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The sequence a1, a2, … , a n, … is such that an = 2an-1 - x [#permalink] Updated on: 24 Dec 2013, 00:16

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The sequence \(a_1\), \(a_2\), … , \(a_n\), … is such that \(a_n = 2a_{n-1} - x\) for all positive integers n ≥ 2 and for certain number x. If \(a_5 = 99\) and \(a_3 = 27\), what is the value of x?

A. 3
B. 9
C. 18
D. 36
E. 45

Originally posted by jet1445 on 04 Jun 2007, 19:23.
Last edited by Bunuel on 24 Dec 2013, 00:16, edited 1 time in total.

Renamed the topic, edited the question and added the OA.

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Re: Sequence Problem [#permalink] 28 Aug 2010, 09:13

udaymathapati wrote:

The sequence a1, a2, … , a n, … is such that an = 2an-1 - x for all positive integers n ≥ 2 and for certain number x. If a5 = 99 and a3 = 27, what is the value of x?
A. 3
B. 9
C. 18
D. 36
E.45

Quite simple to solve this one.

Given: \(a_n = 2a_{n-1} - x\)

\(a_5 = 99\)

\(a_3 = 27\)

\(a_5 = 2a_4 - x = 2(2a_3 - x) - x = 4a_3 - 3x = 99\)

\(4(27) - 3x = 99\)\(3x = 108-99 = 9\)

\(x = 3\)

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Re: PS - Sequence (a1, a2, …) [#permalink] 05 Jun 2007, 01:37

jet1445 wrote:

Q13:
The sequence a1, a2, … , a n, … is such that an = 2an-1 - x for all positive integers n ≥ 2 and for certain number x. If a5 = 99 and a3 = 27, what is the value of x?

A. 3
B. 9
C. 18
D. 36
E. 45

a5= 2*a4 - x = 99

a4 = 2*a3 - x = 2*27 - x

therefore;

a5 = 2*(54 - x ) -x = 99

108 - 3*x = 99

therefore X = 3

This the answer is "A"

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Re: Sequence Problem [#permalink] 28 Aug 2010, 06:30

Good one. Please let me know if some one comes up with a solution that can be worked under 2 mins

given an= 2an-1 - x

a5 = 99 = 2 a4 - X = 2[2a3-X] -X = 4 a3 - 3X
given a3 = 27 ; substituting:

108 - 3X = 99 => X = 3

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Sequence [#permalink] 20 Oct 2010, 05:25

The sequence a1…..a2.......an........is such that an=2an-1-X for all positive integers n>=2 and for certain number X. If a5=99 and a3=27, what is the value of X?

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Re: Sequence [#permalink] 20 Oct 2010, 05:50

monirjewel wrote:

The sequence a1…..a2.......an........is such that an=2an-1-X for all positive integers n>=2 and for certain number X. If a5=99 and a3=27, what is the value of X?

Plug the known values a5=99 and a3 = 27 into the formula:

a5 = 2(a4) - x
99 = 2(a4) - x

a4 = 2(a3) - x
a4 = 2(27)-x = 54-x

Substitute 54-x for a4 in the top equation:
99 = 2(54-x)-x
99=108-3x
3x=9
x=3

On the GMAT, I would recommend that you plug in the answer choices, one of which would say that x=3.

Plug a5 = 99 and x=3 into the formula:
99 = 2(a4) -3
a4 = 51

Plug a4=51, a3=27, and x=3 into the formula:
51 = 2(27) - 3
51 = 51. Success!
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Re: Sequence [#permalink] 20 Oct 2010, 05:56

monirjewel wrote:

The sequence a1…..a2.......an........is such that an=2an-1-X for all positive integers n>=2 and for certain number X. If a5=99 and a3=27, what is the value of X?

A5=99 and A3=27
According to the given nth term, A5=2(A4)-x=2{2(A3)-x}-x=2{(2*27)-x}-x=108-2x-x=108-3x
Hence 108-3x=99
or x=9/3=3

Consider KUDOS if its helpful.

