a.) Find the difference between consecutive terms

$5-1=4$

$9-5=4$

$13-9=4$

$17-13=4$

A common difference of 4 is observed. Hence, it is $bold{arithmetic}$.

b.)

$7-3=4$

$13-7=6$

$17-13=4$

$23-17=6$

$27-23=4$

An alternating difference of $4$ and $6$ is observed. Hence, it is $bold{not;arithmetic}$.

c.)

$6-3=3$

$12-6=6$

$24-12=12$

The difference between consecutive terms varies. Hence, it is $bold{not; arithmetic}$.

d.)

$48-59=-11$

$37-48=-11$

$26-37=-11$

$15-26=-11$

A common difference of -11 is observed. Hence, it is $bold{arithmetic}$

$bold{General;Term:}$ To obtain the general term of an arithmetic sequence, use the following formula

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

where

$t_n=n^{th}$ term

$t_1=$ first term

$d$ = common difference

$bold{Recursive;Formula:}$ The recursive formula of arithmetic sequence is written as

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

a.) $t_1=28$ ; $d=42-28=14$

General term:

$t_n=28+(n-1)(14)$

$t_n=28+14n-14$

$t_n=14+14n$

Recursive formula:

$t_1=28; ;;t_n=t_{n-1}+14$ where $n>1$

b.) $t_1=53$ ; $d=49-53=-4$

General term:

$t_n=53+(n-1)(-4)$

$t_n=53-4n+4$

$t_n=57-4n$

Recursive formula:

$t_1=53; ; ;t_n=t_{n-1}-4$ where $n>1$

c.) $t_1=-1$, $d=(-111)-(-1)=-110$

General term:

$t_n=-1+(n-1)(-110)$

$t_n=-1-110n+110$

$t_n=109-110n$

Recursive formula

$t_1=-1; ; ;t_n=t_{n-1}-110$ where $n>1$

a.) $t_n=14+14n$ , $t_1=28; ;;t_n=t_{n-1}+14$ where $n>1$

b.) $t_n=57-4n$ , $t_1=53; ; ;t_n=t_{n-1}-4$ where $n>1$

c.) $t_n=109-110n$ , $t_1=-1; ; ;t_n=t_{n-1}-110$ where $n>1$

For an arithmetic sequence $t_1, t_2, t_3, t_4 …$

The common difference is

$d= t_2-t_1=t_3-t_2=t_4-t_3=t_n-t_{n-1}$

Thus, the next term can be obtained from the preceding term as

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

$t_{10}=29$

$t_{11}=41$

We shall find the common difference

$d=t_{11}-t_{10}=41-29=12$

Therefore,

$$
t_{12}=t_{11}+d=41+12=53
$$

$$
t_{12}=53
$$

$bold{General;Term:}$ To obtain the general term of an arithmetic sequence, use the following formula

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

where

$t_n=n^{th}$ term

$t_1=$ first term

$d$ = common difference

$t_1=85$

We shall find the common difference

$d=102-85=17$

Thus, the general term is

$t_n=85+(n-1)(17)$

$t_n=85+17n-17$

$$
t_n=17n+68
$$

To find the 15th term, substitute $n=15$ to the general term.

$t_{15}=17(15)+68$

$$
t_n=323
$$

$t_n=17n+68$ ; $t_n=323$

$bold{Identifying;Arithmetic;Sequence}$. To determine whether a given sequence is arithmetic or not, find the difference between two consecutive terms. If the difference is constant, it is arithmetic. Otherwise, it is not arithmetic. In other words, a sequence $t_1, t_2, t_3, t_4 …$ is arithmetic if

$$
t_2-t_1=t_3-t_2=t_4=t_3=t_n-t_{n-1}
$$

$bold{General;Term:}$ To obtain the general term of an arithmetic sequence, use the following formula

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

where

$t_n=n^{th}$ term

$t_1=$ first term

$d$ = common difference

$bold{Recursive;Formula:}$ The recursive formula of arithmetic sequence is written as

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

a.)

(i) Notice that there is a common difference of 3 between consecutive terms. Thus, it is arithmetic.

