Question

WILL MEDAL

Can anyone comment the labeled functions for Arithmetic Series, Arithmetic Sequences, Geometric Series, and Geometric Sequences.?

Answer 1

labeled functions?

Answer 2

Like the functions that solve them with a label that shows where everything should go.
Say if the question was "Identify the 34th term of the arithmetic sequence 2, 7, 12 .."

Answer 3

I need serious help on this Topic

Answer 4

\[a_n=a_1+d(n-1) \text{ is an arithemetic sequence with first term } a_1 \\ \text{ and common difference } d \]

Answer 5

just replace d with 7-2 or 12-7
and then replace a_1 with 2 since it is the first term
then enter in 34 for n
use order of operations to find a_(34)

Answer 6

So it would be \[2_{34}=2_{1}+7-2(34-1)\]

Answer 7

where did you get 2_(34)?

Answer 8

and 2_(1)?

Answer 9

you said replace n with 34

Answer 10

yes but what happen to the a_34?

Answer 11

and why didn't you replace a_1 with 2

Answer 12

you replaced it with 2_1 whatever that means

Answer 13

\[a_n=a_1+d(n-1) \\ a_1 \text{ is the first term } \\ d \text{ is the common difference } \\ a_1 \text{ was given as } 2 \\ \text{ you wanted to know } a_{34} \text{ this is why I said to replace } n \text{ with } 34\]

Answer 14

\[a_{34}=2+5(34-1)\]

Answer 15

7-2 or 12-7
either of these differences will give you the common difference because this an arithmetic sequence
7-2=5
12-7=5
so d=5

Answer 16

anyways just follow order of operations to find a_(34)

Answer 17

Oh okay i see now

Answer 18

What about arithmetic Series.?

Answer 19

here is a sequence of numbers:
\[a_1,a_2,a_3,a_4,...,a_n,...\]
This is an arithmetic sequence if you have:\[a_1,a_1+d,a_1+2d,a_1+3d,a_1+4d,...,a_1+(n-1)d,... \\ \text{ hope you are seeing that I'm using } \\ a_1=a_1 \\ a_2=a_1+d \\ a_3=a_1+2d \\ a_4=a_1+3d \\ ... \\ a_n=a_1+(n-1)d \text{ or can be written as } a_n=a_1+d(n-1) \\ \]
An arithmetic series is just the summing of the terms of an arithmetic sequence.
\[a_1,a_2,a_3,a_4,...,a_n,...\]
This is a geometric sequence if you have: \[a_1,a_1 r,a_1r^2,a_1r^3,a_1r^4,...,a_1r^{n-1},... \\ \text{ I hope you are seeing that I'm using } \\ a_1=a_1 \\ a_2=a_1r \\ a_3=a_1r^2 \\ a_4=a_1r^3 \\ a_5=a_1r^4 \\ ... \\ a_n=a_1r^{n-1} \]
A geometric series is just a summing of the terms of a geometric sequence.

Answer 20

Are you wanting the sum formulas ?

Answer 21

Im still trying to fully comprehend this. I'm not very good at math :/

Answer 22

What is the difference between a series and a sequence.? (Just to add to my notes)

Answer 23

you know what sum means?

Answer 24

A series is a sum of the terms of a sequence.

Answer 25

I was just asking if you knew what sum meant because I basically already said this

Answer 26

Sum is the outcome of adding 2 or more number together.

Answer 27

\[\text{ Sequence of numbers looks like } a_1,a_2,a_3,...,a_n,... \\ \text{ a Series looks like } a_1+a_2+a_3+...+a_n+...\]
notice a series is just as I said the sum of the terms of a sequence

Answer 28

you might also have seen this notation for the series:
\[\sum_{i=1}^{n}a_i\]

Answer 29

Also a series doesn't always have to start at i=1 and end at n

Answer 30

A series can be infinite.
And it can also start at i=2 etc...

Answer 31

Thanks for all your help man, i really appreciate it.

Answer 32

if you want to know the sum formula for an arithmetic series:
\[\sum_{i=1}^{n}(a_1+d(i-1)) \\ \sum_{i=1}^{n}a_1 + \sum_{i=1}^{n}di-\sum_{i=1}^{n}d \\ a_1n+d \frac{n(n+1)}{2}-dn \\ \frac{2a_1n}{2}+\frac{dn(n+1)}{2}-\frac{2dn}{2} \\ \frac{2a_1n+dn^2+dn-2dn}{2} \\ \frac{n(2a_1+dn+d-2d)}{2} \\ \frac{n}{2}(2a_1+dn-d) \\ \frac{n}{2}(a_1+a_1+d(n-1)) \\ \frac{n}{2}(a_1+a_n) \\ \frac{n(a_1+a_n)}{2} \\ \text{ so \in conclusion } \\ \sum_{i=1}^{n}(a_i+d(n-1))=\frac{n(a_1+a_n)}{2}\]

Answer 33

all this is saying if you aren't used to sigma notation is that:
\[(a_1)+(a_1+d)+(a_1+2d)+\cdots +(a_1+(n-1)d)=\frac{n(a_1+a_n)}{2}\]

Answer 34

\[\sum_{i=1}^{n}a_1 r^{i-1}=a_1 \sum_{i=1}^{n} r^{i-1}=a_1 \frac{r^n-1}{r-1}\]

Answer 35

and that is the sum formula for a geometric series

Answer 36

Okay cool :)

Answer 37

and again if you aren't used to sigma notation that just says:
\[(a_1)+(ra_1)+(r^2a_1)+\cdots +(r^na_1)=a_1\frac{r^n-1}{r-1}\]

Answer 38

if the geometric series is infinite though
then you have..
\[\sum_{i=1}^{\infty}a_1r^{i-1}=a_1 \frac{1}{1-r} \text{ which only converges for } |r|<1\]

Answer 39

anyways post a few new questions and try to actually practice with these formulas
this will still be meaningless to you without some practice