BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

OG'17 - A sequence of numbers a1a1, a2a2, a3a3,

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Master | Next Rank: 500 Posts
Posts: 216
Joined: Sun Jul 31, 2016 9:55 pm
Location: Punjab
Thanked: 31 times
Followed by:7 members

OG'17 - A sequence of numbers a1a1, a2a2, a3a3,

by fiza gupta » Wed Nov 16, 2016 11:19 am

Timer

00:00

Your Answer

A

B

C

D

E

Global Stats

A sequence of numbers a1,a2,a3,.... is defined as follows: a1 = 3, a2 = 5, and every term in the sequence after a2 is the product of all terms in the sequence preceding it, e.g, a3=(a1)(a2) and a4=(a1)(a2)(a3). If an=t and n>2, what is the value of a(n+2) in terms of t?

A) 4t
B) t^2
C) t^3
D) t^4
E) t^8

OA:D
Fiza Gupta
Top
Source: — Problem Solving |
User avatar
Legendary Member
Posts: 2663
Joined: Wed Jan 14, 2015 8:25 am
Location: Boston, MA
Thanked: 1153 times
Followed by:128 members
GMAT Score:770

by DavidG@VeritasPrep » Wed Nov 16, 2016 11:51 am
fiza gupta wrote:A sequence of numbers a1,a2,a3,.... is defined as follows: a1 = 3, a2 = 5, and every term in the sequence after a2 is the product of all terms in the sequence preceding it, e.g, a3=(a1)(a2) and a4=(a1)(a2)(a3). If an=t and n>2, what is the value of a(n+2) in terms of t?

A) 4t
B) t^2
C) t^3
D) t^4
E) t^8

OA:D
Say n = 3. A3 = 3*5 = 15, so A3 = t = 15.

We also know that A4 = 3*5*15 = 15*15 (the product of the three previous terms.

To summarize: A1 = 3, A2 = 5; A3 =15 and A4 =15*15

If n = 3, n + 2 would be 5. A5= 3 * 5 * 15 *(15*15); combine the 3 and the 5 to get 15*15*15*15 = 15^4

So if A3 = t = 15, and A5 = 15^4, then the answer is D
Veritas Prep | GMAT Instructor

Veritas Prep Reviews
Save $100 off any live Veritas Prep GMAT Course
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 157
Joined: Mon Aug 16, 2010 7:30 pm
Location: India
Thanked: 65 times
Followed by:3 members

by crackverbal » Wed Nov 16, 2016 11:58 pm
Hi Fiza Gupta,

This is a best question to apply "plugging in" the values and solving the question.

Since plugging in method has already been explained above,

I will give an alternative approach for this question.

We just need to slightly think here,

Given an=t

Which is nothing but, product of all the terms preceding it,

So

a1∗a2∗a3∗.....an−1= t,

In that case an+1 will be,

an+1=(a1∗a2∗a3∗.....an−1)∗(an)=t∗t=t^2

and an+2=(a1∗a2∗a3∗.....an−1)∗(an)∗(an+1)=t∗t∗t^2=t^4

So the answer is D.

Hope this helps �
Join Free 4 part MBA Through GMAT Video Training Series here -
https://gmat.crackverbal.com/mba-throug ... video-2018

Enroll for our GMAT Trial Course here -
https://gmatonline.crackverbal.com/

For more info on GMAT and MBA, follow us on @AskCrackVerbal
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 235
Joined: Wed Oct 26, 2016 9:21 pm
Thanked: 3 times
Followed by:5 members

by Anaira Mitch » Thu Nov 17, 2016 3:42 am
Given an = t
This means a1*a2*a3*.....an−1=t

Therefore an+1=(a1*a2*a3*.....an−1)*(an)= t*t= t^2

and an+2= (a1*a2*a3*.....an−1)*(an)*(an+1) = t*t*t^2 = t^4
Similarly an+3 = (a1*a2*a3*.....an−1)*(an)*(an+1)*an+2 = t*t*t^2*t^4 = t^8

Answer = D

I find above solution more helpful.
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 8085
Joined: Sat Apr 25, 2015 10:56 am
Location: Los Angeles, CA
Thanked: 43 times
Followed by:29 members

by Scott@TargetTestPrep » Fri Nov 18, 2016 7:14 am
fiza gupta wrote:A sequence of numbers a1,a2,a3,.... is defined as follows: a1 = 3, a2 = 5, and every term in the sequence after a2 is the product of all terms in the sequence preceding it, e.g, a3=(a1)(a2) and a4=(a1)(a2)(a3). If an=t and n>2, what is the value of a(n+2) in terms of t?