(ii) $t_1=8$ , $d=3$

General term:

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

$t_n=8+3n-3$

$t_n=3n+5$

(iii) Recursive formula

$t_1=8$ ; $t_n=t_{n-1}+3$ where $n>1$

e.)
(i) A common difference of $d=11$ is observed with $t_1=23$. It is an arithmetic sequence.

(ii) The general term is

$t_n=23+(n-1)(11)$

$t_n=23+11n-11$

$t_n=11n+12$

The recursive formula is

$t_1=23$ ; $t_n=t_{n-1}+11$ where $n>1$

f.) (i) A common difference of $d=dfrac{1}{6}$ is observed with $t_1=dfrac{1}{6}$.

(ii) The general term is

$t_n=dfrac{1}{6}+(n-1)cdot dfrac{1}{6}$

$t_n=dfrac{1}{6}+dfrac{1}{6}n-dfrac{1}{6}$

$t_n=dfrac{n}{6}$

(iii) The recursive formula is

$t_1=dfrac{1}{6}$ ; $t_n=t_{n-1}+dfrac{1}{6}$ ; where $n>1$

a.) $t_n=3n+5$

b.) not arithmetic

c.) not arithmetic

d.) not arithmetic

e.) $t_n=11n+12$ , $t_1=23$ ; $t_n=t_{n-1}+11$ where $n>1$

f.) $t_n=dfrac{n}{6}$ ; $t_1=dfrac{1}{6}$ ; $t_n=t_{n-1}+dfrac{1}{6}$ where $n>1$

$bold{Identifying;Arithmetic;Sequence}$. To determine whether a given sequence is arithmetic or not, find the difference between two consecutive terms. If the difference is constant, it is arithmetic. Otherwise, it is not arithmetic. In other words, a sequence $t_1, t_2, t_3, t_4 …$ is arithmetic if

$$
t_2-t_1=t_3-t_2=t_4=t_3=t_n-t_{n-1}
$$

$bold{General;Term:}$ To obtain the general term of an arithmetic sequence, use the following formula

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

where

$t_n=n^{th}$ term

$t_1=$ first term

$d$ = common difference

$bold{Recursive;Formula:}$ The recursive formula of arithmetic sequence is written as

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

a.)

(i) Notice that there is a common difference of 8 between consecutive terms. Thus, it is arithmetic.

(ii) $t_1=19$ , $d=8$

General term:

$t_n=19+(n-1)(8)$

$t_n=19+8n-8$

$t_n=8n+11$

(iii) Recursive formula

$t_1=19$ ; $t_n=t_{n-1}+8$ where $n>1$

b.)

(i) Notice that there is a common difference of -5 between consecutive terms. Thus, it is arithmetic.

(ii) $t_1=4$ , $d=-5$

General term:

$t_n=4+(n-1)(-5)$

$t_n=4-5n+5$

$t_n=9-5n$

(iii) Recursive formula

$t_1=4$ ; $t_n=t_{n-1}-5$ where $n>1$

c.)

(i) Notice that there is a common difference of 5 between consecutive terms. Thus, it is arithmetic.

(ii) $t_1=21$ , $d=5$

General term:

$t_n=21+(n-1)(5)$

$t_n=21+5n-5$

$t_n=16+5n$

(iii) Recursive formula

$t_1=21$ ; $t_n=t_{n-1}+5$ where $n>1$

d.)

(i) Notice that there is a common difference of $-12$ between consecutive terms. Thus, it is arithmetic.

(ii) $t_4=35$ , $d=-12$

Find the first term.

