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# If the sequence x1, x2, x3, …, xn, … is such that x1 = 3

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Senior Manager
Joined: 23 May 2008
Posts: 353
If the sequence x1, x2, x3, …, xn, … is such that x1 = 3  [#permalink]

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16 Sep 2009, 23:05
2
00:00

Difficulty:

55% (hard)

Question Stats:

76% (01:11) correct 24% (01:44) wrong based on 198 sessions

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If the sequence x1, x2, x3, …, xn, … is such that x1 = 3 and xn+1 = 2xn – 1 for n ≥ 1, then x20 – x19 =

A. 2^19
B. 2^20
C. 2^21
D. 2^20 - 1
E. 2^21 - 1

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-the-sequence-x1-x2-x3-xn-is-such-that-x1-98536.html
Senior Manager
Joined: 23 Jun 2009
Posts: 349
Location: Turkey
Schools: UPenn, UMich, HKS, UCB, Chicago
Re: If the sequence x1, x2, x3, …, xn, … is such that x1 = 3  [#permalink]

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17 Sep 2009, 00:29
x20=2.x19-1
x20-x19=2.x19-1-x19=x19-1................1
x19=2.x18-1..............2
putting in 2 in 1
x20-x19=2.x18-2...........3
x18=2.x17-1-1..............4
putting 4 in 3
x20-x19=2.(2.x17-1)-1-1=4.x17-2-1-1
....
....
x2=2.x1-1
x20-x19=2^18.x1-(2^17+2^16+......+2+1+1)
x20-x19=2^18.3-2^18=2^19
Manager
Joined: 15 Sep 2009
Posts: 113
Re: If the sequence x1, x2, x3, …, xn, … is such that x1 = 3  [#permalink]

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17 Sep 2009, 05:00
I wud go with option A) 2^ 19
Manager
Joined: 27 Oct 2008
Posts: 175
Re: If the sequence x1, x2, x3, …, xn, … is such that x1 = 3  [#permalink]

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19 Sep 2009, 05:07
13
3
If the sequence x1, x2, x3, …, xn, … is such that x1 = 3 and xn+1 = 2xn – 1 for n ≥ 1, then x20 – x19 =
A. 2^19
B. 2^20
C. 2^21
D. 2^20 - 1
E. 2^21 - 1

Soln: the shorter way of doing this is to start from x1,
x(1) = 3
Now since x(n+1) = 2x(n) - 1
x(2) = 5
x(3) = 9
x(4) = 17
we can see the pattern now tat
x(n) = 2^n + 1
therefore
x(20) = 2^20 + 1
x(19) = 2^19 + 1

so x(20) - x(19) = 2^20 - 2^19 = 2^19(2-1) = 2^19

soln is A
Intern
Joined: 25 Oct 2009
Posts: 21
Schools: Wharton, HBS, Stanford
PS q25 from set 1 (31's sets)  [#permalink]

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29 Oct 2009, 10:53
Hi! I've just started my practice phase of GMAT preparation and was very surprised to get wrong answer on the question that I was sured I did right.

If the sequence $$x_1, x_2, x_3,$$ …, $$x_n,$$… is such that $$x_1=3$$ and $$x_{n+1}=2x_n - 1$$ for $$n\geq1$$, then $$x_{20} - x_{19} =$$

A. $$2^{19}$$
B. $$2^{20}$$
C. $$2^{21}$$
D. $$2^{20}-1$$
E. $$2^{21}-1$$

The OA is A, but I answered B and don't understand what I've missed.

P.s. Pls, advise me on how to make powers a little bit smaller?
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Math Expert
Joined: 02 Sep 2009
Posts: 52911
Re: PS q25 from set 1 (31's sets)  [#permalink]

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29 Oct 2009, 16:07
9
3
Vyacheslav wrote:
Hi! I've just started my practice phase of GMAT preparation and was very surprised to get wrong answer on the question that I was sured I did right.

If the sequence $$x_1, x_2, x_3,$$ …, $$x_n,$$… is such that $$x_1=3$$ and $$x_{n+1}=2x_n - 1$$ for $$n\geq1$$, then $$x_{20} - x_{19} =$$

A. $$2^{19}$$
B. $$2^{20}$$
C. $$2^{21}$$
D. $$2^{20}-1$$
E. $$2^{21}-1$$

The OA is A, but I answered B and don't understand what I've missed.

P.s. Pls, advise me on how to make powers a little bit smaller?

We have the sequence $$x_1, x_2, x_3,$$ …, $$x_n,$$… $$x_1=3$$ and $$x_{n+1}=2x_n - 1$$ for $$n\geq1$$.

