Question

what is the 40th term of the arithmetic sequence?
21 18 15 12 9


seems easy, but i forgot how to do this

Answer 1

What pattern do you notice here?

Answer 2

This is easy... in a pattern each number in the pattern is called a "term" the first number in the pattern 21 is the first term.

Answer 3

all subtracted by the next variable, i get 3

Answer 4

to figure this you you must figure out the pattern. to figure it out just find the difference between all the numbers.

Answer 5

every other number starting with the first one gets smaller by 3. some patters will subtract and multiply, but this one just subtracts by 3.

Answer 6

i mean subtract and add, sorry.

Answer 7

so far you have 5 numbers in this pattern. if you want to get to the 40th term just subtract 9 by three and keep doing that 35 times.

Answer 8

thanks, and i really just wanted to know how to input the formula

Answer 9

or an easy way to do it is multiply 3 and 45 together because you have to just keep subtracting 3 from each term.

Answer 10

3 * 45 is 105... so just subtract 105 from 9.

Answer 11

the answer is -96

Answer 12

Example: \(\color{black}{\displaystyle 3,\,7,\,11,\,15,\,17. }\)
What is the \(50{\rm th}\) term of this arithmetic sequence.

Solution: ((Notation: \(a_i\) is the \(i{\rm th}\) term.))
You will note that you add \(4\) to obtain the succeeding term.
So, you know that
\(a_1+4=a_2\)
\(a_1+4+4=a_3\)
\(a_1+4+4+4=a_4\)
and so forth, you get,
\(a_1+\underbrace{4+4+4+...+4}_{\rm (n-1)~times}=a_n\)
or, alternatively, you thus may write
\(a_1+\underbrace{4+4+4+...+4}_{\rm 49~times}=a_{50}\)
\(a_1+4\times 49=a_{50}\)

and we know that \(a_1=3,\) so, \(a_{50}=3+4\times 49=199\)

Answer 13

\[\huge\color{orange}\checkmark\ \]

Answer 14

by 6, i meant the next term not the 40th, whoops XD

Answer 15

So, in general, based on the example above you will observethat for any arithmetic sequence that has a pattern of adding some number \(d\) to the succeeding term, you would have: \(\color{black}{\displaystyle a_1+(n-1)d=a_n }\).

Answer 16

\[\huge\color{red}p \huge\color{orange}u \huge\color{yellow}p \huge\color{green}p \huge\color{blue}i \huge\color{magenta}e \huge\color{red}s \]

Answer 17

Does this formula make sense to you, based on the above example?

Answer 18

ok yes, thanks to both of you :{D

Answer 19

So, if you want to find the 40th term of any sequence, in general, you then have
\(\color{black}{\displaystyle a_{40}=a_1+(40-1)d }\)
\(\color{black}{\displaystyle a_{40}=a_1+39d }\)

Answer 20

So, knowing that
1. The first term is: \(\color{black}{\displaystyle a_1=21 }\)
2. The difference is: \(\color{black}{\displaystyle d=-3 }\) (u subtract 3 (or add -3) each time)

you have
\(\color{black}{\displaystyle a_{40}=21+39(-3) }\)

Answer 21

-96

Answer 22

\(\color{black}{\displaystyle a_{40}=21+39(-3)=7(3)-39(3)=(7-39)(3)=(-32)(3)=-96~~~~ \checkmark }\)

Answer 23

What I did, is just a simple demonstration of multiplying without calculator and without headaches.

Answer 24

Anyway, the takeaway here is that:
\(\color{black}{\displaystyle a_{n}=a_1+(n-1)\times d }\)

where \(a_i\) denotes the \(i{\rm th}\) term, and
\(d\) denotes the "pattern" that you add.

This is your biggest friend for arithmetic sequence.

Answer 25

thanks so much

Answer 26

yw