Question

PLEASE HELP!!
how do i find a1 for the given geometric series??
sn=48455, r=3.2, n=3????

Answer 1

what is the formula for finding a1 in a geometric series???

Answer 2

can you help me hero? i have been trying to get help forever!!

Answer 3

Hang on a sec :D

Answer 4

thank you!! :D

Answer 5

I think the formula is

\(a_n = a_{1}r^{(n-1)}\)

And to get \(a_1\) you simply divide both sides by \(r^{n-1}\) :

\(a_1 = \large\frac{a_n}{a^{n-1}}\)

Answer 6

Whoops

Answer 7

\(a_1 = \large\frac{a_n}{r^{n-1}}\)

Answer 8

but how do i solve for an if i dont have a1?

Answer 9

So all you have to do is plug the numbers in

Answer 10

\(a_1 = \large\frac{48455}{3.2^{(3-1)}}\)

Answer 11

thank you!! i get it now:)

Answer 12

You might want to confirm that with someone else. Maybe @amistre64

Answer 13

ok will do!

Answer 14

I found another formula so I don't know which one it is

Answer 15

\[S_n=a1*\frac{1-r^n}{1-r}\]

Answer 16

Yeah, that's the other one I found

Answer 17

i had it in the cupboard behind the peanut butter cookies

Answer 18

sn=48455, r=3.2, n=3????

Sn probably means the Sum of the first n terms, not the nth term, which would be An.

Answer 19

I should never have found the wrong formula.

Answer 20

ok...so solve for An then?

Answer 21

The correct formula doesn't seem to have A_n in it

Answer 22

Look at what @amistre64 posted.

Answer 23

i already know the formula for sn...i need to solve for a1..which is what is confusing me so much

Answer 24

Alright, so let's first just define some basic terms, okay?
r= the common ratio
\(\large A_n = \text{the nth term}\)
\(\large S_n = \text{the sum of the first n terms}\)

Now, the cool thing is that there is a formula for the sum of the first n terms of a geometric sequence. We'll plug into this formula everything that we know, and that will let us solve for \(a_1\).

Answer 25

The formula we have is:
\(\Large S_n = a_1*\frac{1-r^n}{1-r}\)

The information we have is
\(S_n=48455, r=3.2, n=3\)
So let's plug that information into the formula we have.

Answer 26

some people swap out a negtaive and write (r^n-1)/(r-1)

Answer 27

\(\Large S_3 = 48455 = a_1*\frac{1-3.2^3}{1-3.2}\)
Do you see how I get that? Ask me questions.

Answer 28

okay..so i did that and got -1385.69...correct?..or no good

Answer 29

ok..nvrm ignore my answer..i didnt put the 3 in for sn....continue

Answer 30

You can substitute the 3 in wherever you have n, but that doesn't mean you substitute \(S_n \rightarrow 3\)
Instead, you can substitute and get \(S_n \rightarrow S_3\)

Answer 31

\[S=a*\frac tb\]\[\frac bt(S=a*\frac tb)\to\ S\frac bt = a\]

Answer 32

So I'm not sure what you were trying there, but let's step back and make sure we understand the process.

Step 1: Start with the correct general formula. In this case, it's
\(\Large S_n = a_1 *\frac{1-r^n}{1-r}\)
Step 2: Look at the information we're given and substitute it into the formula, leaving the unknown info as a variable.
Step 3: Do some basic algebra, solve for the unknown. In this case that unknown is \(a_1\).

Answer 33

So let's take this step by step. Do you understand step 1? Does that formula make sense to you, and do you know what all of the variables mean?

Answer 34

yes looks good so far

Answer 35

Great, step 2. Can you substitute in the things you know?

Answer 36

I don't mean to interrupt. I've clearly posted the wrong formulas and things but @mathdummy4, now that you have the right formulas, all you have to do is isolate a_1 since it is what you need to find:

\(a_1 = \large\frac{s^n}{\frac{1-r^n}{1-r}}\)

Answer 37

We're getting there, hero.

Answer 38

That's a shortcut @SmoothMath

Answer 39

When you're working with a student, you have to go at their pace and make sure they understand each step. I know you and I are good with the shortcut, but I'd rather take a steady step by step approach for her sake.

Answer 40

thank you hero, your not interrupting...and based on your formula i got 48419.032, but i do appreciate having someone explain the whole process to me which smoothmath is willing to do

Answer 41

Wrong.

Answer 42

wrong?

Answer 43

Wrong.

Answer 44

....which part the answer or the willing to explain

Answer 45

If you want to do it Hero's way, let him explain it.

Answer 46

woah now ill take all the help i can get!...if you want to keep explaining to me how to do it your way @SmoothMath then by all means continue, i always appreciate it when you teach me how to do this stuff it helps me a lot

Answer 47

I think that good learning takes patience and focus. When other people are jumping in and trying to give you a shortcut or take you right to the answer without considering the process to get there, it hurts your understanding and slows things down.

Answer 48

Maybe it's selfish or rude, but if I'm going to take my time and teach a student, I want to be their only teacher. Multiple teachers just doesn't work.

