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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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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) $$\frac{1}{81}$$

Project PS Butler : Question #85

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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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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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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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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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$$r = 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27}$$ =$$\frac{40}{27}$$

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

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

E
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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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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) $$\frac{1}{81}$$

Project PS Butler : Question #85

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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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[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 = $$1+ \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{81}$$ - ($$1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27}$$) = $$\frac{1}{81}$$ 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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