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If Rn+1-Rn=(-1/2)n for positive integers n, which of the following is [#permalink]
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=>

If \(n = 1\), then R2 – R1 = \(-(\frac{1}{2}) < 0\), and we have R1 > R2.
If \(n = 2\), then R3 – R2 = \(\frac{1}{4} > 0\), and we have R3 > R2.
Furthermore, R3 – R1 = ( R3 – R2 ) + ( R2 – R1 ) = \(\frac{1}{4} + (\frac{-1}{2})\) =\(\frac{-1}{4} < 0\), and we have R3 < R1.
Thus R1 > R3 > R2.

Therefore, A is the answer.
Answer: A

Originally posted by MathRevolution on 08 Apr 2018, 18:16.
Last edited by MathRevolution on 25 Jul 2018, 14:19, edited 1 time in total.
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Re: If Rn+1-Rn=(-1/2)n for positive integers n, which of the following is [#permalink]
MathRevolution wrote:
=>

If \(n = 1\), then R2 – R1 = \(-(\frac{1}{2}) < 0\), and we have R1 > R2.
If \(n = 2\), then R3 – R2 = \(\frac{1}{4} > 0\), and we have R3 > R2.
Furthermore, R3 – R1 = ( R3 – R2 ) + ( R2 – R1 ) = \(\frac{1}{4} + (\frac{-1}{2})\) =\(\frac{-1}{4} < 0\), and we have R3 < R1.
Thus R1 > R3 > R2.

Therefore, C is the answer.
Answer: C


The answer obtained points to Option A and not Option C.

Kindly modify the OA.
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Re: If Rn+1-Rn=(-1/2)n for positive integers n, which of the following is [#permalink]
MathRevolution wrote:
=>

If \(n = 1\), then R2 – R1 = \(-(\frac{1}{2}) < 0\), and we have R1 > R2.
If \(n = 2\), then R3 – R2 = \(\frac{1}{4} > 0\), and we have R3 > R2.
Furthermore, R3 – R1 = ( R3 – R2 ) + ( R2 – R1 ) = \(\frac{1}{4} + (\frac{-1}{2})\) =\(\frac{-1}{4} < 0\), and we have R3 < R1.
Thus R1 > R3 > R2.

Therefore, C is the answer.
Answer: C


Dear Moderator ,
This post needs your attention. The answer obtained is choice A and NOT choice C. Kindly correct the small typo in the post by MathRevolution. Thank you.
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Re: If Rn+1-Rn=(-1/2)n for positive integers n, which of the following is [#permalink]
R2-R1= - 0.5
R3-R2= -1
So, R3-R1= -1.5
There can be a situation when value of R1=3, R2=2.5 R3=1.5 then these values satisfy the above equations. Then R1>R2>R3.
Ans.B
Is this approach right ???





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Re: If Rn+1-Rn=(-1/2)n for positive integers n, which of the following is [#permalink]
BARUAH wrote:
R2-R1= - 0.5
R3-R2= -1
So, R3-R1= -1.5
There can be a situation when value of R1=3, R2=2.5 R3=1.5 then these values satisfy the above equations. Then R1>R2>R3.
Ans.B
Is this approach right ???






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How did you get \(R_3 -R_2= -1\) ?
\(R_3-R_2 = \frac{1}{4}\) because \(R_3-R_2\) = \((\frac{-1}{2})^2\)
Remember the stem \(R _{n+1} -R_n = (\frac{-1}{2})^n\)

So for n= 2 , \((\frac{-1}{2})^2\)

Hope this helps, please feel free to ask if anything is still unclear.
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Re: If Rn+1-Rn=(-1/2)n for positive integers n, which of the following is [#permalink]
(I) R2 - R1 = -1/2 = -0.50
(II) R3 - R2 = +1/4 = +0.25
(III) R3 - R1 = -1/4 = -0.25

We can see that R1 > R3 > R2

If it's not clear, you can change the variables for numbers
R2 = 0.50
R1 = 1.00
R3 = 0.75

(I) R2 - R1 = -1/2 = -0.50
0.5 - 1.0 = -0.50

(II) R3 - R2 = +1/4 = +0.25
0.75 - 0.50 = +0.25

(III) R3 - R1 = -1/4 = -0.25
0.75 - 1.00 = -0.25

1.00 > 0.75 > 0.50
R1 > R3 > R2

The answer is alternative A
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Re: If Rn+1-Rn=(-1/2)n for positive integers n, which of the following is [#permalink]
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