Progressions and series prove this

[utkarshakash]

Homework Statement

If $x_1,x_2,x_3...x_n$ are in H.P. then prove that $x_1x_2+x_2x_3+x_3x_4...+x_{n-1}x_n=(n-1)x_1x_n$

Homework Equations

The Attempt at a Solution

Since $x_1,x_2,x_3...x_n$ are in H.P. therefore
$\frac{1}{x_1},\frac{1}{x_2},\frac{1}{x_3}...,\frac{1}{x_n}$ will be in A.P. Now common difference of this A.P.
$d=\dfrac{\frac{1}{x_n}-\frac{1}{x_1}}{n-1} \\<br /> x_1x_n=\dfrac{x_1-x_n}{d(n-1)}\\<br /> (n-1)x_1x_n=\dfrac{x_1-x_n}{d}$

Looks like I've arrived at the R.H.S. But what about LHS?

 

[Mandelbroth]

Homework Statement

If $x_1,x_2,x_3...x_n$ are in Harmonic Progression. then prove that $x_1x_2+x_2x_3+x_3x_4...+x_{n-1}x_n=(n-1)x_1x_n$

Homework Equations

The Attempt at a Solution

Since $x_1,x_2,x_3...x_n$ are in H.P. therefore
$\frac{1}{x_1},\frac{1}{x_2},\frac{1}{x_3}...,\frac{1}{x_n}$ will be in A.P. Now common difference of this A.P.
$d=\dfrac{\frac{1}{x_n}-\frac{1}{x_1}}{n-1} \\<br /> x_1x_n=\dfrac{x_1-x_n}{d(n-1)}\\<br /> (n-1)x_1x_n=\dfrac{x_1-x_n}{d}$

Looks like I've arrived at the R.H.S. But what about LHS?

This the second one of your problems I've worked on. Just so you know, you need to write out your acronyms. No one has ANY clue what you mean if you don't do it. I was tempted to write something about Hewlett-Packard.

Alright. $x_1x_2+x_2x_3+x_3x_4+...+x_{n-1}x_{n}$ is the same thing as saying $x_{1}(\frac{x_{1}}{1+d}) + (\frac{x_{1}}{1+d})(\frac{x_{1}}{1+2d})+...+(\frac{x_{1}}{1+(n-2)})(\frac{x_{1}}{1+(n-1)})$.

You can probably figure it out from here.

 

[utkarshakash]

This the second one of your problems I've worked on. Just so you know, you need to write out your acronyms. No one has ANY clue what you mean if you don't do it. I was tempted to write something about Hewlett-Packard.

Alright. $x_1x_2+x_2x_3+x_3x_4+...+x_{n-1}x_{n}$ is the same thing as saying $x_{1}(\frac{x_{1}}{1+d}) + (\frac{x_{1}}{1+d})(\frac{x_{1}}{1+2d})+...+(\frac{x_{1}}{1+(n-2)})(\frac{x_{1}}{1+(n-1)})$.

You can probably figure it out from here.

Hey, do you see my title. It says 'Progressions and series prove this'? And I don't understand how can you relate Hewlett-Packard with progressions and series?It's not even a mathematical term. OK I will do expand my acronyms in future.

Returning to my problem I did not understand how did you write $x_2=\dfrac{x_1}{1+d}$ and so on.

 

[utkarshakash]

I get $x_2=\dfrac{x_1}{1+x_1d}$ Looks like you missed the x1 in each of your denominator.

 

[Ray Vickson]

Hey, do you see my title. It says 'Progressions and series prove this'? And I don't understand how can you relate Hewlett-Packard with progressions and series?It's not even a mathematical term. OK I will do expand my acronyms in future.

Returning to my problem I did not understand how did you write $x_2=\dfrac{x_1}{1+d}$ and so on.

Hint: if
$$\frac{1}{x_i} - \frac{1}{x_{i+1}} = r$$ for i = 1,2, ..., n-1 we have
$$x_2 - x_1 = r x_1 x_2, \: x_3 - x_2 = r \, x_2 x_3, \ldots .$$ What does that give you?

RGV

 

[haruspex]

(n-1)x_1x_n=\dfrac{x_1-x_n}{d} [/itex]
Looks like I've arrived at the R.H.S. But what about LHS?

You know that d = 1/x₂-1/x₁, right? So what does that turn into for x₁x₂?

 

[utkarshakash]

Hint: if
$$\frac{1}{x_i} - \frac{1}{x_{i+1}} = r$$ for i = 1,2, ..., n-1 we have
$$x_2 - x_1 = r x_1 x_2, \: x_3 - x_2 = r \, x_2 x_3, \ldots .$$ What does that give you?

RGV

Thanks. It solved my problem.