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M26-26

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M26-26 [#permalink]

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New post 16 Sep 2014, 00:25
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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. \(20*3^{12}\)
[Reveal] Spoiler: OA

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Re M26-26 [#permalink]

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New post 16 Sep 2014, 00:25
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:

\(b_1=1=3^0\);

\(b_2=3=3^1\);

\(b_3=9=3^2\);

\(b_4=27=3^3\);

...

\(b_n=3^{n-1}\);

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


Answer: B
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Re: M26-26 [#permalink]

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New post 07 Nov 2016, 23:12
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.

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Re: M26-26 [#permalink]

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New post 25 Nov 2017, 23:28
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] 25 Nov 2017, 23:28
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