Series...

Got it.. my bad!! A it is..

I just solved it this way:

1° R2-R1 = -1 | So, R1=R2+1 so, R1>R2

2° R3-R2 = 1/2 | So, R3=R2 + 1/2 so, R3>R2

3° R1=R2 + 1 and R3=R2 + 1/2 R1 is 1/2 more than R3 | So R1>R3

So I guess that it´s A the answer...

I think I am missing something here. Do you know what is the source of this q and if there is any element missing from the q?

They way I see it is that all boils down to the value of r1
On simplifying,
r2 = (1/2) + r1
r3 = (1/6) + r1

If r1 is -ve say -(1/2)
r2 = 0 and r3 = -(1/3)
So you have r3<r1<r2

But if r1 is +ve say 1/2
r2 = 1 and r3 = 2/3
Now, r1<r3<r2

I can only conclude that r2 would be the greatest and can eliminate choices A, B and E.

Thanx dendude,

I think that somenthing is wrong with your numbers:

r2 - r1 = -1^n/n
r2= -1^1/1 + r1
r2= -1 + r1 and not r2= (1/2) + r1

keizer Soze wrote:Thanx dendude,

I think that somenthing is wrong with your numbers:

r2 - r1 = -1^n/n
r2= -1^1/1 + r1
r2= -1 + r1 and not r2= (1/2) + r1

You're right thanks for pointing out my blunder
I assumed r2 = (-1)^2 etc....
when in fact it was r(n+1) = (-1)^n and not to the power n+1

With that in mind,
r2 = -1 + r1
r3 = (-1/2) + r1

If r1 is -ve, say -1/2
r2 = -3/2 and r3 = -1
So, r2<r3<r1

If r1 is +ve, say 1/2
r2 = -1/2 and r3 = 0
Also, r2<r3<r1

Clearly, Choice A

Great, we are Ok then!