MHB Understanding Geometric Sequences: Results & Formula

[Casio1]

Just a little help understanding results obtained.

I have found the closed form of a sequence, but am a little unsure if there is a right way or can select either way of using the terms to create the explicit formula.

I have found the common difference from the terms, which is 1.2, in my example I have four terms, -0.4, -1.6, -2.8, -4.

By selecting the term -2.8 I created the formula;

U_n = U₂ + (n - 1)d

I ended up with;

U_n = 1.2n - 4

Then I selected another term, (- 4) and created the formula;

U_n = - 5.2 + 1.2n

So I guess I am really asking, is there a rule that says which term must be selected to create the formula, or if any of the terms are used, are both the formulas correct?

 

[Also sprach Zarathustra]

Just a little help understanding results obtained.

I have found the closed form of a sequence, but am a little unsure if there is a right way or can select either way of using the terms to create the explicit formula.

I have found the common difference from the terms, which is 1.2, in my example I have four terms, -0.4, -1.6, -2.8, -4.

By selecting the term -2.8 I created the formula;

U_n = U₂ + (n - 1)d

I ended up with;

U_n = 1.2n - 4

Then I selected another term, (- 4) and created the formula;

U_n = - 5.2 + 1.2n

So I guess I am really asking, is there a rule that says which term must be selected to create the formula, or if any of the terms are used, are both the formulas correct?

Hello!

d,- difference is not 1.2, but -1.2.

The formula is:

$$ a_n=a_1+(n-1)d $$

 

[soroban]

Hello, Casio!

Given: -0.4, -1.6, -2.8, -4.0
Find the closed form of the sequence.

$\text{The common difference is: }\;\;d = \text{-}1.2 $

$\text{The }n^{th}\text{ term of an arithmetic sequence is: }\;\;a_n \:=\:a_1 + (n-1)d$

$\text{For this sequence: }\;\;a_n \:=\:\text{-}0.4 + (n-1)(\text{-}1.2)$

. . $\text{which simplifies to: }\;\;a_n \:=\:0.8 - 1.2n $

 

[Casio1]

Hello, Casio!


$\text{The common difference is: }\;\;d = \text{-}1.2 $

$\text{The }n^{th}\text{ term of an arithmetic sequence is: }\;\;a_n \:=\:a_1 + (n-1)d$

$\text{For this sequence: }\;\;a_n \:=\:\text{-}0.4 + (n-1)(\text{-}1.2)$

. . $\text{which simplifies to: }\;\;a_n \:=\:0.8 - 1.2n $

Now I could be wrong, but I thought along the lines of;

u_n=u₂+(n-1)d

u_n=-2.8+(n-1)d

u_n=-2.8+1.2n-1.2

u_n= 1.2n-4

Alternatively I also thought;

u_n= -4+1.2n-1.2

u_n= 1.2n - 5.2

or

u_n= -5.2+1.2n

The confusing part for me is in selecting the correct term to define the sequence and hence produce the formula above,which will produce the right solution!

P.S. Not learned the latex for this forum to date!

 

[Prove It]

This is an Arithmetic Sequence, not Geometric...

 

[Also sprach Zarathustra]

This is an Arithmetic Sequence, not Geometric...

Good observation!

 

[Casio1]

Some confusion on my part here on this subject!

I can see the confusion, a SERIES is simply adding the terms in a sequence. An Arithmetic series involves adding the terms of an arithmetic sequence, and a geometric series involves adding the terms of a geometric sequence.

However, nobody has understood my requests to date!

What I wanted was an understanding of why it is possible to have different closed form sequences solutions from the terms and whether you could select anyone of the terms, or there is a standard that says a particular given term is used, i.e. -2.8, -1.6, -0.4, 0.8.

Would it be right to start from left to the right, so use -2.8, which would give a solution Un = 1.2n-4

or could I say do this;

-2.8 -(-4) = 1.2

Then using the term (-4)

-4 + 1.2n - 1.2

1.2n - 5.2 or -5.2 + 1.2n

All the above solutions are correct, I am just after an understanding of whether there is a standard method or way to find the answer, or whether any terms can be used in any order to produce anyone of possible answera?

I hope I have explained it correctly.

 

[HallsofIvy]

The reason no one has understood your requests is because you never asked that question. You, repeatedly, asserted that "a_n= a_2+ (n-1)d" although you were told, repeatedly, that that is not true. You can think of the sequence as starting at a_2 rather than a_1 but then "n" is one less: a_n= a_2+ (n-2)d. In fact, for n> i, a_n= a_i+ (n-i)d.