# A Relation in sequence

1. Apr 29, 2015

### Raghav Gupta

1. The problem statement, all variables and given/known data

If a,b,c,d ∈R+ and a,b,c,d are in H.P. Then
1. a+ d > b+ c
2. a+ c > b+ d
3. a + b > c+d
4. a-b > c-d
2. Relevant equations

Don't know which equation to apply
3. The attempt at a solution
1/a, 1/b , 1/c and 1/d would be in A. P
1/b -1/a = 1/d - 1/c

2. Apr 29, 2015

### BvU

HP is for horsepower, but what is AP ?

3. Apr 29, 2015

### Raghav Gupta

Kidding with me in a maths forum thread. HaHa.
H.P is Harmonic progression and A.P is arithmetic progression.

4. Apr 29, 2015

### BvU

Good. We have established that. What is the problem statement ? Do you have to pick all the statements that are correct, the one and only unique statement that is correct, or perhaps "mark the ones that are not correct"?

But more generally I would investigate the statements using the definition.

5. Apr 29, 2015

### Raghav Gupta

The problem statement is in post 1.
There is only one unique statement that should be correct.
When I take random values using definition all statements are coming correct.

6. Apr 29, 2015

### Staff: Mentor

If the complete problem statement had been in the first post, BvU would not have had to ask these questions.

7. Apr 29, 2015

### BvU

Well, that excludes 1. and 4. because if I add b+d on both sides of 4. I get 1. !

And if I take a,b,c,d = 1/1, 1/0.79, 1/0.58, 1/.37 then 2. and 3. are false !

8. Apr 29, 2015

### Raghav Gupta

Only that much is given in the question of my paper.
And I have told already that there is only one option correct.

9. Apr 29, 2015

### Ray Vickson

What people are trying to get you to understand is that---whenever possible---you should state this as part of the initial message, rather than finally revealing it several messages later in the thread.

Notice that I said "whenever possible", and that means "if you know it yourself for sure". If somebody gives you an incompletely-stated problem, then, of course, you may not be able to state it properly right away. (That was apparently the case with one of your recent "locus" problems, where you were given a problem having a "typo" in it.)

10. Apr 29, 2015

### Raghav Gupta

What you are saying is true.

11. Apr 29, 2015

### Raghav Gupta

But why it was not asked from other posters in other threads, that whether one option is correct or more?
Well if one or more options are correct we can definitely discuss that.

Should I specify more things when I type a question like, what is the name of examination and what is the level of it?

I don't know whether it was for joke or serious matter that what is the full form of H.P and A.P.
I think in maths forums one should know that.
Numbers being in horsepower, does not make sense.

12. Apr 29, 2015

### BvU

No, that horsepower was just ironic. But you really can't expect that all abbreviations are known to everybody. And I really didn't know the terms. It's a math forum under the PF wings, Something like HP or AP can't be googled either. And nevertheless, in this case I was imho () able to provide sensible assistance even so.

If I were you, I wouldn't worry. You do your best to state a clear and complete problem, we do our best to help, and we ask if we are in doubt.

13. Apr 29, 2015

### BvU

Aside from the multiple choice character of this exercise and the confusion that 1 and 4 are equivalent: can you show (prove) that 1 and 4 are correct when you start from the definition of a harmonic progression ?

The given information that $a,\ b,\ c,\ d \in R^+$ is necessary here (otherwise the first wikipedia example is already a counterexample).

14. Apr 29, 2015

### Raghav Gupta

I think you mean the second example in wikipedia
No, I am not able to bring the inequality sign.
Getting 1/b - 1/a = 1/d - 1/c

15. Apr 29, 2015

### BvU

Sorry, yes, second example (has = instead of >).

My interpretation of the definition is that we can write a, b, c, d as $${1\over x},\ {1\over x + y},\ {1\over x+2y},\ {1\over x+3y}$$ so that a + d and b + c become $${1\over x} + {1\over x+3y} \quad {\rm and} \quad {1\over x + y} + {1\over x+2y}$$
and when I work these out I get the same numerators and the denominator in the second one is a little different from the denominator in a+d

16. Apr 29, 2015

### Staff: Mentor

Everybody knows that!

17. Apr 30, 2015

### Raghav Gupta

Thanks, since x, x+y, x+2y , x + 3y are positive we will get a bigger denominator in b + c ( I studying the basics, eh? )

Means a+ d > b+ c
Obviously then on rearranging,
a - b > c - d

18. Apr 30, 2015

### Raghav Gupta

Sorry, but I do not know that.
But I know now.

19. Apr 30, 2015

### BvU

Yes. The difference in the denominators is a mere 2y2, which is positive. (Note that y itself may well be negative). And we need the fact that all a, b, c, d are > 0 to establish the > sign in statements 1 and 4.