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The sum of the first n terms of a geometric sequence is 255, and the
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14 Dec 2023, 01:14
14 Dec 2023, 01:14
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Data Insights (DI) Butler 2023-24 [Question #150, Date: Dec-11-2023] [Click here for Details]
The sum of the first n terms of a geometric sequence is 255, and the sum of the reciprocals of these \(n\) terms is \(\frac{255}{128}\). The ratio, \(r\), of the sequence is a positive integer greater than 1, and the first term of the sequence is 1. The sum \(S_{n}\), of the first n terms of a geometric sequence is given by \(S_{n}\) = \(\frac{a_{1} [1-r^n]}{1-r}\) where \(a_{1}\) is the first term of the sequence.
In the table, select the number that equals the ratio, r, of the sequence, and the number that equals the number, n, of the terms. Make only one selection in each column.
| r | n | |
| 2 | ||
| 4 | ||
| 6 | ||
| 8 | ||
| 10 | ||
| 12 |
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Official Answer
r: 2
n: 8
Re: The sum of the first n terms of a geometric sequence is 255, and the
[#permalink]
14 Dec 2023, 04:00
14 Dec 2023, 04:00
Sum of first n terms = Sn = a1(1-r^n)/(1-r)
a1(1-r^n)/(1-r) = 255
(1-r^n)/(1-r) = 255
When we put r=2 and n=8 it satisfies the equation. Therefore, r=2 and n=8.
a1(1-r^n)/(1-r) = 255
(1-r^n)/(1-r) = 255
When we put r=2 and n=8 it satisfies the equation. Therefore, r=2 and n=8.
Re: The sum of the first n terms of a geometric sequence is 255, and the
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23 Feb 2024, 02:27
23 Feb 2024, 02:27
Sajjad1994 wrote:
Data Insights (DI) Butler 2023-24 [Question #150, Date: Dec-11-2023] [[url=https://gmatclub.com/forum/data-insights-di-butler-419150.html]Click here for Details]
The sum of the first n terms of a geometric sequence is 255, and the sum of the reciprocals of these \(n\) terms is \(\frac{255}{128}\). The ratio, \(r\), of the sequence is a positive integer greater than 1, and the first term of the sequence is 1. The sum \(S_{n}\), of the first n terms of a geometric sequence is given by \(S_{n}\) = \(\frac{a_{1} [1-r^n]}{1-r}\) where \(a_{1}\) is the first term of the sequence.
In the table, select the number that equals the ratio, r, of the sequence, and the number that equals the number, n, of the terms. Make only one selection in each column.
Sum of first n terms of geometric sequence = \(S_{n}\) = \(\frac{a_{1} [1-r^n]}{1-r}\) = 255 , and a1 = 1, r is the common ratio and n is the number of terms;
Sequence of the recriporcal of these terms will have a ratio of 1/r an therefore,
Sum of recriprocal of these terms = \(S_{1/n}\) = 1* (1-\(\frac{1}{r^n}\))/(1-\(\frac{1}{r}\))= \(\frac{255}{128}\)
Rearranging numerator and denominator,
\(S_{1/n}\) = \(\frac{1*[1-r^n]}{1-r}\)* \(\frac{1}{r^{n-1}}\) = \(\frac{255}{128}\)
Subsitute value of \(\frac{1*[1-r^n]}{1-r}\) as 255 from first expression
==> 255*\(\frac{1}{r^{n-1}}\) = \(\frac{255}{128}\)
==> \(r^{n-1}\) = \(2^7\)
=> r = 2 and n=8