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Re: PS - Sequence (a1, a2, …) [#permalink] 11 Sep 2011, 09:03

a(n) = 2*a(n-1) -x

a5 = 99
a3=27

a5 = 2a4-x
a4 = 2a3-x

=>99 = 2(2a3-x)-x

99 = 4a3-3x = 4*27-3x

=>x=3

Answer is A.

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Re: PS - Sequence (a1, a2, …) [#permalink] 12 Sep 2011, 04:14

A is the answer

a4= 54-x

=> a5 = 2 (54 -x) -x = 99

=> 108 - 3x = 99
=> x= 3

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Re: The sequence a1, a2, … , a n, … is such that an = 2an-1 - x [#permalink] 04 Jul 2014, 07:51

whiplash2411 wrote:

udaymathapati wrote:

The sequence a1, a2, … , a n, … is such that an = 2an-1 - x for all positive integers n ≥ 2 and for certain number x. If a5 = 99 and a3 = 27, what is the value of x?
A. 3
B. 9
C. 18
D. 36
E.45

Quite simple to solve this one.

Given: \(a_n = 2a_{n-1} - x\)

\(a_5 = 99\)

\(a_3 = 27\)

\(a_5 = 2a_4 - x = 2(2a_3 - x) - x = 4a_3 - 3x = 99\)

\(4(27) - 3x = 99\)\(3x = 108-99 = 9\)

\(x = 3\)

QUESTION : How exactly did you get from 2(2a3 - x) to 4a3 * 3x, wouldn't it be 4a3 - 2x?

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Re: The sequence a1, a2, … , a n, … is such that an = 2an-1 - x [#permalink] 04 Jul 2014, 07:59

sagnik242 wrote:

whiplash2411 wrote:

udaymathapati wrote:

The sequence a1, a2, … , a n, … is such that an = 2an-1 - x for all positive integers n ≥ 2 and for certain number x. If a5 = 99 and a3 = 27, what is the value of x?
A. 3
B. 9
C. 18
D. 36
E.45

Quite simple to solve this one.

Given: \(a_n = 2a_{n-1} - x\)

\(a_5 = 99\)

\(a_3 = 27\)

\(a_5 = 2a_4 - x = 2(2a_3 - x) - x = 4a_3 - 3x = 99\)

\(4(27) - 3x = 99\)\(3x = 108-99 = 9\)

\(x = 3\)

QUESTION : How exactly did you get from 2(2a3 - x) to 4a3 * 3x, wouldn't it be 4a3 - 2x?

\(a_4=2a_3-x\) --> \(a_5 = 2a_4 - x\) --> \(a_4=2(2a_3-x)-x=4a_3-2x-x=4a_3-3x\).
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Re: The sequence a1, a2, … , a n, … is such that an = 2an-1 - x [#permalink] 21 Jan 2017, 02:28

jet1445 wrote:

A. 3
B. 9
C. 18
D. 36
E. 45

\(a_5= 2*a_4 - x = 99\)

\(a_4 = 2*a_3 - x = 2*27 - x\)

\(a_5 = 2*(54 - x ) -x = 99\)

108 - 3*x = 99

Therefore X = 3

Hence option A is correct.

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Re: The sequence a1, a2, … , a n, … is such that an = 2an-1 - x [#permalink] 01 Aug 2017, 11:23

I did not understand the relation at first, but giving another shot worked well, thanks.
a5=99
a3=27

a5=2a4−x
=2(2a3−x)−x
=4a3−3x=99a5
=2a4−x
=2(2a3−x)−x
=4a3−3x=99

4(27)−3x
=994(27)−3x
=993x
=108−99
=93x
=108−99
=9

x=3
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The sequence a1, a2, … , a n, … is such that an = 2an-1 - x [#permalink] 06 Mar 2018, 03:43

jet1445 wrote:

A. 3
B. 9
C. 18
D. 36
E. 45

Hello Bunuel,

following the sequence formula \(a_n = a_{n-1} + d = a_1 + (n-1)d\) mentioned here

https://gmatclub.com/forum/math-sequenc ... ml#p790381

can i now convert this \(a_n = 2a_{n-1} - x\) in a more friendly version to understand how it reads and what it wants :)