$35=t_1+(4-1)(-12)$

$$
35=t_1-36
$$

$t_1=71$

General term:

$t_n=71+(n-1)(-12)$

$t_n=71-12n+12$

$t_n=83-12n$

(iii) Recursive formula

$t_1=71$ ; $t_n=t_{n-1}-12$ where $n>1$

a.) $t_n=8n+11$

b.) $t_n=9-5n$

c.) $t_n=5n+16$

d.) $t_n=83-12n$

$bold{Identifying;Arithmetic;Sequence}$. To determine whether a given sequence is arithmetic or not, find the difference between two consecutive terms. If the difference is constant, it is arithmetic. Otherwise, it is not arithmetic. In other words, a sequence $t_1, t_2, t_3, t_4 …$ is arithmetic if

$$
t_2-t_1=t_3-t_2=t_4=t_3=t_n-t_{n-1}
$$

$bold{Recursive;Formula:}$ The recursive formula of arithmetic sequence is written as

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

a.) From the recursive formula, the first term is 13 and the common difference $d=14$

The first 5 terms are

$t_1=13$

$t_2=13+14=27$

$t_3=27+14=41$

$t_4=41+14=55$

$$
t_5=55+14=69
$$

b.) The recursive formula results in constant ratio instead of constant difference.

$t_1=5$

$t_2=3(5)=15$

$t_3=3(15)=45$

$15-5neq 45-15$

Thus, it is not an arithmetic sequence.

c.) With this recursive formula,

$t_1=4$

$t_2=4+2-1=5$

$t_3=5+3-1=7$

$5-4neq7-5$

Thus, it is not an arithmetic sequence.

d.) With this recursive formula

$t_1=1$

$t_2=2(1)-2+2=2$

$t_3=2(2)-3+2=3$

$t_4=2(3)-4+2=4$

$t_5=2(4)-5+2=5$

A common difference of $d=1$ is observed, thus it is an arithmetic sequence.

The general term is

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

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

$t_n=n$

The first 5 terms are 1,2,3,4,5

$bold{Identifying;Arithmetic;Sequence}$. To determine whether a given sequence is arithmetic or not, find the difference between two consecutive terms. If the difference is constant, it is arithmetic. Otherwise, it is not arithmetic. In other words, a sequence $t_1, t_2, t_3, t_4 …$ is arithmetic if

$$
t_2-t_1=t_3-t_2=t_4=t_3=t_n-t_{n-1}
$$

$bold{General;Term:}$ To obtain the general term of an arithmetic sequence, use the following formula

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

where

$t_n=n^{th}$ term

$t_1=$ first term

$d$ = common difference

$bold{Recursive;Formula:}$ The recursive formula of arithmetic sequence is written as

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

a.)

Notice that there is a common difference of 5 between consecutive terms. Thus, it is arithmetic.

(i) $t_1=35$ , $d=5$

General term:

$t_n=35+(n-1)(5)$

$t_n=35+5n-5$

$t_n=5n+30$

(ii) Recursive formula

$t_1=35$ ; $t_n=t_{n-1}+35$ where $n>1$

(iii) We shall find the eleventh term $t_{11}$

$$
t_{11}=5(11)+30=85
$$

b.)

Notice that there is a common difference of $-11$ between consecutive terms. Thus, it is arithmetic.

(i) $t_1=31$ , $d=-11$

General term:

$t_n=31+(n-1)(-11)$

$t_n=31-11n+11$

$t_n=42-11n$

(ii) Recursive formula

$t_1=31$ ; $t_n=t_{n-1}+31$ where $n>1$

(iii) We shall find the eleventh term $t_{11}$

$$
t_{11}=42-11(11)=-79
$$

c.)

Notice that there is a common difference of $-12$ between consecutive terms. Thus, it is arithmetic.

(i) $t_1=-29$ , $d=-12$

General term:

$t_n=-29+(n-1)(-12)$

$t_n=-29-12n+12$

$t_n=-17-12n$

(ii) Recursive formula

$t_1=-29$ ; $t_n=t_{n-1}-12$ where $n>1$

(iii) We shall find the eleventh term $t_{11}$

$$
t_{11}=-17-12(11)=-149
$$

d.)

Notice that there is a common difference of $0$ between consecutive terms. Thus, it is arithmetic.

(i) $t_1=11$ , $d=0$

General term:

$t_n=11+(n-1)(0)$

$t_n=11$

(ii) Recursive formula

$t_1=11$ ; $t_n=t_{n-1}$ where $n>1$

(iii) We shall find the eleventh term $t_{11}$

$$
t_{11}=11
$$

e.)

