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![In a sequence of integers, [m]a_1 = 40[/m] and [m]a_n < a_{n](/forum/styles/gmatclub_light/imageset/icon_post_target_unread.svg "In a sequence of integers, [m]a_1 = 40[/m] and [m]a_n < a_{n"){kind=icon}
17 Apr 2026, 23:53
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
28% (02:26) correct
72%
(02:21)
wrong
based on 76
sessions
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Not Attempted Yet
In a sequence of integers, \(a_1 = 40\) and \(a_n < a_{n-1}\) for all integers \(n > 1\).
From which of the following can it be concluded that \(a_{20}\) is positive?
I. \(a_{25} = \frac{a_{24}}{2}\)
II. The sum of the first 42 terms is positive.
III. \(a_{29} > a_{27} - a_{28}\)
(A) I only
(B) II only
(C) III only
(D) I and III
(E) I, II, and III
From which of the following can it be concluded that \(a_{20}\) is positive?
I. \(a_{25} = \frac{a_{24}}{2}\)
II. The sum of the first 42 terms is positive.
III. \(a_{29} > a_{27} - a_{28}\)
(A) I only
(B) II only
(C) III only
(D) I and III
(E) I, II, and III
21 Apr 2026, 07:55
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I would like to know why this is not the correct answer:
I. a25 = a24/2 => a1+24r = (a1+23r)/2 => 2a1+48r = a1+23r => 25r= -a1 (a1 = 40) => r = -40/25 = -8/5
Since we want to know if a20 > 0? (using a21 which is lower than a20 for easier caculation) a21 = a1+20r = 40 +20*-8/5 = 40-32=8 then a20 is positive.
II. Sn = (a1+an)n/2 => (a1+a42)*42/2 > 0 => a1+a42 > 0 => 2a1+41r >0 => r > -80/41 if this we know that r > -2 (-80/41 aprox. -2) a20 = 40 -2*19 = 40 - 38 = 2 so a20 must be positive since r must be lower than -2 to a20 to be negative.
III. a29>a27-a28 => a29> a1+26r -(a1+27r) => a29 > -r since we know that an<a(n-1) than r must be negative so a29> -r is positive so a20 is also positive.
Thanks for reading!
Edited:
Answer: the exercise never says the sequence is a arithmetic progression!
I. a25 = a24/2 => a1+24r = (a1+23r)/2 => 2a1+48r = a1+23r => 25r= -a1 (a1 = 40) => r = -40/25 = -8/5
Since we want to know if a20 > 0? (using a21 which is lower than a20 for easier caculation) a21 = a1+20r = 40 +20*-8/5 = 40-32=8 then a20 is positive.
II. Sn = (a1+an)n/2 => (a1+a42)*42/2 > 0 => a1+a42 > 0 => 2a1+41r >0 => r > -80/41 if this we know that r > -2 (-80/41 aprox. -2) a20 = 40 -2*19 = 40 - 38 = 2 so a20 must be positive since r must be lower than -2 to a20 to be negative.
III. a29>a27-a28 => a29> a1+26r -(a1+27r) => a29 > -r since we know that an<a(n-1) than r must be negative so a29> -r is positive so a20 is also positive.
Thanks for reading!
Edited:
Answer: the exercise never says the sequence is a arithmetic progression!
kevincan
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