If you notice there is a specific pattern int it:
$$x_1=3=2^1+1$$
$$x_2=5=2^2+1$$
$$x_3=9=2^3+1$$
...
$$x_n=2^n+1$$

So, $$x_{20}=2^{20}+1$$ and $$x_19=2^{19}+1$$.

$$x_{20}-x_{19}=2^{20}+1-2^{19}-1=2^{20}-2^{19}=2^{19}$$

Hope it helps.
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Joined: 01 Apr 2008
Posts: 764
Name: Ronak Amin
Schools: IIM Lucknow (IPMX) - Class of 2014
Re: PS q25 from set 1 (31's sets)  [#permalink]

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31 Oct 2009, 04:19
I was trying to do it by another way:

x20 - x19 = x19 - 1.

Now, x19 = 2 (x18) - 1 = 2 ( 2 (x17 - 1 ) ) - 1 and so on... we get 2^18(x1 - 1 ) - 1
x19 = 2^19 - 1

Hence, required ans is 2^19 - 1 - 1...I am missing something for sure
Intern
Joined: 13 Sep 2011
Posts: 4
Re: PS q25 from set 1 (31's sets)  [#permalink]

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17 May 2012, 22:00
Bunuel wrote:
Vyacheslav wrote:
Hi! I've just started my practice phase of GMAT preparation and was very surprised to get wrong answer on the question that I was sured I did right.

If the sequence $$x_1, x_2, x_3,$$ …, $$x_n,$$… is such that $$x_1=3$$ and $$x_{n+1}=2x_n - 1$$ for $$n\geq1$$, then $$x_{20} - x_{19} =$$

A. $$2^{19}$$
B. $$2^{20}$$
C. $$2^{21}$$
D. $$2^{20}-1$$
E. $$2^{21}-1$$

The OA is A, but I answered B and don't understand what I've missed.

P.s. Pls, advise me on how to make powers a little bit smaller?

We have the sequence $$x_1, x_2, x_3,$$ …, $$x_n,$$… $$x_1=3$$ and $$x_{n+1}=2x_n - 1$$ for $$n\geq1$$.

If you notice there is a specific pattern int it:
$$x_1=3=2^1+1$$
$$x_2=5=2^2+1$$
$$x_3=9=2^3+1$$
...
$$x_n=2^n+1$$

So, $$x_{20}=2^{20}+1$$ and $$x_19=2^{19}+1$$.

$$x_{20}-x_{19}=2^{20}+1-2^{19}-1=2^{20}-2^{19}=2^{19}$$

Hope it helps.

Hi, Bunnel
Your explanation has always been a sigh of high relief to me. Owing to my sluggish expertise, I couldn't figure out as to how you were able to determine that there is a pattern and that is this:

We have the sequence x_1, x_2, x_3, …, x_n,… x_1=3 and x_{n+1}=2x_n - 1 for n\geq1.

Could you please throw light on as to how we should make out the pattern/serious in such questions?

Thank you so much.
Math Expert
Joined: 02 Sep 2009
Posts: 52911
Re: PS q25 from set 1 (31's sets)  [#permalink]

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18 May 2012, 00:13
smileforever41 wrote:
Bunuel wrote:
Vyacheslav wrote:
Hi! I've just started my practice phase of GMAT preparation and was very surprised to get wrong answer on the question that I was sured I did right.

If the sequence $$x_1, x_2, x_3,$$ …, $$x_n,$$… is such that $$x_1=3$$ and $$x_{n+1}=2x_n - 1$$ for $$n\geq1$$, then $$x_{20} - x_{19} =$$

A. $$2^{19}$$
B. $$2^{20}$$
C. $$2^{21}$$
D. $$2^{20}-1$$
E. $$2^{21}-1$$

The OA is A, but I answered B and don't understand what I've missed.

P.s. Pls, advise me on how to make powers a little bit smaller?

We have the sequence $$x_1, x_2, x_3,$$ …, $$x_n,$$… $$x_1=3$$ and $$x_{n+1}=2x_n - 1$$ for $$n\geq1$$.

If you notice there is a specific pattern int it:
$$x_1=3=2^1+1$$
$$x_2=5=2^2+1$$
$$x_3=9=2^3+1$$
...
$$x_n=2^n+1$$

So, $$x_{20}=2^{20}+1$$ and $$x_19=2^{19}+1$$.

$$x_{20}-x_{19}=2^{20}+1-2^{19}-1=2^{20}-2^{19}=2^{19}$$

Hope it helps.

Hi, Bunnel
Your explanation has always been a sigh of high relief to me. Owing to my sluggish expertise, I couldn't figure out as to how you were able to determine that there is a pattern and that is this:

We have the sequence x_1, x_2, x_3, …, x_n,… x_1=3 and x_{n+1}=2x_n - 1 for n\geq1.

