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# M26-26

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Math Expert
Joined: 02 Sep 2009
Posts: 56366

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16 Sep 2014, 01:25
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45% (medium)

Question Stats:

71% (01:38) correct 29% (02:09) wrong based on 214 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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Joined: 02 Sep 2009
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16 Sep 2014, 01:25
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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
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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.
Status: Preparing for GMAT
Joined: 25 Nov 2015
Posts: 1031
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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Joined: 25 Jul 2017
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23 Aug 2018, 00:28
Bunuel wrote:
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. $$20*3^{12}$$

After observation, we can easily write the general formula to denote the nth term here in this sequence as (3)^n-1. Where n is the nth term.

Now 12th, 13th, 14th & 15th term are 3^11, 3^12,3^13 & 3^14 respectively.
Now question asks (3^14+3^12) - (3^13 + 3^11)
=> 3^12 (3^2+1) - 3^11 (3^2+1)
=> 3^12*10 - 3^11*10
=> 3^11*10( 3^1 -1)
=> 3^11*10*2
=> 20*3^11

Re: M26-26   [#permalink] 23 Aug 2018, 00:28
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