Question

can anybody tell me if 3.9,3.99,3.999,......... is a sequence or not please its really important

Answer 1

Well, if r = 0.1, you could identify
\[a^{n+1}=a^n+r^n*0.9=a^n+0.1^n*0.9\]
with
\[a_0=3\]
right?

Answer 2

that means its a sequence isnt it

Answer 3

I gave the answer as you seemed to be in a hurry :)

Answer 4

I'm afraid I dont know the exact definition of a sequence, could anyone confirm?

Answer 5

its ok then

Answer 6

I should have made the sequence number a subscript for clarity:
\[a_{n+1}=a_n+r^n*0.9\]

Answer 7

Well the math checks out as far as I can see. At least with the parameters mentioned. That formula would turn out some weird numbers(at least weird compared to the stated ones) if you went left on the number line xP

Answer 8

thanks TimSmit

Answer 9

Hm that is true, only for positive numbers this :) But when you take
\[n \in \mathbb{N}\]
its ok right?

Answer 10

yes thats what I wanted to know thanks for helping me

Answer 11

or to be more clear
\[n \in \mathbb{N}_0\]

Answer 12

No problem, I hope I didn't give the answer too soon.

Answer 13

Tim sorry sorry the formula you gave me isnt giving me the sequence which I asked for

Answer 14

I know it's not that important but shouldn't
\[a _{n+1}=a _{n}+r ^{n}∗0.9\]
be
\[a _{n+1}=a _{n}+r ^{n}∗9\]

Answer 15

Yes. I was substituting that automatically for some reason so I didn't even notice it.

Answer 16

Unless \[a _{0}\] should be 3.9 then 0.9 is correct?

Answer 17

I start counting from 0 most of the times :) Matter of preference I guess...You are right if starting from 1.

Answer 18

Yes. a0 is 3.9 according to mahendra

Answer 19

Using a1=3.9 and starting at 1 it also checks out I think:
\[a_{n+1}=a_n + 0.1^n*0.9\]
\[3.99=3.9 + 0.1^1*0.9\]
\[3.999=3.99 + 0.1^2*0.9\]
etc

Answer 20

my concern is whether there are restrictions on r's value.

Answer 21

Can r be an "equation" instead of just a "number"?

Answer 22

this is the final answer to my question and Tim you were asking what is sequence any series whose next value can be derived from previous one is called a sequence here n is the only variable and its value only belongs to the set of natural no.

Answer 23

r is just a constant

Answer 24

So you're saying r should be a constant and your list of numbers is not a sequence?

Answer 25

no with r as a constant the list of no. we get is a sequence because with r as a constant every next term of the list can be derived from the previous one which is according to the definition of sequence

Answer 26

but r changes so r is not a constant...

Answer 27

the value of r is not changing its value is constant equal to 0.1 value of r to the power n is changing because n is variable

Answer 28

It is a sequence. Note that we can re-write the terms as:\[\bf \left\{ 3.9.3.99,3.999.. \right\}=\left\{ 3.9,3.9+0.09,3.99+0.009.. \right\}\]\[\bf = \left\{ 3.9+0.09\left( \frac{ 1 }{ 10 } \right)^0,3.99+0.09\left( \frac{ 1 }{ 10 } \right)^1,3.999+0.09\left( \frac{ 1 }{ 10 } \right)^2.. \right\}\]Hence the sequence is defined as:\[\bf a_n=a_{n-1}+0.09\left( \frac{ 1 }{ 10 } \right)^{n-1}\]Given that \(\bf a_1=3.9+0.09(1/10)^0=3.99\).

@m4mahendra20

Answer 29

Sorry, let me make a correction:

I mean to say; "Given that \(\bf a_1=3.9+0.09(1/10)^0=3.9\)"*

I accidentally wrote 3.99 in last statement...

Answer 30

So it's an arithmetic sequence?

Answer 31

i cant really say that the only thing i can say that it is a sequence but perhaps it is not arithmetic

Answer 32

You are adding someting to each preceding term so one would think it would be arithmetic instead of geometric. The only problem is that d (difference, we used r) is not a constant and I don't think you can do that for an arithmetic sequence.
\[a _{n}=a+(n-1)d\]
but in your case d changes.