Question

Find an equation for the nth term of the arithmetic sequence.

-15, -6, 3, 12, ...

Answer 1

Find a pattern. What are you doing to each number to produce the next in the sequence?

Answer 2

plus 9

Answer 3

-15 + 9(n - 1)

Answer 4

yeah, that type has constant change through one term to next

Answer 5

so i'm right?

Answer 6

yeah

you can test it, put in values of n, does it generate that sequence

Answer 7

thank you!! can u help with another one

Answer 8

Find an equation for the nth term of the arithmetic sequence.

a14 = -33, a15 = 9

Answer 9

sure

Answer 10

that type of sequence, like i said has a cponstant change from one term to the next. like the +9 in the last one..

Answer 11

\[a _{14}=-33 \] and \[a _{15} = 9\]

Answer 12

here they gave you consecutive terms, and told you the sequence is arithmetic

Answer 13

the constant change from one term to the next is 42

Answer 14

yes

Answer 15

so \[a _{n}\]= -579 + 42(n - 1) ??

Answer 16

or is it an = -579 + 42(n + 1)

Answer 17

f the initial term is a1 and the common difference is d, then the nth term of the sequence is given by;

an = a1 + (n-1)d

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By taking the above result further, the nth term can be given also as;

an = am + (n-m)d, where am is a random term in the sequence such that n > m.

Answer 18

is my answer right then

Answer 19

an = am + (n-m)*d


an = -33 + (n - 15) *42

Answer 20

that isnt one of my answer choices

Answer 21

that more general form for any 2 terms, not just 1 and 2

Answer 22

yeah fix it up ,

an = initial term a0 + (n -1)*d

Answer 23

an = -33 + (n-14)*42

an = -33 + (n-1-13)*42

an = -33 + (n-1)*42 - (13*42)

an = -579 + (n-1)*42

Answer 24

the initial term a0 is -579

Answer 25

thx

Answer 26

welcome