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805+ (Hard)|   Sequences|      
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DisciplinedPrep
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The second term of a GP is 1000 and the common ratio is r = 1/n, where 11 Nov 2019, 23:41
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95% (hard)

Question Stats:

42% (02:41) correct 58% (03:10) wrong based on 72 sessions

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The second term of a GP is 1000 and the common ratio is r = 1/n, where n is a natural number. \(P_n\) is the product of n terms of this GP series. If \(P_6\)>\(P_5\) and \(P_6\)>\(P_7\), find the sum of all possible values of n.

A. 4
B. 5
C. 9
D. 13
E. 15
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C
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chetan2u
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The second term of a GP is 1000 and the common ratio is r = 1/n, where 12 Nov 2019, 00:35
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DisciplinedPrep
The second term of a GP is 1000 and the common ratio is r = 1/n, where n is a natural number. \(P_n\) is the product of n terms of this GP series. If \(P_6\)>\(P_5\) and \(P_6\)>\(P_7\), find the sum of all possible values of n.

A. 4
B. 5
C. 9
D. 13
E. 15
Show more

Let the first number be a, then second number = \(a*\frac{1}{n}\), \(3^{rd}=a*\frac{1}{n^2}\) and so on..

\(P_5=a*\frac{a}{n}*\frac{a}{n^2}...*\frac{a}{n^4}=\frac{a^5}{n^{10}}\)...
\(P_6=a*\frac{a}{n}*\frac{a}{n^2}...*\frac{a}{n^5}=\frac{a^6}{n^{15}}\)...
\(P_7=a*\frac{a}{n}*\frac{a}{n^2}...*\frac{a}{n^6}=\frac{a^7}{n^{21}}\)...

Thus
(I) \(P_6\)>\(P_5\) ........\(\frac{a^6}{n^{15}}>\frac{a^5}{n^{10}}.......a>n^5\) and
(II) \(P_6\)>\(P_7\)........\(\frac{a^6}{n^{15}}>\frac{a^7}{n^{21}}.......n^6<a\)

Combined..
\(n^5<a<n^6\)..Divide by n....\(n^4<\frac{a}{n}<n^5\).
a/n is nothing but the second term = 1000...
\(n^4<1000<n^5\)...

Only possible values 4 and 5.. 4+5=9

C
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The second term of a GP is 1000 and the common ratio is r = 1/n, where 12 Nov 2019, 01:50
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chetan2u
DisciplinedPrep
The second term of a GP is 1000 and the common ratio is r = 1/n, where n is a natural number. \(P_n\) is the product of n terms of this GP series. If \(P_6\)>\(P_5\) and \(P_6\)>\(P_7\), find the sum of all possible values of n.

A. 4
B. 5
C. 9
D. 13
E. 15

Let the first number be a, then second number = \(a*\frac{1}{n}\), \(3^{rd}=a*\frac{1}{n^2}\) and so on..

\(P_5=a*\frac{a}{n}*\frac{a}{n^2}...*\frac{a}{n^4}=\frac{a^5}{n^{10}}\)...
\(P_6=a*\frac{a}{n}*\frac{a}{n^2}...*\frac{a}{n^5}=\frac{a^6}{n^{15}}\)...
\(P_7=a*\frac{a}{n}*\frac{a}{n^2}...*\frac{a}{n^6}=\frac{a^7}{n^{21}}\)...

Thus
(I) \(P_6\)>\(P_5\) ........\(\frac{a^6}{n^{15}}>\frac{a^5}{n^{10}}.......a>n^5\) and
(II) \(P_6\)>\(P_7\)........\(\frac{a^6}{n^{15}}>\frac{a^7}{n^{21}}.......n^6<a\)

Combined..
\(n^5<a<n^6\)..Divide by n....\(n^4<\frac{a}{n}<n^5\).
a/n is nothing but the second term = 1000...
\(n^4<1000<n^5\)...

Only possible values 4 and 5.. 4+5=9

C
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Sir,

Should the highlighted text not be 2,3,4 summing to 9 ?
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The second term of a GP is 1000 and the common ratio is r = 1/n, where 12 Nov 2019, 02:08
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ShankSouljaBoi
chetan2u
DisciplinedPrep
The second term of a GP is 1000 and the common ratio is r = 1/n, where n is a natural number. \(P_n\) is the product of n terms of this GP series. If \(P_6\)>\(P_5\) and \(P_6\)>\(P_7\), find the sum of all possible values of n.

A. 4
B. 5
C. 9
D. 13
E. 15

Let the first number be a, then second number = \(a*\frac{1}{n}\), \(3^{rd}=a*\frac{1}{n^2}\) and so on..

\(P_5=a*\frac{a}{n}*\frac{a}{n^2}...*\frac{a}{n^4}=\frac{a^5}{n^{10}}\)...
\(P_6=a*\frac{a}{n}*\frac{a}{n^2}...*\frac{a}{n^5}=\frac{a^6}{n^{15}}\)...
\(P_7=a*\frac{a}{n}*\frac{a}{n^2}...*\frac{a}{n^6}=\frac{a^7}{n^{21}}\)...

Thus
(I) \(P_6\)>\(P_5\) ........\(\frac{a^6}{n^{15}}>\frac{a^5}{n^{10}}.......a>n^5\) and
(II) \(P_6\)>\(P_7\)........\(\frac{a^6}{n^{15}}>\frac{a^7}{n^{21}}.......n^6<a\)

Combined..
\(n^5<a<n^6\)..Divide by n....\(n^4<\frac{a}{n}<n^5\).
a/n is nothing but the second term = 1000...
\(n^4<1000<n^5\)...

Only possible values 4 and 5.. 4+5=9

C


Sir,

Should the highlighted text not be 2,3,4 summing to 9 ?
Show more

No.
2...2^4=16 and 2^5=32<1000
3.....3^4=243 but 3^5=729<1000
4....4^4=256 and 4^5=1024>1000...yes
5....5^4=625 and 5^5=3125>1000..yes
6...6^4>1000....no

Only 4 and 5
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The second term of a GP is 1000 and the common ratio is r = 1/n, where 02 Nov 2023, 23:15
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geometric progression is ar, ar^2, ar^3 . So how come P5 is a^5/n^10?
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The second term of a GP is 1000 and the common ratio is r = 1/n, where 03 Nov 2023, 02:27
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deepak.brijendra
geometric progression is ar, ar^2, ar^3 . So how come P5 is a^5/n^10?
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P5 is not the fifth term but product of first five terms.
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The second term of a GP is 1000 and the common ratio is r = 1/n, where 03 May 2025, 19:38
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DisciplinedPrep
The second term of a GP is 1000 and the common ratio is r = 1/n, where n is a natural number. \(P_n\) is the product of n terms of this GP series. If \(P_6\)>\(P_5\) and \(P_6\)>\(P_7\), find the sum of all possible values of n.

A. 4
B. 5
C. 9
D. 13
E. 15
Show more
Since one of the learners as ked so I am responding here on thread that typical GP questions such as this haven't been witnessed in GMAT so therefore this type of questions may be best be avoided and the time may be used better for DI section sections prep.
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