Answer 49

ok well from this point on @SmoothMath shall be my only teacher for this problem...everyone else that comes in to contribute shall be shunned..and put to death!....better? i do understand your point and i think that patience and focus are attributes that every good teacher should possess...now are you still willing to walk me through how to do this? it would really help..i have a test tomorrow..

Answer 50

Yes =)
This is a very important skill in general. You are using a known formula, plugging in the variables you know, and then using algebra to solve for the unknown variable.

Answer 51

Step 1: Figure out the right formula
Step 2: Plug in the things you know
Step 3: Solve for the thing you don't know

Answer 52

reminds me of AP chem lol...my favorite...anyway continue

Answer 53

Okay, so you said that you understand the step 1 here. We decided to use the formula
\(\Large S_n = a_1\frac{1-r^n}{1-r}\)
And hopefully we know what each of those variables means. If you don't, you should ask.

Answer 54

well i dont know what the r represents

Answer 55

Thankyou for saying so =)
The idea with a geometric sequence is that to get to the next number, we multiply by some number. Here's an example of a really simple geometric sequence.
1, 2, 4, 8, 16, 32, 64
Do you see how it's doubling every time? So the number we multiply by to get the next number is 2. Now, that number multiply by is called the common ratio, and we usually give it the variable name r
So for that sequence,
r=2

Answer 56

In your problem, r=3.2
So what that means is that I get the next number by multiplying by 3.2.

Answer 57

o ok! well thats easy...continue

Answer 58

Do you understand the difference between
\(A_n \text{ and } S_n\)?

Answer 59

an is nth term itself sn is sum of n terms...right?

Answer 60

Right, exactly. So in my example sequence,
1,2,4,8,16
\(\large A_3\) would be 4
and
\(\large S_3\) would be 7

Answer 61

looks good!

Answer 62

Okay, so we're good with step 1.
Step 2 is to look at the info you know and plug into the formula.
The formula we have is
\(\Large S_n = A_1*\frac{1-r^n}{1-r}\)
And the info we're given is
\(S_n = 48455, r=3.2, n=3\)

Answer 63

okay...so it would be S(3)=A(1)* 1-3.2^3/1-3.2

Answer 64

Very good =)
And we can do a little bit better than that, even. Since you're given Sn = 48455 when n=3.

Answer 65

so just 48455=A(1)* 1-3.2^2/1-3.2?

Answer 66

Exactly.
Now the last step is just algebra, but this is the kind of algebra problem that a lot of students struggle with a little bit. Do you know how to solve for A1 here?

Answer 67

well wishfully i would like to solve 1-3.2^2/1-3.2 and then divide my answer by both sides but im thinking thats not gonna fly..so im assuming i have to divide both sides by 1-3.2^2/1-3.2?...

Answer 68

The first thing you said totally works. Do \(\Large \frac{1-3^2}{1-3}\). That'll make your problem a lot simpler.

Answer 69

The second thing you said totally works too =) It's just not as simple, I'd say.

Answer 70

Oh, oops. I said that wrong. Do \(\Large \frac{1-3.2^3}{1-3.2}\)

Answer 71

lol well usually im all for the easy! but i just went ahead and did the long way...and i got 2383.66...is that right?

Answer 72

You made a mistake somewhere, which is why I like the first way. You probably won't make the same mistake if you simplify it first.

Answer 73

ok let me try again...one sec

Answer 74

okay i did 1-3.2^3/1-3.2=-34.968...48455/-34.968= -1385.695493

Answer 75

I think you're making a mistake in the order of operations somehow.

Answer 76

Do the top and the bottom separately, then divide.

Answer 77

Okay, yeah, I see what you did. You're probably putting it into your calculator without the right parenthesis.

Answer 78

You got -34.968, which is what you get if you do

\(\Large 1-\frac{3.2^2}{1}-3.2\)

Answer 79

\(3.2^3\) I mean.

Answer 80

Whenever you put a fraction into your calculator, you have to make sure you put the parenthesis around the correct things so that your calculator divides correctly.

Answer 81

alrighty well i did what you said and i tried doing them separately and i got 14.44...then that divided by 48455 and got...

Answer 82

14.44 is right =)

Answer 83

355.609

Answer 84

oops wait

Answer 85

3355.609

Answer 86

So you get
48455 = \(a_1*14.44\)

Answer 87

Good =) Exactly.

Answer 88

Okay, so I think that we can conclude that the part you need to work on the most in this whole thing is just the basic algebra.

Answer 89

yes!! thank you so much!! this is WAY easier than i thought it was!!

Answer 90

and ya i agree

Answer 91

You're fine with setting up the formula, I think.

Answer 92

You're welcome =)

Answer 93

your going to be an amazing math teacher!!! thank you thank you thank you again! lol

Answer 94

@SmoothMath, I never wanted to interrupt, but I was the one helping her originally. I had the formula wrong and I was trying to make up for it. I found the correct formula, and then you jumped in.

Answer 95

You wouldn't allow me to make up for my mistake. That's what I don't appreciate.

Answer 96

Yeah, well. I don't consider what you were doing to be an explanation. If you were teaching, I'd leave you be.

Answer 97

@SmoothMath, I would have provided a better explanation, but you interfered.