Notice that there is a common difference of $1/5$ between consecutive terms. Thus, it is arithmetic.

(i) $t_1=1$ , $d=1/5$

General term:

$t_n=1+(n-1)left(dfrac{1}{5}right)$

$t_n=1+dfrac{1}{5}n-dfrac{1}{5}$

$t_n=dfrac{1}{5}n+dfrac{4}{5}$

(ii) Recursive formula

$t_1=1$ ; $t_n=t_{n-1}+dfrac{1}{5}$ where $n>1$

(iii) We shall find the eleventh term $t_{11}$

$$
t_{11}=dfrac{1}{5}(11)+dfrac{4}{5}=3
$$

f.)

Notice that there is a common difference of $0.17$ between consecutive terms. Thus, it is arithmetic.

(i) $t_1=0.4$ , $d=0.17$

General term:

$t_n=0.4+(n-1)(0.17)$

$t_n=0.4+0.17n-0.17$

$t_n=0.17n+0.23$

(ii) Recursive formula

$t_1=0.4$ ; $t_n=t_{n-1}+0.17$ where $n>1$

(iii) We shall find the eleventh term $t_{11}$

$$
t_{11}=0.17(11)+0.23=2.1
$$

a.) $t_{11}=85$

b.) $t_{11}=-79$

c.) $t_{11}=-149$

d.) $t_{11}=11$

e.) $t_{11}=3$

f.) $t_{11}=2.1$

$bold{General;Term:}$ To obtain the general term of an arithmetic sequence, use the following formula

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

where

$t_n=n^{th}$ term

$t_1=$ first term

$d$ = common difference

$bold{Identifying;Arithmetic;Sequence;from;General;Term:}$ If the general term is a linear function, then it is arithmetic. A linear function is a polynomial such that the exponent of the variable is 1, usually written in the form

$y=mx+b$

where $m$ = slope and $b$ = y-intercept

a.) This is a linear function with $m=-2$ and $b=8$

The first 5 terms are

$t_1=8-2(1)=6$

$t_2=8-2(2)=4$

$t_3=8-2(3)=2$

$t_4=8-2(4)=0$

$t_5=8-2(5)=-2$

The common difference is $-2$

c.) This is a linear function with $m=dfrac{1}{4}$ and $b=dfrac{1}{2}$. Thus, the first five terms are

$t_1=dfrac{1}{4}cdot 1+dfrac{1}{2}=dfrac{3}{4}$

$t_2=dfrac{1}{4}cdot 2+dfrac{1}{2}=1$

$t_3=dfrac{1}{4}cdot 3 +dfrac{1}{2}=dfrac{5}{4}$

$t_4=dfrac{1}{4}cdot 4+dfrac{1}{2}=dfrac{3}{2}$

$t_5=dfrac{1}{4}cdot 5 +dfrac{1}{2}=dfrac{7}{4}$

The common difference is $dfrac{1}{4}$

a.) arithmetic with $d = -2$

b.) not arithmetic

c.) arithmetic with $d =dfrac{1}{4}$

d.) not arithmetic

27,34, 41 …

$d=34-27=41-34=7$

Thus, it is arithmetic sequence.

$bold{Identifying;Arithmetic;Sequence}$. To determine whether a given sequence is arithmetic or not, find the difference between two consecutive terms. If the difference is constant, it is arithmetic. Otherwise, it is not arithmetic. In other words, a sequence $t_1, t_2, t_3, t_4 …$ is arithmetic if

$$
t_2-t_1=t_3-t_2=t_4=t_3=t_n-t_{n-1}
$$

Since $t_1=27$ and $d=7$, the general term is

$t_n=27+(n-1)(7)$

$t_n=27+7n-7$

$$
t_n=7n+20
$$

$bold{General;Term:}$ To obtain the general term of an arithmetic sequence, use the following formula

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

where

$t_n=n^{th}$ term

$t_1=$ first term

$d$ = common difference

a.) At 10th row,

$$
t_{10}=7(10)+20=90
$$

Use the general term to calculate the $n^{th}$ term.

