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The sum of the terms of a geometric progression is 2047.

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The sum of the terms of a geometric progression is 2047. [#permalink]

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New post 11 Aug 2017, 09:17
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The sum of the terms of a geometric progression is 2047. Find the common ratio.

(1) The first and last terms of the series are 1 and 1024 respectively.

(2) Last but one term of the series is 512.

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Re: The sum of the terms of a geometric progression is 2047. [#permalink]

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New post 11 Aug 2017, 13:45
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Gnpth wrote:
The sum of the terms of a geometric progression is 2047. Find the common ratio.

(1) The first and last terms of the series are 1 and 1024 respectively.

(2) Last but one term of the series is 512.


In case there are people reading this question and worrying that they haven't learned about geometric progressions and common ratios, you need not worry. The GMAT does not expect you to be familiar with these terms. Likewise, you don't need to know the formula for the sum of a geometric series/sequence/progression.

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Brent
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The sum of the terms of a geometric progression is 2047. [#permalink]

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New post 07 Feb 2018, 04:17
Gnpth wrote:
The sum of the terms of a geometric progression is 2047. Find the common ratio.

(1) The first and last terms of the series are 1 and 1024 respectively.

(2) Last but one term of the series is 512.




Summation of n terms of a G.P = sum = a\((\frac{ 1- r^n}{1-r})\)

Where a is the first term and n is the number of terms whose sum is being taken and r is the common ratio.



2047 = \(1 (\frac{ 1- r^n}{1-r})\)

Also nth term of a G.P.= \(ar^{(n-1)}\) where a is the first term and r is the common ratio.
So last term ( nth term) = 1024 = \(ar^{(n-1)}\)
We know a (first term) is 1
\(1024=r^{(n-1)}\)
\(2^{10}\) = 1024 here n-1= 10 so n=11
\(4^5\) = 1024 here n-1= 5 so n=6
\(32^2\) = 1024 here n-1= 2 so n=3

Out of these only n=11 will give sum of 2047 hence r= 2 is the required common ratio.

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The sum of the terms of a geometric progression is 2047.   [#permalink] 07 Feb 2018, 04:17
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