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# M26-26

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Math Expert
Joined: 02 Sep 2009
Posts: 46284

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16 Sep 2014, 01:25
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Difficulty:

65% (hard)

Question Stats:

56% (01:42) correct 44% (02:07) wrong based on 32 sessions

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In an infinite sequence 1, 3, 9, 27, ... each term after the first is three times the previous term. What is the difference between the sum of 13th and 15th terms and the sum of 12th and 14th terms of the sequence?

A. $$10*3^{11}$$
B. $$20*3^{11}$$
C. $$10*3^1$$
D. $$40*3^{11}$$
E. $$20*3^{12}$$

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Math Expert
Joined: 02 Sep 2009
Posts: 46284

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16 Sep 2014, 01:25
Official Solution:

In an infinite sequence 1, 3, 9, 27, ... each term after the first is three times the previous term. What is the difference between the sum of 13th and 15th terms and the sum of 12th and 14th terms of the sequence?

A. $$10*3^{11}$$
B. $$20*3^{11}$$
C. $$10*3^1$$
D. $$40*3^{11}$$
E. $$20*3^{12}$$

You don't need to know geometric progression formula to solve this question. All you need is to find the pattern:

$$b_1=1=3^0$$;

$$b_2=3=3^1$$;

$$b_3=9=3^2$$;

$$b_4=27=3^3$$;

...

$$b_n=3^{n-1}$$;

$$(b_{13}+b_{15})-(b_{12}+b_{14})=3^{12}+3^{14}-3^{11}-3^{13}=3^{11}(3+3^3-1-3^2)=20*3^{11}$$

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Joined: 07 Aug 2016
Posts: 8
GMAT 1: 610 Q43 V31

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08 Nov 2016, 00:12
What if we were to solve this using geometric progression formula?

I got the answer wrong because I had (r-1) in denominator i.e 3-1=2.

I don't understand where am I going wrong.
Director
Status: Preparing for GMAT
Joined: 25 Nov 2015
Posts: 663
Location: India
GPA: 3.64

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26 Nov 2017, 00:28
La1yaMalhotra wrote:
What if we were to solve this using geometric progression formula?

I got the answer wrong because I had (r-1) in denominator i.e 3-1=2.

I don't understand where am I going wrong.

The G.P. formula for nth term is $$ar^{n-1}$$ where a is first term, r is common ratio and n is nth term.
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Re: M26-26   [#permalink] 26 Nov 2017, 00:28
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# M26-26

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