6)=90$

b.) This is a geometric sequence with $r=dfrac{320}{640}=dfrac{1}{2}$ and $t_1=640$.

General term: $t_n=640left(dfrac{1}{2}right)^{n-1}$

Recursive formula: $t_1=640$ , $t_n=dfrac{1}{2}cdot t_{n-1}$ where $n>1$

Sixth term: $t_6=640left(dfrac{1}{2}right)^{6-1}=640cdot dfrac{1}{2^5}=20$

c.) This is a geometric sequence with $r=-dfrac{46}{23}=-2$ and $t_1=23$.

General term: $t_n=23left(-2right)^{n-1}$

Recursive formula: $t_1=23$ , $t_n=-2cdot t_{n-1}$ where $n>1$

Sixth term: $t_6=23cdot (-2)^{6-1}=-736$

d.) This is a geometric sequence with $r=dfrac{900}{3;000}=dfrac{3}{10}$ and $t_1=3;000$.

General term: $t_n=3;000cdot left(dfrac{3}{10}right)^{n-1}$

Recursive formula: $t_1=3;000$ , $t_n=dfrac{3}{10}cdot t_{n-1}$ where $n>1$

Sixth term: $t_6=3;000cdot left(dfrac{3}{10}right)^{6-1}=7.29$

e.) This is an arithmetic sequence with $d=5-3.8=1.2$ and $t_1=3.8$.

General term: $t_n=3.8+(n-1)(1.2)=15n$

Recursive formula: $t_1=3.8$ , $t_n=t_{n-1}+1.2$ where $n>1$

Sixth term: $t_6=1.2(6)+2.6=9.8$

f.) This is a geometric sequence with $r=dfrac{1/3}{1/2}=dfrac{2}{3}$ and $t_1=dfrac{1}{2}$.

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

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

Sixth term: $t_6=dfrac{1}{2}cdot left(dfrac{2}{3}right)^{6-1}=dfrac{16}{243}$

$bold{Identifying;Sequences}$ For a general sequence with terms

$t_1,;t_2,;t_3,;t_4,;…$

If $t_2=t_1=t_3-t_2=t_4-t_3=d implies$ $bold{arithmetic; sequence}$

If $dfrac{t_2}{t_1}=dfrac{t_3}{t_2}=dfrac{t_4}{t_3}=rimplies$ $bold{geometric; sequence}$

If neither of the conditions above are satisfied, the sequence is neither arithmetic nor geometric sequence.

$bold{General;Term;of;a;Sequence}$

For an arithmetic sequence with first term $t_1$ and common difference $d$, the general term is

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

For a geometric sequence, with common ratio $r$, the general term is

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

$bold{Recursive;Formula}$. Recursive formula contains the first term and a rule on how to get the next term from the previous term.

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

For geometric sequence: $t_1;;, t_n=rcdot t_{n-1}$ where $n>1$

a.) A common ratio of $r=5$ is observed, thus it is geometric sequence.

The first five terms are:

$t_1=5^1=5$

$t_2=5^2=25$

$t_3=5^3=125$

$t_4=5^4=625$

$$
t_5=5^5=3125
$$

b.) The general term is $t_n=dfrac{3}{4^n+3}$. The first five terms are:

$t_1=dfrac{3}{4+3}=dfrac{3}{7}$

$t_2=dfrac{3}{4^2+3}=dfrac{3}{19}$

$t_3=dfrac{3}{4^3+3}=dfrac{3}{67}$

$t_4=dfrac{3}{4^4+3}=dfrac{3}{359}$

$t_5=dfrac{3}{4^5+3}=dfrac{3}{1027}$

Since $dfrac{t_2}{t_1} neq dfrac{t_3}{t_2}$, it is not a geometric sequence.

c.) The recursive formula defines a common difference $d=-12$ between consecutive terms, hence, it is arithmetic sequence.

The first five terms are:

$t_1=5$

$t_2=5-12=-7$

$t_3=-7-12=-19$

$t_4=-19-12=-31$

$$
t_5=-31-12=-43
$$

d.) The recursive formula indicates a constant ratio of $r=-2$ between consecutive terms, hence, it is geometric sequence.

$t_1=-2$

$t_2=-2(-2)=4$

$t_3=-2(4)=-8$

$t_4=-2(-8)=16$

$$
t_5=-2(16)=-32
$$

e.) We shall solve the first five terms given the recursive formula

$t_1=8$

$t_2=11$

$t_3=2(11)-8=14$

$t_4=2(14)-11=17$

$t_5=2(17)-14=20$

Notice that it resemble an arithmetic sequence with $t_1=8$ and $d=3$ and a general term

$t_n=8+(n-1)(3)=3n+5$

Since the price decreases by 10$%$ each week, there is a common ratio of $r=0.9$ between each consecutive week. Thus, we can model it as a geometric sequence.