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

$181=27+(n-1)(7)$

$n=dfrac{1}{7}(181-27)+1=23$

The opera house has 23 rows.

We need to find $n$ such that $t_n=181$.
Use the formula for general term.

a.) $t_{10}=90$

b.) $n=23$

$bold{General;Term:}$ To obtain the general term of an arithmetic sequence, use the following formula

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

where

$t_n=n^{th}$ term

$t_1=$ first term

$d$ = common difference

Her salary starts at $$9.25$ and each month, it goes up by $$0.15/hr$. We shall find when would her salary reach double her starting salary, that is, $2cdot 9.25 = $18.50$.

In this case, $t_1=9.25$, $d=0.15$, we will find $n$

$18.50=9.25+(n-1)(0.15)$

$n=dfrac{1}{0.15}(18.50-9.25)+1=62dfrac{2}{3}$

Therefore, her salary will be at least double on her 63$^{rd}$ month.

$63^{rd}$ month

The initial amount $$5,000$ in the account increases each year at $3.5%$, that is $5,000cdot 0.035=175$ per year. We shall find on which year does the account value reach at least $7,800$.

Here, we can apply the formula for arithmetic sequence.

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

In this case,

$t_1=5,000$ , $t_n=7,800$ , $d=175$, and we need to find $n$.

$7;800=5;000+(n-1)(175)$

$n=dfrac{1}{175}(7800-5000)+1=17$

Thus, his account will reach $7;800$ on the $17^{th}$ year or after $16$ years

$bold{Identifying;Arithmetic;Sequence}$. To determine whether a given sequence is arithmetic or not, find the difference between two consecutive terms. If the difference is constant, it is arithmetic. Otherwise, it is not arithmetic. In other words, a sequence $t_1, t_2, t_3, t_4 …$ is arithmetic if

$$
t_2-t_1=t_3-t_2=t_4=t_3=t_n-t_{n-1}
$$

$bold{General;Term:}$ To obtain the general term of an arithmetic sequence, use the following formula

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

where

$t_n=n^{th}$ term

$t_1=$ first term

$d$ = common difference

a.)

Notice that there is a common difference of 2 between consecutive terms. Thus, it is arithmetic sequence with $t_1=7$ , $d=2$

General term:

$t_n=7+(n-1)(2)$

$t_n=7+2n-2$

$t_n=2n+5$

If 63 is the last term, we shall find how many terms are present

$63=2(n)+5$

$n=dfrac{1}{2}(63-5)=29$

Thus, the sequence contains 29 terms.

b.)

Notice that there is a common difference of $-5$ between consecutive terms. Thus, it is arithmetic sequence with $t_1=-20$ , $d=-5$

General term:

$t_n=-20+(n-1)(-5)$

$t_n=-15-5n$

If $-205$ is the last term, we shall find how many terms are present

$-205=-15-5n$

$n=dfrac{1}{5}(205-15)=38$

Thus, the sequence contains 38 terms.

c.)

Notice that there is a common difference of $-4$ between consecutive terms. Thus, it is arithmetic sequence with $t_1=31$ , $d=-4$

General term:

$t_n=31+(n-1)(-4)$

$t_n=35-4n$

If $-25$ is the last term, we shall find how many terms are present

$-25=35-4n$

$n=dfrac{1}{4}(35+25)=15$

Thus, the sequence contains 15 terms.

d.)

Notice that there is a common difference of $7$ between consecutive terms. Thus, it is arithmetic sequence with $t_1=9$ , $d=7$

General term:

$t_n=9+(n-1)(7)$

$t_n=7n+2$

If $100$ is the last term, we shall find how many terms are present

$100=7n+2$

$n=dfrac{1}{7}(100-2)=14$

Thus, the sequence contains 14 terms.

e.)

Notice that there is a common difference of $7$ between consecutive terms. Thus, it is arithmetic sequence with $t_1=-33$ , $d=7$

General term:

$t_n=-33+(n-1)(7)$

$t_n=7n-40$

If $86$ is the last term, we shall find how many terms are present

$86=7n-40$

$n=dfrac{1}{7}(86+40)=18$

Thus, the sequence contains 18 terms.

f.)

