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If r= 1 + 1/3 + 1/9 + 1/27 and s= 1 + 1/3(r), then s exceeds r by

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If r= 1 + 1/3 + 1/9 + 1/27 and s= 1 + 1/3(r), then s exceeds r by  [#permalink]

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New post 18 Dec 2018, 10:26
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Question Stats:

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If \(r = 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27}\) and \(s = 1 + \frac{1}{3}r\), then \(s\) exceeds \(r\) by

(A) \(\frac{1}{3}\)
(B) \(\frac{1}{6}\)
(C) \(\frac{1}{9}\)
(D) \(\frac{1}{27}\)
(E) \(\frac{1}{81}\)

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If r= 1 + 1/3 + 1/9 + 1/27 and s= 1 + 1/3(r), then s exceeds r by  [#permalink]

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New post Updated on: 31 Jul 2019, 17:40
1
we can rewrite R and S as the following:
\(R = 3^0 + 3^{-1} + 3^{-2} + 3^{-3}\)

\(S = 3^0 + 3^{-1}R\)
\(S = 3^0 + 3^{-1}(3^0 + 3^{-1} + 3^{-2} + 3^{-3}) = 3^0 + 3^{-1} + 3^{-2} + 3^{-3} + 3^{-4}\)

upon subtraction,
\(S-R\) = \((3^0 + 3^{-1} + 3^{-2} + 3^{-3} + 3^{-4})\) \(-(3^0 - 3^{-1} - 3^{-2} - 3^{-3})\) \(= 3^{-4}\)

which is E
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Originally posted by MahmoudFawzy on 18 Dec 2018, 12:57.
Last edited by MahmoudFawzy on 31 Jul 2019, 17:40, edited 1 time in total.
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Re: If r= 1 + 1/3 + 1/9 + 1/27 and s= 1 + 1/3(r), then s exceeds r by  [#permalink]

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New post 19 Dec 2018, 08:52
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r= 1 + 1/3 + 1/9 + 1/27
Apply G.P summation formula, sum1 = (a*(1-r^n))/(1-r)
where a = 1; r = 1/3; n = 3
Therefore sum1 = 13/9
s= 1 + 1/3(r)
i.e s = 1 + 1/3 + 1/9 + 1/27 + 1/81
Apply G.P summation formula, sum2 = (a*(1-r^n))/(1-r)
where a = 1; r = 1/3; n = 3
Therefore sum2 = 40/27
Do sum1 - sum2 = 1/27

Option (D)
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If r= 1 + 1/3 + 1/9 + 1/27 and s= 1 + 1/3(r), then s exceeds r by  [#permalink]

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New post 19 Dec 2018, 09:10
AmanAngrish wrote:
r= 1 + 1/3 + 1/9 + 1/27
Apply G.P summation formula, sum1 = (a*(1-r^n))/(1-r)
where a = 1; r = 1/3; n = 3
Therefore sum1 = 13/9
s= 1 + 1/3(r)
i.e s = 1 + 1/3 + 1/9 + 1/27 + 1/81
Apply G.P summation formula, sum2 = (a*(1-r^n))/(1-r)
where a = 1; r = 1/3; n = 3
Therefore sum2 = 40/27
Do sum1 - sum2 = 1/27

Option (D)


your concept is very interesting and correct,
however, I think there is a small mistake in the application.

I think n=4 for sum1 (as 4 elements are added),
and n=5 for sum 2 (as 5 elements are added).

sum1 would be \(40/27\)
sum2 would be \(121/81\)
and by subtraction, the result would be \(1/81\)
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If r= 1 + 1/3 + 1/9 + 1/27 and s= 1 + 1/3(r), then s exceeds r by  [#permalink]

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New post 19 Dec 2018, 12:01
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\(r = 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27}\) =\(\frac{40}{27}\)


\(s = 1 + (\frac{1}{3} * \frac{40}{27})\) = \(\frac{121}{81}\)


\(\frac{121}{81}\) - \(\frac{40}{27}\) = \(\frac{1}{81}\)

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Re: If r= 1 + 1/3 + 1/9 + 1/27 and s= 1 + 1/3(r), then s exceeds r by  [#permalink]

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New post 12 Feb 2019, 08:08
HKD1710 wrote:
If \(r = 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27}\) and \(s = 1 + \frac{1}{3}r\), then \(s\) exceeds \(r\) by

(A) \(\frac{1}{3}\)
(B) \(\frac{1}{6}\)
(C) \(\frac{1}{9}\)
(D) \(\frac{1}{27}\)
(E) \(\frac{1}{81}\)

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So i thought of some other method, it did not work

\(r = 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27}\)

\(s = 1 + \frac{1}{3}r\)

Just substitute r in the second expression

s = 1 + 1/3 * \(1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27}\)

So it will give us 1/81

E
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Re: If r= 1 + 1/3 + 1/9 + 1/27 and s= 1 + 1/3(r), then s exceeds r by  [#permalink]

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New post 31 Jul 2019, 07:21
[quote="HKD1710"]If \(r = 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27}\) and \(s = 1 + \frac{1}{3}r\), then \(s\) exceeds \(r\) by

(A) \(\frac{1}{3}\)
(B) \(\frac{1}{6}\)
(C) \(\frac{1}{9}\)
(D) \(\frac{1}{27}\)
(E) \(\frac{1}{81}\)


\(s = 1 + \frac{1}{3}r\) = \(1+ \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{81}\)

Now, s-r = \(1+ \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{81}\) - (\(1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27}\)) = \(\frac{1}{81}\)
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Re: If r= 1 + 1/3 + 1/9 + 1/27 and s= 1 + 1/3(r), then s exceeds r by   [#permalink] 31 Jul 2019, 07:21
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