----> \(a_n = 2a_{n-1} - x\) = ---- > \(a_1+2(n-1)x\) am i correct ? please confirm

Why \(a_n = a_{n-1} + d= a_1 + (n-1) d\) why in the first formula we add \(d\) whereas in next formula we multiply by \(d\)

Hello niks18 perhaps you can help

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Re: The sequence a1, a2, … , a n, … is such that an = 2an-1 - x [#permalink] 06 Mar 2018, 08:27

dave13 wrote:

jet1445 wrote:

A. 3
B. 9
C. 18
D. 36
E. 45

Hello Bunuel,

following the sequence formula \(a_n = a_{n-1} + d = a_1 + (n-1)d\) mentioned here

https://gmatclub.com/forum/math-sequenc ... ml#p790381

can i now convert this \(a_n = 2a_{n-1} - x\) in a more friendly version to understand how it reads and what it wants :)

----> \(a_n = 2a_{n-1} - x\) = ---- > \(a_1+2(n-1)x\) am i correct ? please confirm

Why \(a_n = a_{n-1} + d= a_1 + (n-1) d\) why in the first formula we add \(d\) whereas in next formula we multiply by \(d\)

Hello niks18 perhaps you can help

Hi dave13

The question does not mention that the sequence is an AP series. so you cannot blindly apply AP formula to every sequence.

Regarding you second query \(a_n = a_{n-1} + d= a_1 + (n-1) d\), you need to understand how the formula is derived and what is the meaning of different notations in this formula. The derivation of this formula should be available in GMAT quant books or high school maths book.

For your understanding if I have an AP series as: 1,2,3,4,5,6 then can you identify what is \(a_n , a_{n-1} , d , a_1, n\) for this series.
Once you have identified these elements then try and plug in the values of the elements in the above formula to understand how it works.

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Re: The sequence a1, a2, … , a n, … is such that an = 2an-1 - x [#permalink] 06 Mar 2018, 12:12

niks18 wrote:

dave13 wrote:

jet1445 wrote:

A. 3
B. 9
C. 18
D. 36
E. 45

Hello Bunuel,

following the sequence formula \(a_n = a_{n-1} + d = a_1 + (n-1)d\) mentioned here

https://gmatclub.com/forum/math-sequenc ... ml#p790381

can i now convert this \(a_n = 2a_{n-1} - x\) in a more friendly version to understand how it reads and what it wants

----> \(a_n = 2a_{n-1} - x\) = ---- > \(a_1+2(n-1)x\) am i correct ? please confirm

Why \(a_n = a_{n-1} + d= a_1 + (n-1) d\) why in the first formula we add \(d\) whereas in next formula we multiply by \(d\)

Hello niks18 perhaps you can help

Hi dave13

The question does not mention that the sequence is an AP series. so you cannot blindly apply AP formula to every sequence.

thanks Niks, i wlll dive in these sequences

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The sequence a1, a2, … , a n, … is such that an = 2an-1 - x [#permalink] 07 Mar 2018, 09:20

jet1445 wrote:

A. 3
B. 9
C. 18
D. 36
E. 45

Hello niks18

ok I learnt something new today

I ve found out that there are actually two kind of sequence formals for finding nth term, i wish Bunuel added this information to his post about sequences, it would be very helpful for such dummies like me

Explicit formula

\(a_n = d(n-1)+a_1\)

Recursive formula

\(a_n = a_{n-1} + d\)

\(a_1\) is first term

\(a_n\) is next term

\(a_{n-1}\) is previous term

Now what is the difference between recursive and explicit formulas ? the difference is that recursive is better for finding the first few terms, whereas explicit is great for finding far off terms in sequence.

i have one question what does it mean "for all positive integers n ≥ 2" ?

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The sequence a1, a2, … , a n, … is such that an = 2an-1 - x [#permalink] 07 Mar 2018, 09:39

dave13 wrote:

jet1445 wrote:

A. 3
B. 9
C. 18
D. 36
E. 45

Hello niks18

ok I learnt something new today

Explicit formula

\(a_n = d(n-1)+a_1\)

Recursive formula

\(a_n = a_{n-1} + d\)

\(a_1\) is first term

\(a_n\) is next term

\(a_{n-1}\) is previous term

i have one question what does it mean "for all positive integers n ≥ 2" ?

