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

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

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

00:00

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
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by [email protected] » 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
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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

a1âˆ—a2âˆ—a3âˆ—.....anâˆ’1= t,

In that case an+1 will be,

an+1=(a1âˆ—a2âˆ—a3âˆ—.....anâˆ’1)âˆ—(an)=tâˆ—t=t^2

and an+2=(a1âˆ—a2âˆ—a3âˆ—.....anâˆ’1)âˆ—(an)âˆ—(an+1)=tâˆ—tâˆ—t^2=t^4

Hope this helps ï�Š
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by Anaira Mitch » Thu Nov 17, 2016 3:42 am
Given an = t
This means a1*a2*a3*.....anâˆ’1=t

Therefore an+1=(a1*a2*a3*.....anâˆ’1)*(an)= t*t= t^2

and an+2= (a1*a2*a3*.....anâˆ’1)*(an)*(an+1) = t*t*t^2 = t^4
Similarly an+3 = (a1*a2*a3*.....anâˆ’1)*(an)*(an+1)*an+2 = t*t*t^2*t^4 = t^8

I find above solution more helpful.

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by [email protected] » 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

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by [email protected] » 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) = 15Â²
term5 = (term4)(term3)(term2)(term1) = (15Â²)(15)(5)(3) = 15â�´
term6 = (term5)(term4)(term3)(term2)(term1) = (15â�´)(15Â²)(15)(5)(3) = 15â�¸

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 = tÂ²
term_n+2 = tâ�´

Cheers,
Brent

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### 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.