Could you please throw light on as to how we should make out the pattern/serious in such questions?

Thank you so much.

Well, there is no some kind of general technique to recognize a pattern. The first thing you should do is just write down several terms and look for a pattern.
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Joined: 11 Oct 2011
Posts: 4
Re: If the sequence x1, x2, x3, …, xn, … is such that x1 = 3  [#permalink]

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19 May 2012, 19:27
Benelux, you may need to show how you factored out 2^19 and multiplied (2-1). Knowing it and seeing it was the difference for me.
SVP
Joined: 06 Sep 2013
Posts: 1696
Concentration: Finance
Re: PS q25 from set 1 (31's sets)  [#permalink]

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05 Oct 2013, 12:47
I also find it hard to spot patterns specially under time constraints. Sometimes, I write down a couple of items and try to see the difference between each member but it doesn't seem to work that well.
If anyone has any suggestions/techniques/tips that could make pattern recognition easier I would be more than grateful

Cheers!
Intern
Joined: 30 Jan 2014
Posts: 3
Re: If the sequence x1, x2, x3, …, xn, … is such that x1 = 3  [#permalink]

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07 Feb 2014, 20:33
No frills or shortcuts, just straight algebra.

For $$x_n$$ = 2$$x_n$$ - 1:

$$x_1$$= 3

+2

$$x_2$$= 5

+4

$$x_3$$= 9

+8

$$x_4$$= 17

+16

$$x_5$$= 33,

Notice the difference between each entry of the sequence is a geometric progression: 2, 4, 8, 16.

$$x_1$$ x $$r^{n-1}$$

2 x $$2^{n-1}$$

Thus,

= $$x_{20}$$ - $$x_{19}$$

= (2 x $$2^{19}$$) - (2 x $$2^{18}$$)

= 2 x $$2^{18}$$ (2 - 1)

= $$2^{19}$$ (1)

= $$2^{19}$$
Manager
Joined: 30 Mar 2013
Posts: 108
Re: If the sequence x1, x2, x3, …, xn, … is such that x1 = 3  [#permalink]

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24 May 2014, 05:52
srivas wrote:
If the sequence x1, x2, x3, …, xn, … is such that x1 = 3 and xn+1 = 2xn – 1 for n ≥ 1, then x20 – x19 =
A. 2^19
B. 2^20
C. 2^21
D. 2^20 - 1
E. 2^21 - 1

Soln: the shorter way of doing this is to start from x1,
x(1) = 3
Now since x(n+1) = 2x(n) - 1
x(2) = 5
x(3) = 9
x(4) = 17
we can see the pattern now tat
x(n) = 2^n + 1
therefore
x(20) = 2^20 + 1
x(19) = 2^19 + 1

so x(20) - x(19) = 2^20 - 2^19 = 2^19(2-1) = 2^19

soln is A

SorryI dont get the last part: 2^20 - 2^19 = 2^(20-19)...which is 2^1?
Math Expert
Joined: 02 Sep 2009
Posts: 52911
Re: If the sequence x1, x2, x3, …, xn, … is such that x1 = 3  [#permalink]

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24 May 2014, 05:57
usre123 wrote:
srivas wrote:
If the sequence x1, x2, x3, …, xn, … is such that x1 = 3 and xn+1 = 2xn – 1 for n ≥ 1, then x20 – x19 =
A. 2^19
B. 2^20
C. 2^21
D. 2^20 - 1
E. 2^21 - 1

Soln: the shorter way of doing this is to start from x1,
x(1) = 3
Now since x(n+1) = 2x(n) - 1
x(2) = 5
x(3) = 9
x(4) = 17
we can see the pattern now tat
x(n) = 2^n + 1
therefore
x(20) = 2^20 + 1
x(19) = 2^19 + 1

so x(20) - x(19) = 2^20 - 2^19 = 2^19(2-1) = 2^19

soln is A

SorryI dont get the last part: 2^20 - 2^19 = 2^(20-19)...which is 2^1?

$$2^{20}-2^{19}$$ --> factor out $$2^{19}$$:

$$2^{20}-2^{19}=2^{19}(2-1)= 2^{19}$$.

Hope it's clear.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-the-sequence-x1-x2-x3-xn-is-such-that-x1-98536.html
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Re: If the sequence x1, x2, x3, …, xn, … is such that x1 = 3  [#permalink]

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31 Jan 2019, 05:48
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Re: If the sequence x1, x2, x3, …, xn, … is such that x1 = 3   [#permalink] 31 Jan 2019, 05:48
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