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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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Joined: 21 Jun 2014
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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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18 Dec 2018, 09:26
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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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Manager
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Joined: 27 Oct 2018
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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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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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Intern
Joined: 14 Jan 2018
Posts: 2
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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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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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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VP
Joined: 09 Mar 2016
Posts: 1287
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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19 Dec 2018, 11:01
$$r = 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27}$$ =$$\frac{40}{27}$$

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

$$\frac{121}{81}$$ - $$\frac{40}{27}$$ = $$\frac{1}{81}$$

E
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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