Question

Identify the 27th term of an arithmetic sequence where a1 = 38 and a17 = -74.

Answer 1

HELP

Answer 2

\(\large\color{black}{ a_n=a_1+(n-1)d }\) is the formula you need.

Answer 3

Choose one answer.
A. -20.5
B. -151
C. -22.75
D. -144

i tihkn its b

Answer 4

Plug in \(\large\color{black}{ a_{17} }\) and \(\large\color{black}{ a_{1} }\) that you are given,

into \(\large\color{black}{ a_{n}=a_1+(n-1)d }\)

Answer 5

You are plugging:
~ \(\large\color{black}{ 38 }\) for \(\large\color{black}{ a_{1} }\)
~ \(\large\color{black}{ -74 }\) for \(\large\color{black}{ a_{n} }\)
~ \(\large\color{black}{ 17 }\) for \(\large\color{black}{ n }\)

and solving for \(\large\color{black}{ d }\) .

Answer 6

is it b?

Answer 7

wait, I don't get you, what is b?

Answer 8

oh you mean the answer is b?

Answer 9

yes

Answer 10

I haven't found out yet...

Answer 11

how did you get this answer?

Answer 12

-74=38(17-1)d

Answer 13

that cant be b

Answer 14

that expression is not an answer, it is just a set up to find the d (based on the given 1st and 17th terms)

Answer 15

after you have found \(\large\color{black}{ d }\),
then use:

\(\large\color{black}{ a_{27}=38+d(27-1) }\)

but with your given \(\large\color{black}{ d }\) (difference).

Answer 16

if you got any questions about what I said so far, ask them please...

Answer 17

-74=608d ?

Answer 18

oh there is a plus that you missed between 38:
to solve for d you should use:
\(\large\color{black}{ -74=38\color{blue}{+}d(17-1) }\) (with a plus, which you missed)

Answer 19

solve for d (using my just now posted expression)

Answer 20

i dont know what do do im just very confused

Answer 21

where are you confused?
you don't get the entire thing, or just some part of it?

Answer 22

i dont get anything im homeschooled and dont get taught this

Answer 23

okay, lets start of from knowing the notation:

for example you have the following sequence.
\(\large\color{black}{ 1,~3~,5~,7~,9~,11~... }\)

you can tell that you add +2 to get the the next term.
that means that the difference is \(\large\color{black}{ 2 }\) .

we use a notation to denote that, like this: \(\large\color{blue}{ d=2 }\)

Answer 24

does this make sense so far?

Answer 25

yes

Answer 26

@Directrix

Answer 27

okay, now the \(\large\color{black}{ a_1 }\) is just a way to denote your first term.
(as, \(\large\color{black}{ a_n }\) is any \(\large\color{black}{ \rm~\color{red}{n}th~term }\) )

Answer 28

I disconnected sorry

Answer 29

kinda yes

Answer 30

\(\large\color{black}{ a_1 }\) is a notation for the \(\large\color{black}{ 1 }\)st term in a sequence.
\(\large\color{black}{ a_2 }\) is a notation for the \(\large\color{black}{ 2 }\)nd term in a sequence.
\(\large\color{black}{ a_3 }\) is a notation for the \(\large\color{black}{ 3 }\)rd term in a sequence.
\(\large\color{black}{ a_4 }\) is a notation for the \(\large\color{black}{ 4 }\)th term in a sequence.

.....

Answer 31

yes, so the formula for any \(\large\color{red}{ \rm n }\)th term in a sequence
(when you are adding/subtracting the same number to the terms)

is: \(\large\color{blue}{ a_n=a_1+(n-1)\times d }\)

Answer 32

we will see why it makes sense:
lets say, again, we have:

\(\large\color{black}{ 1~, 3~ ,5~ ,7~ ,9 ~,11~ ... }\)
you can tell
that to find \(\large\color{black}{ a_2 }\) (the 2nd term) you add 2 to the first term once.

Answer 33

@SolomonZelman i don't think that crusty understands this at all, maybe you could just give them the answer?

Answer 34

and so, will you:

add \(\large\color{black}{ 2 }\) once to the \(\large\color{black}{ a_1 }\) to find \(\large\color{black}{ a_2 }\)
add \(\large\color{black}{ 2 }\) twice to the \(\large\color{black}{ a_1 }\) to find \(\large\color{black}{ a_3 }\)
add \(\large\color{black}{ 2 }\) 3 times to the \(\large\color{black}{ a_1 }\) to find \(\large\color{black}{ a_4 }\)
add \(\large\color{black}{ 2 }\) 4 times to the \(\large\color{black}{ a_1 }\) to find \(\large\color{black}{ a_5 }\)

Answer 35

again disconnected.

Answer 36

again, back to your sequence:

Answer 37

i g2g soon

Answer 38

you are given that:

\(\large\color{black}{a_1=38 }\)
\(\large\color{black}{a_{17}=-74 }\)

Answer 39

there is our formula:
\(\large\color{black}{a_{n}=a_1+(n-1)d }\)

plugging in your terms:
\(\large\color{black}{a_{17}=38+(17-1)d }\)

Answer 40

can you solve for \(\large\color{black}{d }\) (the difference, -i.e. the number you add/subtract
every time to get to the next term)

Answer 41

go ahead...:)

Answer 42

I forgot to plug in 17th term:

\(\large\color{black}{-74=38+(17-1)d }\) like this,>> NOW, solve for d.

Answer 43

hmm aint it C. -22.75 sorry im in kind of a rush

Answer 44

nope