Question

Which of the following equations would generate the sequence 4,7,10,13...

A. an=4n+3
B. an=4+3(n-1)
C. an=3n+4
D. an=4+7(n-1)

Answer 1

Ah! It probably looks like \(a_n=4n+3\) and so on, right? Well first, do you see a pattern in the numbers?

Answer 2

One way is to put values of n =0,1,2 ,3............and check which equation gives the correct sequence

Answer 3

no eric you are wrong

Answer 4

@atlas has a good idea, too!

Answer 5

I am wrong?

Answer 6

well i know it increases by 3 so i thought it was A also but when i did 4(2)+3 i got 11 instead of 10

Answer 7

so thats where im stuck

Answer 8

Then is \(a_n\) really \(a\times n\)? I was NOT supposing I gave the answer. I was questioning the formatting, and started with the first option and said, "and so on."

Answer 9

Another way is to see that each number in the sequence increases by 3 and the first number is 4.
so sequence is : 4, 4+3, 4+3+3, 4+3+3 and so on

Answer 10

Please, @atlas , don't immediately assume someone is wrong without questioning them or what you think is right - it leads to a less educational environment, in my opinion... But It was probably just a miscommunication! :)

Answer 11

So you see it should be 4+multiple of 3 i.e 4+3n

Answer 12

I am sorry theEric: I hope you see your mistake

Answer 13

it isnt a x n it is a sub n

Answer 14

So now we have two ways to look at it!

1. Use the options and see which one works.
2. Try to look at the pattern and form an expression for \(a_n\).

Both are just as good!

Answer 15

thank you both! :)

Answer 16

so 4+3n would create the same numbers in the arithmetic sequence?

Answer 17

Of course...........why don't you try out and see

Answer 18

an=4+3(n-1)

Answer 19

this is the ans

Answer 20

4 7 10 13
3 3 3


\(a_n = 4+3(n-1)\)

Answer 21

put n=1,2,3,4,5,6,7.........

Answer 22

@gorv: yeah it depends on whether you start from n=0 or n=1

Answer 23

I am a digital guy so I start counting from 0 :P

Answer 24

lol mee too a binary guy :)
n starts from 1, unless otherwise specified

Answer 25

yeah buddyyy

Answer 26

yeah if you count from n=1 -> the answer is 1+3n as others have pointed out :P

Answer 27

sorry i was eating lunch but thank you so much guys! :)

Answer 28

so the answer would be B. an=4+3(n-1)? :o

Answer 29

@atlas

Answer 30

yeah

Answer 31

thank you, just wanted to confirm :)

Answer 32

@ganeshie8 is right

Answer 33

I agree! :)

Answer 34

woohoo! @ganeshie8 is awesome & so are you guys! I wish i could give all of you guys medals :( lol

Answer 35

Haha, glad to help!

Answer 36

so for next time you dont start at 0?

Answer 37

i can see theEric was trying to help u by giving u a wrong example, and then proving to u why it wudnt work... and then we entered and spoiled the thing :)

Answer 38

hehe all of you guys were a great help! i really appreciate it :)

Answer 39

I still count from 0
It is a habit . :P

Answer 40

Eric helped me see why my first choice was wrong, because i had intially thought A was right :)

Answer 41

aww thnks, never start at 0, unless its specified explicitly in the question. sequences start from 1. in general humans count from 1, and machines count from 0 ! :P

Answer 42

Haha, I gave no example, sorry!

Miscommunication!
"Ah! It probably looks like an=4n+3 and so on, right? Well first, do you see a pattern in the numbers?"
That was as compared to "A. an=4n+3".

It is not \(an=4n+3\). It is \(a_n=4n+3\). The an=4n+3 was just the lack of formatting!

But you can test to eliminate options like that :)
Thanks!

Answer 43

makes sense! thanks @ganeshie8 :)

Answer 44

:) Take care all! Sorry for the confusion that I introduced!

Answer 45

All arithmetic expressions follow the same pattern :
first term + difference * (n-1)
4 + 3 * (n-1)

Answer 46

Check with this : 1,6,11,16.21.......... ??

Answer 47

ahh start wid n=1
4, 7, 10, 13, 16,....

Answer 48

I see you guys are still talking about this, so I'll share a link with you. It might be a nice part of your discussion, I don't know!

http://www.algebralab.org/lessons/lesson.aspx?file=Algebra_ArithSeq.xml