Hi dave13

first of all do not mix AP formula with this question. The sequence mentioned in this question does not state that it is an AP series, hence you cannot apply AP formula here.
"for all positive integers n ≥ 2" means that n is either greater than or equal to 2. for values of n less than 2, this sequence does not hold.

you need not go into so much of technical definitions. Here you only need to understand the application. I had given an example in my earlier post of an AP series which you can use to understand the application; 1,2,3,4,5,6

Here first term, \(a_1=1\); common difference \(d=1\) and last term \(a_n=6\) and number of terms \(n=6\). So if I have to find \(a_n\) i.e the last term, then I can find it through either of the two formula mentioned by you above

so \(a_n=a_1+(n-1)*d=1+(6-1)*1=1+5=6\)

\(a_n=a_{n-1}+d=5+1=6\). Note here \(a_{n-1}\) means a term just before the last term hence the notation \(n-1\)

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Re: The sequence a1, a2, … , a n, … is such that an = 2an-1 - x [#permalink] 08 Mar 2018, 01:59

niks18 wrote:

dave13 wrote:

jet1445 wrote:

A. 3
B. 9
C. 18
D. 36
E. 45

Hello niks18 :)

ok I learnt something new today :)

Explicit formula

\(a_n = d(n-1)+a_1\)

Recursive formula

\(a_n = a_{n-1} + d\)

\(a_1\) is first term

\(a_n\) is next term

\(a_{n-1}\) is previous term

i have one question what does it mean "for all positive integers n ≥ 2" ? :-)

Hi dave13

so \(a_n=a_1+(n-1)*d=1+(6-1)*1=1+5=6\)

\(a_n=a_{n-1}+d=5+1=6\). Note here \(a_{n-1}\) means a term just before the last term hence the notation \(n-1\)

Hello there niks18 :) thanks for explanation. highly appreciate. you know i think i am starting to undesrstand some nuances :) i ve read this post

https://magoosh.com/gmat/2012/sequences-on-the-gmat/

it is about sequences

here is an extract from the link "
Explicit Series
A explicit series is a series in which the general rule for finding each term can be stated, either verbally or mathematically. Sometimes the GMAT will give you the general rule for a sequence in algebraic form:

if a sequence is defined by \(a_n = n^2-n\) for n ≥ 2 what is the difference between 4th and 3rd terms

so what can i conclude is that if we dont know the distance between two terms, all we have to do is plug in the index number to find the value of each term. Here, we plug in n = 3 to find the third term, and plug in n = 4 to find the fourth term.

the same rule applied to the question we discussed right? because in all above posts everyone just plug in 5 (as fifth term) into \({n-1}\) to know the fifth term.

--- > \(a_5 = 2a_{5-1} - x = 99\)

a5= 2*a4 - x = 99

Am i understanding now correctly ?

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Re: The sequence a1, a2, … , a n, … is such that an = 2an-1 - x [#permalink] 08 Mar 2018, 10:30

dave13 wrote:

Hello there niks18 thanks for explanation. highly appreciate. you know i think i am starting to undesrstand some nuances i ve read this post

https://magoosh.com/gmat/2012/sequences-on-the-gmat/

it is about sequences

if a sequence is defined by \(a_n = n^2-n\) for n ≥ 2 what is the difference between 4th and 3rd terms

the same rule applied to the question we discussed right? because in all above posts everyone just plug in 5 (as fifth term) into \({n-1}\) to know the fifth term.

--- > \(a_5 = 2a_{5-1} - x = 99\)

a5= 2*a4 - x = 99

Am i understanding now correctly ?

yes your understanding is correct. A function has a dependent and independent variable. you plug in the independent variable to get the dependent variable.

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Re: The sequence a1, a2, … , a n, … is such that an = 2an-1 - x [#permalink]

08 Mar 2018, 10:30

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