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

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fiza gupta Master | Next Rank: 500 Posts
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OG'17 - A sequence of numbers a1a1, a2a2, a3a3,

Post Wed Nov 16, 2016 11:19 am
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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Post 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) = 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

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Post 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

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Post 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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crackverbal Master | Next Rank: 500 Posts
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Post 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 

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Anaira Mitch Master | Next Rank: 500 Posts
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Post 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.

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