Notice that there is a common difference of $-9$ between consecutive terms. Thus, it is arithmetic sequence with $t_1=28$ , $d=-9$

General term:

$t_n=28+(n-1)(-9)$

$t_n=37-9n$

If $-44$ is the last term, we shall find how many terms are present

$-44=37-9n$

$n=dfrac{1}{9}(37+44)=9$

Thus, the sequence contains 9 terms.

a.) $n=29$

b.) $n=38$

c.) $n=15$

d.) $n=14$

e.) $n=18$

f.) $n=9$

Once the common difference $d$ is obtained, the terms in the sequence can be obtained as

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

We shall determine how much must be added to $t_4$ to obtain $t_8$

$t_8=t_1+(8-1)d$

$t_4=t_1+(4-1)d$

Find the difference between $t_8$ and $t_4$

$t_8 – t_4=(t_1-t_1)+(7-3)d = 4d$

Thus, to obtain $t_8$ from $t_4$, we must add $4d$.

Similarly, to obtain $t_{100}$ from $t_{8}$, we must add

$$
t_{100}-t_8=(t_1-t_1)+(99-7)d=92d
$$

$4d$ and $92d$

Once the common difference $d$ is obtained, the terms in the sequence (general term) can be obtained as

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

In this case, $t_{93}=539$ and $t_{50}=238$, using the formula above

$539=t_1+(93-1)d$

$238=t_1+(50-1)d$

Subtract the 1st and 2nd equation.

$$
539-238=(t_1-t_1)+(92-49)d implies d=dfrac{539-238}{92-49}=7
$$

We shall find $t_1$ from either of the two equations.

$t_1=539-92(7)=-105$

Thus, the general term is

$t_n=-105+(n-1)(7)$

$$
t_n=-112+7n
$$

$d=7$ ; $t_n=-112+7n$

a) We must create three different sequences containing 20 and 50.

sequence 1: 10, 20, 30, 40, 50, 60, 70 …

sequence 2: 15, 20, 25, 30, 35, 40, 45, 50, 55, 60 …

sequence 3: 60, 50, 40, 30, 20, 10, 0, $-$10 …

b) sequence 1: $d=10$

sequence 2: $d=5$

sequence 3: $d=-10$

The common difference must be a factor of both 20 and 50. In other words, both of them should be divisible by the common difference.

In this case, $t_1=13$. If $37$ and $73$ are part of the arithmetic sequence, we still don’t know which term are they. Let’s assume $t_j=37$ and $t_k=73$. We shall write the general term for each case

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

$37=13+(j-1)(d)implies 24=(j-1)(d)implies (j-1)=dfrac{24}{d}$

$73=13+(k-1)(d)implies 60=(k-1)(d)implies (k-1)=dfrac{60}{d}$

Notice that $j$ and $k$ should be a positive integer, so we shall find possible values of $d$ such that $dfrac{24}{d}$ and $dfrac{60}{d}$ are integers.

In that case, $d$ could be $1,2,3,4,6, 12$

Now, we shall find $t_{100}$ for each possible values of $d$

$d=1implies t_{100}=13+(100-1)(1)=112$

$d=2implies t_{100}=13+(100-1)(2)=211$

$d=3implies t_{100}=13+(100-1)(3)=310$

$d=4implies t_{100}=13+(100-1)(4)=409$

$d=6implies t_{100}=13+(100-1)(6)=607$

$$
d=12implies t_{100}=13+(100-1)(12)=1201
$$

$$
112,; 211, ;310,;409,;607,;1201
$$

In the expression for general term of arithmetic sequence

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

We shall determine whether we would still get an arithmetic sequence if the $nth$ term of the new sequence is replaced by the $t_n^{th}$ of the original arithmetic sequence.

We shall let $k_n$ be the new arithmetic sequence.

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

$k_n=t_1+[(t_1+(n-1)(d))-1](d)$

$k_n=t_1+t_1d+(n-1)(d^2)-d$

$$
k_n=(t_1+t_1d-d)+(n-1)d^2
$$