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805+ (Hard)|   Sequences|      
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Bunuel
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In a sequence of numbers in which each term after the first is found 15 May 2020, 10:05
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In a sequence of numbers in which each term after the first is found by multiplying the previous one by 1/n, where n is a positive integer, the second term is 1,000. If \(P_n\) is the product of the first n terms of the sequence, \(P_6 > P_5\) and \(P_6 > P_7\), what is the sum of all possible values of n?

A. 4
B. 5
C. 7
D. 9
E. 13


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D
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urmi760
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In a sequence of numbers in which each term after the first is found Updated on: 26 May 2020, 12:45
Originally posted by urmi760 on 15 May 2020, 10:28.
Last edited by urmi760 on 26 May 2020, 12:45, edited 1 time in total.
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answer D

n^4<1000 and n^5>1000

only numbers 4 and 5 satisfy the given condition, so sum of all number is nine.
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In a sequence of numbers in which each term after the first is found 15 May 2020, 10:33
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Suppose first term = a

\(\frac{a}{n} =1000\)

a=1000n

\(P_6>P_5\)

\(\frac{a^6}{n^{15}} > \frac{a^5}{n^{10}}\)

Since a,n>0

\(a>n^5\).......(1)

\(P_6>P_7\)

\(\frac{a^6}{n^{15}} > \frac{a^7}{n^{21}}\)

\(a<n^6\).........(2)

\(n^5<a<n^6\) [From (1) and (2)]

\(n^5<1000n<n^6\)

\(n^4<1000<n^5\)

n can be 4 or 5

Sum of all possible values of n = 4+5=9


Bunuel
In a sequence of numbers in which each term after the first is found by multiplying the previous one by 1/n, where n is a positive integer, the second term is 1,000. If \(P_n\) is the product of the first n terms of the sequence, \(P_6 > P_5\) and \(P_6 > P_7\), what is the sum of all possible values of n?

A. 4
B. 5
C. 7
D. 9
E. 13


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In a sequence of numbers in which each term after the first is found 15 May 2020, 18:55
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In a sequence of numbers in which each term after the first is found by multiplying the previous one by 1/n, where n is a positive integer, the second term is 1,000. If Pn is the product of the first n terms of the sequence, P6>P5 and P6>P7, what is the sum of all possible values of n?

T2=1000
so,T1=1000n
also,T3=1000/n
T4=1000/n^2
and ... so on

P6>P5
or, T1*T2*T3*T4*T5*T6 > T1*T2*T3*T4*T5
or, T6 > 1 [ we can divide both side by T1*T2*T3*T4*T5 because each term is positive being n is a positive integer ]

or,1000/n^4 >1
or,n^4 <1000----(1)

also, P6>P7
so, 1> T7
or,T7<1
so, 1000/n^5 <1
or,n^5>1000 ---(2)

from (1) and (2),
we can say, n cannot be 1, 2, 3 (3^5=243)
if n= 4 (4^4 = 2^8 = 256 , 256*4 >1000 (mind calculation 4*250 = 1000)
n=5 (5^4 = 25^2 = 625 , 625 *5 obviously >100)
n=6 (6^4; 6^3 = 216 , 216*6 >1000 mind calculation 200*6 =1200 )
no need to check further.
only possible value of n = 4 and 5
sum of all possible values = 4+5 =9

correct answer D
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In a sequence of numbers in which each term after the first is found 17 May 2020, 00:04
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In a sequence of numbers in which each term after the first is found by multiplying the previous one by 1/n, where n is a positive integer, the second term is 1,000. If PnPn is the product of the first n terms of the sequence, P6>P5 and P6>P7, what is the sum of all possible values of n?

..., 1000, \(\frac{1000}{n}\), \(\frac{1000}{n^{2}}\), \(\frac{1000}{n^{3}}\), \(\frac{1000}{n^{4}}\), \(\frac{1000}{n^{5}}\).

P6 > P5 ---> \(\frac{1000^{5}}{n^{10}} > \frac{1000^{4}}{n^{6}}\)

\(\frac{1000}{n^{4}} > 1\)
Since n is positive integer, n could be 1,2,3,4,5

P6 > P7 ---> \(\frac{1000^{5}}{n^{10}} > \frac{1000^{6}}{n^{15}}\)

\(1 > \frac{1000}{n^{5}}\)
Since n is positive integer, n could be 4,5,6,7,.........

Taken together,
n could be only 4 and 5

---> 4+5 = 9

Answer (D).
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In a sequence of numbers in which each term after the first is found 23 Oct 2023, 08:43
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Bunuel
In a sequence of numbers in which each term after the first is found by multiplying the previous one by 1/n, where n is a positive integer, the second term is 1,000. If \(P_n\) is the product of the first n terms of the sequence, \(P_6 > P_5\) and \(P_6 > P_7\), what is the sum of all possible values of n?

A. 4
B. 5
C. 7
D. 9
E. 13


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a2=1000 => a1/2=a2, so a1= 2000; similarly, a3= 1000/3; a4= 1000/12; a5= 1000/60, so P1= a1; P2= a1*a2= 2000*1000; P3=2000*1000*1000/3; I believe this is what the question says about Pn, by this logic, the condition where P7<P6<P5, can not arise. I am either lacking in the understanding of the question and am missing something, or the question has been framed incorrectly, where am I going wrong? Can someone please help me out here??
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In a sequence of numbers in which each term after the first is found 23 May 2025, 14:19
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