Question

Hi...can somebody help me understand this question better?
The 1st 4 terms of the sequence {a } are____.
Please look at the picture of the problem I posted...Thanks
n
Given a1 oe 125 and an oe 5" an 1for n  1

lim
n n
Ä_
a oe ______

Answer 1

Pease ignore everything after the word "Thanks"

Answer 2

I feel lke this question is missing information or something!

Answer 3

I mean, should An only have numbers and some "n's" in it, not another An?

Answer 4

the equation gives you the relationship between any term of the sequence and its immediate predecessor.
a2, for example, denotes the second term of the sequence.
\[\large a_2=(1/5)a_{2-1}=(1/5)a_1\]
you have a1. Calculate a2. Then a3, a4, a5 in the same way.

Answer 5

hmmm...

Answer 6

so do i replace a1 with 125 to make it look like (1/5)125=25 ? Making the first first term in the sequence 25 ?

Answer 7

a1 IS the first term. So 25 (a2) would be the second term.

Answer 8

what do you mean?

Answer 9

if a2 = 25, why is a1 bigger at 125?

Answer 10

why do you think the previous term can't be bigger?

Answer 11

honestly I'm not sure...I'm pretty mixed up right now unfortunately :(

Answer 12

usually they give you a series called An and then say to calculate the first however many terms...then I just plop in a 1 for all the n's for example, then plop in a 2 for all the n's for example and so on...this questions seems backwards to me and confusing

Answer 13

A term CAN be bigger than the preceding term. No problemo.
This sequence will look like this:
125, 25, 5, 1, 0.2 ..... (each term is one-fifth the previous term)

You may be familiar with sequences where a term is bigger than it's previous term. Something like 2, 4, 8, 16, 32.....(each term is 2 times it's previous term).

The fact is, both these are examples of sequences.

Answer 14

*smaller (first line)

Answer 15

ok...so if I understand correctly, to actually get the first term a1 to be 125, that (1/5) portion of An must have been multiplied by 625 ?
So it would be like 125=(1/5)x where x = A_n-1 ?

Answer 16

So then A_zero would be 625 ?

Answer 17

Absolutely!

Answer 18

But, you don't have to find a0 for this particular question. Just sayin.
But I think you've understood the concept now.

Answer 19

ok...I think I get it...so then once I find a2, it will be (1/5) times 25 = 5 right?

Answer 20

and that would be the 3rd number in the sequence?

Answer 21

Yup. 5 will be a3.

Answer 22

then a4 = 1

Answer 23

yep

Answer 24

whew...thanks raj

Answer 25

you're welcome :)

Answer 26

one last questions though...

Answer 27

just to put my mind at ease...that is sort of a backwards question than say this one 15.bmp right?

Answer 28

what do you think the answer is again?

Answer 29

to which problem?

Answer 30

the one you just posted

Answer 31

dunno...one sec so i can figure it out...brb

Answer 32

the 1st 4 terms should be 1/3, 6/13, 9/17, and 4/7
and the lim as n -> infinity = zero, since the denominator will always grow bigger and faster than the numerator...is that correct?

Answer 33

wait...that cant be right...

Answer 34

the sequence is increasing...hmmm

Answer 35

Aha! that's what I thought when I was learning this stuff too.
But it's not true.
Here's how you get the right answer:
you take \(\large n^{2}\) common from the denominator and cancel it with the \(\large n^{2}\) of the numerator. NOW substitute n=\(\infty\).
Tell me what you get.

Answer 36

one sec...

Answer 37

is the limit 3/4?

Answer 38

BOOYA

Answer 39

i remember that now...if the degree of the num and denom are the same, the limit will be the ratio of their coefficients! haha

Answer 40

Yeah! that works too

Answer 41

so then what the heck is the limit of my first question then? yikes!

Answer 42

the first question, the An isn't normal looking...how do you take the limit of that (1/5)an-1 ?

Answer 43

idk if they've taught you this but it's given by
\[\frac{a}{1-r}\].

Answer 44

can you solve it for an?

Answer 45

wait....

Answer 46

its a geometric series...i need to find a and r...5/25=1/5...so the r must = (1/5)...that means a must equal 1...so a/1-r = 1/(1/5) = 5/4 and (5/4)>0 so the series converges...i think

Answer 47

i mean 1/(1-(1/5)=5/4

Answer 48

a is 125 (the first term of the sequence)

Answer 49

oh yeah...a=125

Answer 50

thanks i forgot that...so then 125/(1-(1/5)=???

Answer 51

yup

Answer 52

so what does a/(1-r) mean again? is that the limit of a geometric series?

Answer 53

yeah the limit when \(n\rightarrow \infty\)

Answer 54

geez...ok...looks like i have some more review to do...so then when writing out the limit...would you write lim as n ->infinity of a sub n-1 ...would that be correct?

Answer 55

i dont know why but that just doesn't feel right

Answer 56

it's like a n-1 isn't a real equation or function...do you know what i mean?

Answer 57

hmm..i see where you have a problem.
You see, that formula is used to calculate the limit of the SEQUENCE.
If you wanna write it, you write it like this:
It's not the limit of a_(n-1). That equation for a_(n-1) is just a rule you can use to work out the terms of the sequence.
But the whole sequence IN GENER
AL is represented as
\[\Large f(n) \equiv a_{n} = ar^{n-1}.\]
I hope you understand that the limit is the limit of the SEQUENCE.