A) 4t
B) t^2
C) t^3
D) t^4
E) t^8

OA:D
We are given a sequence in which every term in the sequence after a2 is the product of all terms in the sequence preceding it. So:

a(n+1) = a(n) x a(n-1) x ... x a(2) x a(1)

By the same reasoning, we have:

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

We can substitute a(n-1) x... x a(2) x a(1) in the a(n+1) equation for a(n), so we have a(n+1) = a(n) x a(n).

However, recall that a(n) = t, so a(n+1) = t x t = t^2. By the same reasoning, we have:

a(n+2) = a(n+1) x a(n) x a(n-1) x ... x a(2) x a(1)

However, a(n) x a(n-1) x .... x a(2) x a(1) = a(n+1) and a(n+1) = t^2, so:

a(n+2) = a(n+1) x a(n+1) = t^2 x t^2 = t^4

Answer: D

Scott Woodbury-Stewart
Founder and CEO
[email protected]

Image

See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews

ImageImage
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pm
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1268 members
GMAT Score:770

by Brent@GMATPrepNow » Wed Nov 15, 2017 12:49 pm
fiza gupta wrote:A sequence of numbers a1,a2,a3,.... is defined as follows: a1 = 3, a2 = 5, and every term in the sequence after a2 is the product of all terms in the sequence preceding it, e.g, a3=(a1)(a2) and a4=(a1)(a2)(a3). If an=t and n>2, what is the value of a(n+2) in terms of t?

A) 4t
B) t^2
C) t^3
D) t^4
E) t^8

OA:D
Let's list a few terms....
term1 = 3
term2 = 5
term3 = (term2)(term1) = (5)(3) = 15 (term2)(term1)
term4 = (term3)(term2)(term1) = (15)(5)(3) = 15²
term5 = (term4)(term3)(term2)(term1) = (15²)(15)(5)(3) = 15�
term6 = (term5)(term4)(term3)(term2)(term1) = (15�)(15²)(15)(5)(3) = 15�

At this point, we can see the pattern.

Continuing, we get....
term7 = 15^16
term8 = 15^32

Each term in the sequence is equal to the SQUARE of term before it

If term_n =t and n > 2, what is the value of term_n+2 in terms of t?
So, term_n = t
term_n+1 = t²
term_n+2 = t�

Answer: D

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top

GMAT/MBA Expert

User avatar
Elite Legendary Member
Posts: 10392
Joined: Sun Jun 23, 2013 6:38 pm
Location: Palo Alto, CA
Thanked: 2867 times
Followed by:511 members
GMAT Score:800

Re: OG'17 - A sequence of numbers a1a1, a2a2, a3a3,

by [email protected] » Sun Apr 25, 2021 9:14 pm
Hi All,

While this is a wordy sequence question, the information is given to us in a logical order, so we just have to get the information on the pad and follow the ‘instructions’ of the sequence to answer the specific question that is asked.

We’re told the first two terms of a sequence:
1st term = 3
2nd term = 5

and then we’re told that each term after the 2nd term is the PRODUCT of ALL of the terms that came before it…
3rd term = (3)(5) = 15
4th term = (3)(5)(15) = (15)(15) = 225
5th term = (3)(5)(15)(225) = (225)(225)
Etc.

We’re told that the Nth term = T and N > 2. We’re asked for the value of the (N+2)th term in this sequence IN TERMS OF T. Now that we know how the sequence works, we can use TEST IT to get to the correct answer.

IF…. N = 3, then we already know that the 3rd term = T = (3)(5) = 15. We’re asked for the value of the (3+2) = 5th term, which we know is (225)^2.

225 = (15)(15); by extension… (225)^2 = (15)(15)(15)(15) = 15^4. The value of T is 15, so 15^4 is the equivalent of T^4.

Final Answer: D

GMAT Assassins aren’t born, they’re made,
Rich
Contact Rich at [email protected]
Image
Top
Post new topic Post Reply
• Page 1 of 1