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# Sequence P is defined by pn = p(n-1)+ 3, p1= 11, sequence Q is defined

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Sequence P is defined by pn = p(n-1)+ 3, p1= 11, sequence Q is defined  [#permalink]

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17 Sep 2016, 10:46
6
00:00

Difficulty:

95% (hard)

Question Stats:

51% (03:11) correct 49% (02:51) wrong based on 85 sessions

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Sequence P is defined by $$p_n = p_{n-1}+ 3$$, $$p_1= 11$$, sequence Q is defined as $$q_n=q_{n-1}– 4$$, $$q_3= 103$$. If $$p_k > q_{k+2}$$, what is the smallest value k can take?

A. 6
B. 9
C. 11
D. 14
E. 15

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Re: Sequence P is defined by pn = p(n-1)+ 3, p1= 11, sequence Q is defined  [#permalink]

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17 Sep 2016, 20:32
GMATantidote wrote:
Sequence $$P$$ is defined by $$p_n$$ = $$p_{n-1}$$ $$+ 3$$, $$p_1$$ $$= 11$$, sequence $$Q$$ is defined as $$q_n$$ = $$q_{n-1}$$ $$– 4$$, $$q_3$$ $$= 103$$. If $$p_k$$ > $$q_{k+2}$$, what is the smallest value $$k$$ can take?

A. 6
B. 9
C. 11
D. 14
E. 15

Set P = {11, 14, 17,.....}
Set Q = {110, 107, 103,....}

For Arithmetic progression $$a(n) = a(1) + (n-1) d$$
Started with C) 11 does not satisfy $$p_k$$ > $$q_{k+2}$$
and D) 14, $$p_k$$ = $$q_{k+2}$$ So answer is E, no need to calculate.
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Re: Sequence P is defined by pn = p(n-1)+ 3, p1= 11, sequence Q is defined  [#permalink]

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12 Oct 2016, 19:24
GMATantidote wrote:
Sequence P is defined by $$p_n = p_{n-1}+ 3$$, $$p_1= 11$$, sequence Q is defined as $$q_n=q_{n-1}– 4$$, $$q_3= 103$$. If $$p_k > q_{k+2}$$, what is the smallest value k can take?

A. 6
B. 9
C. 11
D. 14
E. 15

I am really struggling trying to obtain the correct answer. I can't seem to set up the math even with the GMAT Math Book. Could someone possibly detail it out, nothing too in-depth but enough to show why 15 is correct? I have Pn = a1 + (n-1)d but don't know how to apply the darn formula...

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Sequence P is defined by pn = p(n-1)+ 3, p1= 11, sequence Q is defined  [#permalink]

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16 Oct 2016, 12:11
mikeyg51 wrote:
GMATantidote wrote:
Sequence P is defined by $$p_n = p_{n-1}+ 3$$, $$p_1= 11$$, sequence Q is defined as $$q_n=q_{n-1}– 4$$, $$q_3= 103$$. If $$p_k > q_{k+2}$$, what is the smallest value k can take?

A. 6
B. 9
C. 11
D. 14
E. 15

I am really struggling trying to obtain the correct answer. I can't seem to set up the math even with the GMAT Math Book. Could someone possibly detail it out, nothing too in-depth but enough to show why 15 is correct? I have Pn = a1 + (n-1)d but don't know how to apply the darn formula...

Sequence P has a common difference of 3 and is (11, 14, 17......)

Sequence Q has a common difference of -4 and is (111, 107, 103.....)

Nth term of any arithmetic sequence can be written as a(n) = a1 + (n-1)d where a1 is first term and d is common difference

Taking values from options starting with C k = 11

That gives P(11) = P1 + (11-1)3 = 11 + 10*3 = 41

Q(13) = Q1 + (13-1)*(-4) = 111 + 12*(-4) = 111 - 48 = 63

P(11) < Q(13) so it doesn't help.

We need to try the same with 14.

P(14) = P1 + (14-1)*3 = 11 + 13*3 = 11 + 39 = 50

Q(16) = Q1 + (16-1)*(-4) = 111 + 15*(-4) = 111 - 60 = 51

Still P(14) < Q(16) so this can't be the value of K.

You can either solve for K = 15 and check or mark E as the answer as that's the only value greater than 14.

HTH
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Re: Sequence P is defined by pn = p(n-1)+ 3, p1= 11, sequence Q is defined  [#permalink]

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21 Dec 2017, 22:47
2
algebraic approach:
Q3 = 103 => Q1 = 111

Pn : 11 + (n - 1)3 , n >1
Qn : 111 - (n - 1)4, n > 1

to find min value of k such that Pk > Q(k+2)
11 + (k-1) * 3 > 111 - (k+2-1)*4
=> 11 + 3k - 3 > 111 - 4k - 4
=> 7k > 99 => k > 14 => Answer (E), k = 15

Kudos if you like my approach, need to unlock my Gmat Club Tests
Re: Sequence P is defined by pn = p(n-1)+ 3, p1= 11, sequence Q is defined   [#permalink] 21 Dec 2017, 22:47
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