GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Nov 17Live Online Classes with Top GRE Expert Rich Zwelling
08:00 PM EST
-10:00 PM EST
Get live instruction from a top-scoring, industry-recognized GRE teachers + 6 months of access to TTP Self-Study included! Tues/Thur Sept 7, 2026---Feb. 2, 2027, 1:00pm-3:00pm EST
Dec 14Live Online Classes with Top GRE Expert Michael Butville
08:00 PM EST
-10:00 PM EST
Get live instruction from a top-scoring, industry-recognized GRE teachers + 6 months of access to TTP Self-Study included! Mon/Wed Dec. 14, 2026---Feb. 22, 2027, 8:00pm-10:00pm EST
GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2960
Given Kudos: 1289
GPA: 3.39
The first term in sequence Q equals 1, and for all positive integers
[#permalink]
15 Apr 2021, 00:56
15 Apr 2021, 00:56
1
Expert Reply
5
Bookmarks
00:00
Question Stats:
85% (02:38) correct
15% (01:24) wrong based on 20 sessions
Hide Show timer Statistics
The first term in sequence Q equals 1, and for all positive integers n equal to or greater than 2, the nth term in sequence Q equals the absolute value of the difference between the nth smallest positive perfect cube and the (n-1)st smallest positive perfect cube. The sum of the first seven terms in sequence Q is
(A) 91
(B) 127
(C) 216
(D) 343
(E) 784
(A) 91
(B) 127
(C) 216
(D) 343
(E) 784
The first term in sequence Q equals 1, and for all positive integers
[#permalink]
15 Apr 2021, 03:06
15 Apr 2021, 03:06
Quote:
the nth term in sequence Q equals the absolute value of the difference between the nth smallest positive perfect cube and the (n-1)st smallest positive perfect cube.
Translate this. Q = 1, |8-1|, |27-8|, ... so on.
Observe two things here. First, the absolute value sign makes no difference because the difference will always be positive.
So you may even imagine it as
Q = 1, 8-1, 27-8, ... so on.
Second, when you sum all the values, pairs such as 1 and -1, and 8 and -8 get cancelled out, leaving only the nth smallest positive perfect cube of the last (or the nth) term in the sum.
Quote:
The sum of the first seven terms in sequence Q is
Now you can imagine that this will only leave out \(7^3\) as it is the last term in the sum.
Hence, D is the correct answer.
Retired Moderator
Joined: 02 Jan 2020
Status:GRE Quant Tutor
Posts: 1141
Given Kudos: 9
Location: India
Concentration: General Management
Schools: XLRI Jamshedpur, India - Class of 2014
GMAT 1: 700 Q51 V31
GPA: 2.8
WE:Engineering (Computer Software)
The first term in sequence Q equals 1, and for all positive integers
[#permalink]
08 Jul 2021, 07:09
08 Jul 2021, 07:09
1
The first term in sequence Q equals 1 and
the nth term in sequence Q equals the absolute value of the difference between the nth smallest positive perfect cube and the (n-1)st smallest positive perfect cube
This simply means that
\(T_n\) = | \(n^3\) - \((n-1)^3\) |
So, \(T_2\) = | \(2^3\) - \((2-1)^3\) | = | \(2^3\) - \(1^3\) |
As the result is positive so the absolute value sign doesn't really matter here. So we can open the Absolute values to have
\(T_2\) = \(2^3\) - \(1^3\)
Similarly, \(T_3\) = \(3^3\) - \(2^3\) and so on...
\(T_7\) = \(7^3\) - \(6^3\)
Sum of First seven terms is given by
\(T_1\) + \(T_2\) + \(T_3\) + \(T_4\) + \(T_5\) + \(T_6\) + \(T_7\)
= 1 + \(2^3\) - \(1^3\) + \(3^3\) - \(2^3\) + .... + \(7^3\) - \(6^3\)
So, alternate terms will cancel out. Ex \(2^3\) (First part in \(T_2\)) and -\(2^3\)second part in \(T_3\) and so on..
So, we will be left with
= 1 - \(1^3\) + \(7^3\)
= \(7^3\) = 343
So, answer will be D
Hope it helps!
To learn more about Absolute value and Sequences watch the following videos
the nth term in sequence Q equals the absolute value of the difference between the nth smallest positive perfect cube and the (n-1)st smallest positive perfect cube
This simply means that
\(T_n\) = | \(n^3\) - \((n-1)^3\) |
So, \(T_2\) = | \(2^3\) - \((2-1)^3\) | = | \(2^3\) - \(1^3\) |
As the result is positive so the absolute value sign doesn't really matter here. So we can open the Absolute values to have
\(T_2\) = \(2^3\) - \(1^3\)
Similarly, \(T_3\) = \(3^3\) - \(2^3\) and so on...
\(T_7\) = \(7^3\) - \(6^3\)
Sum of First seven terms is given by
\(T_1\) + \(T_2\) + \(T_3\) + \(T_4\) + \(T_5\) + \(T_6\) + \(T_7\)
= 1 + \(2^3\) - \(1^3\) + \(3^3\) - \(2^3\) + .... + \(7^3\) - \(6^3\)
So, alternate terms will cancel out. Ex \(2^3\) (First part in \(T_2\)) and -\(2^3\)second part in \(T_3\) and so on..
So, we will be left with
= 1 - \(1^3\) + \(7^3\)
= \(7^3\) = 343
So, answer will be D
Hope it helps!
To learn more about Absolute value and Sequences watch the following videos