On the first week, the price already decreased by $10%$ implying $t_1=$10;000cdot (0.9)=$9;000$

We shall on which week will the price reach not more than $$100$.

Thus, we shall $n$ such that $t_nleq 100$

Using the general term of geometric sequence

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

We shall substitute the given values

$9000cdot (0.9)^{n-1}leq 100$

Notice that when

$n=43implies t_n=9000cdot (0.9)^{43-1}=107.753$

$n=44implies t_n=9000cdot (0.9)^{44-1}=96.977$

Thus, the price will be less than or equal to $$100$ on the $boxed{bold{44^{th}; week}}$

Note that the value $ngeq 44$ can also be obtained directly using the concept of logarithms which will be introduced to you in the future.

on the $44^{th}$ week

We shall find the pattern in this sequence by taking 1st differences.

$t_2-t_1=8$

$t_3-t_2=20$

$t_4-t_3=38$

$t_5-t_4=62$

$t_6-t_5=92$

No apparent pattern is observed so we shall take 2nd differences.

$(t_3-t_2)-(t_2-t_1)=20-8=12$

$(t_4-t_3)-(t_3-t_2)=38-20=18$

$(t_5-t_4)-(t_4-t_3)=62-38=24$

$(t_6-t_5)-(t_5-5_4)=92-62=30$

Now a pattern can be observed. It appears the second differences resemble an arithmetic sequence with $d=6$.

Knowing that $t_6=221$, we expect that

$(t_7-t_6)-(t_6-t_5)=30+6 implies (t_7-221)-(92)=36implies t_7=349$

$(t_8-t_7)-(t_7-t_6)=36+6implies (t_8-349)-(349-221)=42 implies t_8=519$

$$
(t_9-t_8)-(t_8-t_7)=42+6implies (t_9-519)-(519-349)=48implies t_9=737
$$

$$
349,;519,;737
$$

We need to find patterns in the sequence by inspection. Each element in the sequence is a sum of two terms. The first term appears to be a geometric sequence with first term $t_1=x$ and common ratio $r=dfrac{x^2}{x}=x$ while the second term appears to be an arithmetic sequence with first term $t_1=y$ and common difference $d=2y-y=y$.

Knowing that the general term for geometric sequence and arithmetic sequence are

$t_n=t_1cdot r^{n-1}$ and $t_n=t_1+(n-1)d$ respectively, the general term for the original sequence is therefore

$t_n=xcdot x^{n-1}+y+(n-1)(y)$

$$
t_n=x^n+ny
$$

$$
t_n=x^n+ny
$$

a.) Observe that for the first few large cubes, the pattern is

$1^3,;2^3;,3^3;$ thus, the first few terms are $1,;8,;27, …$

b.) A cube with size $n$ requires $ntimes n times n =n^3$ units. Thus, this becomes

$$
4^3,;5^3,;6^3implies 64,;125,;216,…
$$

c.) It is clear from the pattern that the general term of the sequence is just the cube of $n$, thus

$$
t_n=n^3
$$

d.) Using the general term in part(c), the number of unit cubes on the $15^{th}$ cube $(n=15)$ is

$$
t_{15}=15^3=3375
$$

a.) 1, 8 , 27, …

b.) 64, 125, 216, …

c.) $t_n=n^3$

d.) $t_{15}=3375$

a.) We shall find a pattern in the given sequences. We can try examining the first differences.

$t_2-t_1=-1$

$t_3-t_2=3$

$t_4-t_3=2$

$t_5-t_4=5$

The pattern for the first differences $t_n$ appears to be $t_n=t_{n-1}+t_{n-2}$ similar to a Fibonacci sequence.

Thus, knowing that $t_5=12$ and we expect that

$t_6=t_5+t_4=12+7=19$

$t_7=t_6+t_5=19+12=31$

$t_8=t_7+t_6=31+19=50$

$t_9=t_8+t_7=50+31=81$

$t_{10}=t_9+t_8=81+50=131$

$t_{11}=t_{10}+t_{9}=131+81=212$

$t_{12}=t_{11}+t_{10}=212+131=343$

$t_{13}=t_{12}+t_{11}=343+212=555$

$t_{14}=t_{13}+t_{12}=555+343=898$

$$
t_{15}=t_{14}+t_{13}=898+555=1;453
$$

b.) As mentioned in part (a), the recursive formula is

$t_n=t_{n-1}+t_{n-2}$ where $n>2$

a.) 19, 31, 50, 81, 131, 212, 343, 555, 898, 1453

b.) $t_1=3$, $t_2=2$, $t_n=t_{n-1}+t_{n-2}$ where $n>2$