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# The first term in sequence Q equals 1, and for all positive integers

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Math Expert
Joined: 02 Sep 2009
Posts: 50670
The first term in sequence Q equals 1, and for all positive integers  [#permalink]

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09 Sep 2015, 00:05
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13
00:00

Difficulty:

55% (hard)

Question Stats:

67% (02:56) correct 33% (02:35) wrong based on 137 sessions

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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

Kudos for a correct solution.

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Joined: 04 Sep 2014
Posts: 13
GMAT 1: 690 Q46 V38
GMAT 2: 720 Q49 V39
Re: The first term in sequence Q equals 1, and for all positive integers  [#permalink]

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09 Sep 2015, 00:45
1
1
note: This is my first post for GMAT club so please forgive me if my notation is weird (or non existent)...

we have that Q1=1 (by definition)

for Qn, where n=>2, the formula is Qn= | Q(nth smallest perfect cube) - Q(second smalls perfect cube) |

in order to make things a little easier... lets find some positive perfect cubes... which are:
1^3 : 1 =1 × 1 × 1 (smallest perfect cube)
2^3: 8 =2 × 2 × 2 (2nd smallest perfect cube)
3^3: 27 =3 × 3 × 3
4^3: 64 =4 × 4 × 4
5^3: 125 =5 × 5 × 5
6^3: 216 =6 × 6 × 6
7^3: 343 =7 × 7 × 7
8^3: 512 =8 × 8 × 8
9^3: 729 =9 × 9 × 9
10^3 : 1000 =10 × 10 × 10 (10th smallest perfect cube)

so to complete the Q2, we need: | Q(2nd smallest perfect cube) - Q(1st smallest cube) | = | 8 - 1 | = 7
Q3: | Q(3rd smallest perfect cube) - Q(3rd smallest perfect cube) = | 27 - 8 | = 19....
.
.
.
Q7: |Q(7th smallest perfect cube) - Q(6th smallest perfect cube) = |343-216 | = 127

which would mean answer b? no that's a trap answer... that's the 7th term of the sequence.. not the sum of the 1st seven terms...

you can get all the terms and add them if you'd like.. but to quote Sweet Brown "aint nobody got time for that".....

instead lets do the sum of the sequences to see of there is a pattern....

Q1 =1..... which happens to be 1x1x1...
Q1 + Q2 = 1 + 7 = 8 .... happens to be 2x2x2
Q1 + Q2 + Q3 = 1 +7 +19= 27... happens to be 3x3x3....

imma guess that
Q1 + ....... + Q7= whatever to be 7x7x7.... which is 343.....=> D
Manager
Joined: 10 Aug 2015
Posts: 103
Re: The first term in sequence Q equals 1, and for all positive integers  [#permalink]

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09 Sep 2015, 01:53
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Solution : The question means $$Q_n = n^3 - (n-1)^3$$ for n>=2 and $$Q_1 = 1$$
Sum of n terms = $$n^{3}$$ = 7*7*7 = 343

Option, D
Math Expert
Joined: 02 Sep 2009
Posts: 50670
Re: The first term in sequence Q equals 1, and for all positive integers  [#permalink]

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13 Sep 2015, 08:51
Bunuel wrote:
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

Kudos for a correct solution.

MANHATTAN GMAT OFFICIAL SOLUTION:

To solve this problem, we must translate the verbal instructions for the construction of the sequence. The first term is easy: $$Q_1 = 1$$.

Next, we have this difficult wording: “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.”

So we’re dealing with all the positive integers beyond 1. Let’s take as an example n = 2. The instructions become these: “the second term equals the absolute value of the difference between the second (nth) smallest positive perfect cube and the first (that is, n-1st) smallest positive perfect cube.”

We know we need to consider the positive perfect cubes in order:
1^3 = 1 = smallest positive perfect cube (or “first smallest”).
2^3 = 8 = second smallest positive perfect cube.

The absolute value of the difference between these cubes is 8 – 1 = 7. Thus $$Q_2 = 8 – 1 = 7$$.
Likewise, $$Q_3 = |3^3 – 2^3| = 27 – 8 = 19$$, and so on.
Now, rather than figure out each term of Q separately, then add up, we can save time if we notice that the cumulative sums “telescope” in a simple way. This is what telescoping means:

The sum of $$Q_2$$ and $$Q_1 = 7 + 1 = 8$$. We can also write (8 – 1) + 1 = 8. Notice how the 1’s cancel.
The sum of $$Q_3$$, $$Q_2$$ and $$Q_1 = 19 + 7 + 1 = 27$$. We can also write (27 – 8 ) + (8 – 1) + 1 = 27. Notice how the 8’s and the 1’s cancel.

At this point, we hopefully notice that the cumulative sum of $$Q_1$$ through $$Q_n$$ is just the nth smallest positive perfect cube.

So the sum of the first seven terms of the sequence is 7^3, which equals 343.

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The first term in sequence Q equals 1, and for all positive integers  [#permalink]

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07 Sep 2017, 10:17
2
1
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
1 terms is 1
2 term will be ( 2^3 - 1^3),
3 term will be ( 3^3 - 2^3),
4 term will be ( 4^3 - 3^3),
5 term will be ( 5^3 - 4^3),
6 term will be ( 6^3 - 5^3),
7 term will be ( 7^3 - 6^3).

if we add all the terms = ( 7^3 - 6^3 +6^3 - 5^3+5^3 - 4^3+ 4^3 - 3^3+ 3^3 - 2^3+ 2^3 - 1^3 +1) we are left with 7 ^3

i have not taken absolute value because we have ( grater positive value)^3 - ( smaller postie value )^3 which will anyway give postie value.
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Abhimanyu

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Joined: 15 Jan 2017
Posts: 359
Re: The first term in sequence Q equals 1, and for all positive integers  [#permalink]

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05 Dec 2017, 14:19
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

Method I used:

A1= 1 --> 1^3= 1 Thus 1 is smallest cube

Next smallest cube =8 (it is 2^3).

Now the series can be endless, and the pattern of nth term will remain the same.

So A(n) =|1 - 8| = |-7| = difference of 7 {likely to repeat itself}
So, the difference must a power of 7. Which 343.
Ans D

For those extremely sleepy and not paying attention the answer as multiple of 9 was also present to confuse (216). But either way, great question. Took just about 2 minutes when I got it

Kudos, if my explanation was useful for you :D. If not I would love to learn an alternate way!
Re: The first term in sequence Q equals 1, and for all positive integers &nbs [#permalink] 05 Dec 2017, 14:19
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