A sequence of numbers a1, a2, a3,…. is defined as follows: a1 = 3, a2

Question

A sequence of numbers  a_1 ,  a_2 ,  a_3 ,.... is defined as follows:  a_1 = 3 ,  a_2 = 5 , and every term in the sequence after  a_2  is the product of all terms in the sequence preceding it, e.g,  a_3 = (a_1)(a_2)  and  a4 = (a_1)(a_2)(a_3) . If  a_n =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

BrentGMATPrepNow · Accepted Answer

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 ) =   15⁴  
term6 = (term5)(term4)(term3)(term2)(term1) = (  15⁴  )( 15² )( 15 )( 5 )( 3 ) =   15⁸

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 = t⁴

Answer: D

Cheers,
Brent

JD01 · Answer

Given  a_n = t 
This means  a_1 * a_2 * a_3* ..... a_{n-1} = t

Therefore  a_{n+1} = (a_1 * a_2 * a_3* ..... a_{n-1}) * (a_n) = t*t = t^2

and  a_{n+2} = (a_1 * a_2 * a_3* ..... a_{n-1}) * (a_n) * (a_{n+1}) = t*t*t^2 = t^4

Answer =  D

Sash143 · Answer

Say n = 3  . Hence we must find value of a5

a1 = 3, 
a2 = 5,
a3 = 5*3 = 15 = t  
a4 = 15*15
a5= 15*15*15*15 = t^4

varundixitmro2512 · Answer

IMO D

An=t so that means product from A1 to An-1 is t. therefore An+1 = A1x.....An-1 x An=txt= t^2
proceeding in same way An+2 will be t^4

Kurtosis · Answer

An = (A1)(A2)......(An-1) = t
An+1 = t * An = t^2
An+2 = t * An * An+1 = t * t * t^2 = t^4

Answer: D

adiagr · Answer

a_n  =  (a_1)(a_2)........(a_{n-1} )  =  t

a_{n+1}   =  (a_1)(a_2)........(a_n)  =   (a_1)(a_2)........(a_{n-1})  *  a_n   = t *t =   t^2

a_{n+2}  =  (a_1)(a_2)........(a_{n+1})  =   (a_1)(a_2)........(a_{n-1})   *   a_n   *  a_{n+1}

= t * t *  t^2  =   t^4  
 
D is the answer.

BillyZ · Answer

a_3 = a_n = a_1*a_2 = t 
 a_4 = a_{n + 1} = a_1*a_2*a_3 = t*a_3 = t*t = t^2 
 a_5 = a_{n + 2} = a_1*a_2*a_3*a_4 = t*t*(t^2) = t^4

JeffTargetTestPrep · Answer

We are given a sequence in which every term in the sequence after a(2) 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

shm111 · Answer

Observe/Create a sequence:
A3 = A1*A2
A4 = A3*A1*A2 = A3*A3
A5 = A4*A4
.
.
.
An = t
An+1 = An*An = t*t = t^2
An+2 = An+1*An+1 = t^2*t^2 = t^4

D.

EMPOWERMathExpert · Answer

For problems like this, it’s best that we observe the first few terms of the sequence to see if there’s any noticeable pattern among the results.

Since it’s given that  a_1=3  and  a_2=5 , this means that

a_3=(a_1 )(a_2 )=3(5)=15=15^1 
 a_4=(a_1 )(a_2 )(a_3 )=(3)(5)(15)=225=15^2 
 a_5=(a_1 )(a_2 )(a_3 )(a_4 )=(3)(5)(15)(15^2 )= 15^4 =(15^2 )^2 
 a_6=(3)(5)(15)(15^2 )(15^4 )= 15^8=(15^4 )^2

As can be seen, as the sequence progresses, the next term would be the square of the term before it.

So, if we start at  a_n=t , the next term would be  t^2  and term after that would be  (t^2 )^2=t^4 .

a_n=t 
 a_{(n+1)}=t^2 
 a_{(n+2 )}=(t^2 )^2=t^4

Hence, the final answer is  B .

newdimension · Answer

BrentGMATPrepNow  would this pattern hold for all problems with similar statements? The value would only change depending on a1 and a2.

BrentGMATPrepNow · Answer

That's correct. 
If a1 = x and a2 = y, then....

a3 = xy
a4 = (xy)^2
a5 = (xy)^4
a6 = (xy)^8
.
.
.
etc

MHIKER · Answer

The shorter is to understand the logic.
 a_1 = 3

a_2 = 5

a_3 = 15 , Given that  a_n =t  and  n > 2 , we see 3>2, so can write  a_3 = a_n=t

∴ a_3=a_n=15=t

a_4=a_{n+1}=15*5*3=15*15=t^2

a_5=a_{n+2}=15*15*15*5*3=15*15*15*15=t^4

Thus,  a_{n+2}=t^4

The answer is D

EMPOWERgmatRichC · Answer

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

Kinshook · Answer

Given: A sequence of numbers  a_1 ,  a_2 ,  a_3 ,…. is defined as follows:  a_1 = 3 ,  a_2 = 5 , and every term in the sequence after  a_2  is the product of all terms in the sequence preceding it, e.g,  a_3 = (a_1)(a_2)  and  a4 = (a_1)(a_2)(a_3) .
Asked: If  a_n =t  and  n > 2 , what is the value of  a_{n+2}  in terms of t?

a_n = t = (a_1)(a_2)(a_3)..(a_{n-1}) 
 a_{n+1} = (a_1)(a_2)(a_3)..(a_{n-1})(a_n) = t*t = t^2 
 a_{n+2} = (a_1)(a_2)(a_3)..(a_{n-1})(a_n)(a_{n+1}) = t^2*t^2 = t^4

IMO D

avigutman · Answer

Video solution from Quant Reasoning: Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1 https://www.youtube.com/watch?v=4h_emTrzFro

CrackverbalGMAT · Answer

lets analyze the example given
 a1= 3, a2 =5

a3= 3*5 = 15 
 a4= 3*5 * 15= a3*a3 = a3^2 = 225 
 a5= 3 * 5 * 15 * 225 = a4 * a4 = a3^2 * a3^2 = a3 ^4

Similarly,
an=t
 an+1 = an* an= t*t = t^2 
 an+2 = an+1 * an+1= t^2 * t^2 = t ^ 4

Option D is the correct answer

Thanks,
Clifin J Francis,
GMAT SME

TarunKumar1234 · Answer

As a_n = a_1*a_2*a_3*.....*a_{n-1} = t So, a_{n+2} = t*a_n*a_{n+1} =t * t* t* a_n = t^4 I think D.

PolarisTestPrep · Answer

Hey guys,

If a1 = 3 and a2 = 5,

a3 = a1 * a2 = 3 * 5 = 15

Let's say 15 = x

a3 = x

a4 = a1 * a2 * a3 = 3 * 5 * 15 = 15 * 15 = x * x =  x^2

a5 = a1 * a2 * a3 * a4 = 3 * 5 * 15 * (15 * 15) = 15 * 15 * 15 * 15 =  15^4  =  x^4

If a(n) = t

Let's say y * z = t

In that case a(n) = y * z = t

a(n+1) = (y * z) * a(n) = t * t

a(n + 2) = (y * z) * a(n) * a(n+1) = (y * z) * t * (t * t) = t * t * t * t =  t^4

The answer is D

A sequence of numbers a1, a2, a3,…. is defined as follows: a1 = 3, a2

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A sequence of numbers a1, a2, a3,…. is defined as follows: a1 = 3, a2

Prep Toolkit

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My Rewards