GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 17 Jun 2019, 00:51

### GMAT 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

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# A certain series is defined by the following recursive rule: Sn=K(Sn-1

Author Message
TAGS:

### Hide Tags

Manager
Joined: 16 May 2011
Posts: 172
Concentration: Finance, Real Estate
GMAT Date: 12-27-2011
WE: Law (Law)
A certain series is defined by the following recursive rule: Sn=K(Sn-1  [#permalink]

### Show Tags

06 Jun 2011, 14:02
1
14
00:00

Difficulty:

55% (hard)

Question Stats:

67% (02:41) correct 33% (02:40) wrong based on 202 sessions

### HideShow timer Statistics

A certain series is defined by the following recursive rule: Sn=K(Sn-1) , where k is a constant. If the 1st term of this series is 64 and the 25th term is 192, wha is the 9th term?

A. ROOT 2
B. ROOT 3
C. 64*ROOT 3
D. 64*3^1/3
E. 64*3^24
Retired Moderator
Joined: 20 Dec 2010
Posts: 1748
Re: A certain series is defined by the following recursive rule: Sn=K(Sn-1  [#permalink]

### Show Tags

06 Jun 2011, 15:20
2
4
dimri10 wrote:
:lol: A certain series is defined by the following recursive rule: Sn=K(Sn-1) , where k is a constant. If the 1st term of this series is 64 and the 25th term is 192, wha is the 9th term?
1. ROOT 2
2. ROOT 3
3. 64*ROOT 3
4> 64*3^1/3
5.64*3^24

For GP:
$$A_n=A_1*r^{(n-1)}$$

Here, r=k
$$A_{25}=A_1*k^{(25-1)}=A_1*k^{24}$$

$$192=64*k^{24}$$

$$192=64*(k^{8})^{3}$$

$$(k^8)^3=\frac{192}{64}=3$$

Taking cube-root on both sides,
$$k^8=\sqrt[3]{3}$$

Now,
$$A_9=A_1*k^8$$

$$A_9=64\sqrt[3]{3}$$

Ans: "D"
_________________
##### General Discussion
Intern
Joined: 24 Feb 2011
Posts: 28
Re: A certain series is defined by the following recursive rule: Sn=K(Sn-1  [#permalink]

### Show Tags

06 Jun 2011, 18:52
1
Fluke what does GP stand for? And what is the name of the formula you are referencing?
Senior Manager
Joined: 24 Mar 2011
Posts: 335
Location: Texas
Re: A certain series is defined by the following recursive rule: Sn=K(Sn-1  [#permalink]

### Show Tags

06 Jun 2011, 21:06
Sn = k Sn-1

S25 = 192 = k S24
= k^24 * 64
And S1 = 64 = k S0

==> k^24 = 192/64 = 3.. (1)

Now, S9 = k^8 * 64
From (1), (k^8)^3 = 3
k^8 = 3^\frac{1}{3}
S9 = 3^\frac{1}{3} * 64

Manager
Joined: 19 Apr 2011
Posts: 82
Re: A certain series is defined by the following recursive rule: Sn=K(Sn-1  [#permalink]

### Show Tags

07 Jun 2011, 00:32
$$S1 = xk = 64 S25 = xk^25 = 192 xkk^24 = 192 k^24 = 192/64 = 3 k = 3^(1/24) S9 = xk^9=xkk^8 = 64*(3^(1/24*8) = 64*(3^(1/3))$$
Retired Moderator
Joined: 20 Dec 2010
Posts: 1748
Re: A certain series is defined by the following recursive rule: Sn=K(Sn-1  [#permalink]

### Show Tags

07 Jun 2011, 01:01
olivite wrote:
Fluke what does GP stand for? And what is the name of the formula you are referencing?

GP is Geometric Progression, where the ratio between two consecutive terms is always same.

According to the question, this series is a Geometric Series.

As,
$$A_2=A_1*k \hspace{3} OR \hspace{3} \frac{A_2}{A_1}=k$$
$$A_3=A_2*k \hspace{3} OR \hspace{3} \frac{A_3}{A_2}=k$$
$$A_4=A_3*k \hspace{3} OR \hspace{3} \frac{A_4}{A_3}=k$$
$$A_5=A_4*k \hspace{3} OR \hspace{3} \frac{A_5}{A_4}=k$$

For such series, the $$n^{th}$$ term can be found using following formula:

$$A_n=A_1*k^{(n-1)}$$

Where,
$$k=$$ratio between two consecutive terms
$$A_1=$$first term of the series
$$n=$$index of the term we are trying to find.

Thus,
$$25^{th}$$ term of the series, $$A_{25}=A_1*k^{(25-1)}$$
$$9^{th}$$ term of the series, $$A_{9}=A_1*k^{(9-1)}$$
_________________
Manager
Joined: 08 Sep 2010
Posts: 114
Re: A certain series is defined by the following recursive rule: Sn=K(Sn-1  [#permalink]

### Show Tags

11 Jun 2011, 10:55
Ans...D

No need for any GP formula here
The rule is that nth term is K times the (n-1)th term.

1st = 64
2nd = k.64
3rd = k^2.64
.
.
.
9th term = k^8 *64
.
.
.
so 25th = k^24*64

Using this solve for k and substitute k in the equation for the 9th term
_________________
My will shall shape the future. Whether I fail or succeed shall be no man's doing but my own.

If you like my explanations award kudos.
Director
Joined: 01 Feb 2011
Posts: 637
Re: A certain series is defined by the following recursive rule: Sn=K(Sn-1  [#permalink]

### Show Tags

11 Jun 2011, 11:05
s(n) = k s(n-1)

=> its GP = a + ar + ar^2.....+ar^(n-1)

first term = 64 = a
nth term in a GP is given by ar^(n-1)

25th term is 192 = 64* r^24 => r^24 = 3 => r = 3^(1/24)

9th term = ar^8 = 64* 3^(8/24) = 64*3^(1/3)

Intern
Joined: 28 Aug 2018
Posts: 27
Location: India
Schools: LBS '21 (A)
GMAT 1: 650 Q49 V31
GPA: 3.16
Re: A certain series is defined by the following recursive rule: Sn=K(Sn-1  [#permalink]

### Show Tags

15 Nov 2018, 00:19
S1 = 64. We have to find out S9 which is equal to k^(8)*64

S25 = k^(24)*S1
Therefore, 192 = k^(24)*64
k = 3^(1/24)

S9 = 3^(8/24)*64 which is 3^(1/3)*64. Option D
Senior Manager
Joined: 12 Sep 2017
Posts: 263
Re: A certain series is defined by the following recursive rule: Sn=K(Sn-1  [#permalink]

### Show Tags

27 Jan 2019, 13:20
dimri10 wrote:
A certain series is defined by the following recursive rule: Sn=K(Sn-1) , where k is a constant. If the 1st term of this series is 64 and the 25th term is 192, wha is the 9th term?

A. ROOT 2
B. ROOT 3
C. 64*ROOT 3
D. 64*3^1/3
E. 64*3^24

How do we know that we are talking about a GP?

Kind regards!
Re: A certain series is defined by the following recursive rule: Sn=K(Sn-1   [#permalink] 27 Jan 2019, 13:20
Display posts from previous: Sort by