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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, 09:26
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A
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D
E

Difficulty:

  45% (medium)

Question Stats:

68% (01:40) correct 32% (01:55) wrong based on 60 sessions

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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 18 Dec 2018, 11:57
we can rewrite R and S as the following:
R = 3^0 + 3^-1 + 3^-2 + 3^-3 and S = 3^0 + 3^-1 * R
so, 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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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, 07:52
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, 08:10
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, 11:01
\(r = 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27}\) =\(\frac{40}{27}\)


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


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

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If r= 1 + 1/3 + 1/9 + 1/27 and s= 1 + 1/3(r), then s exceeds r by &nbs [#permalink] 19 Dec 2018, 11:01
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