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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 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 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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Scott Woodbury Stewart Founder & CEO
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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

_________________
Brent Hanneson – Founder of GMATPrepNow.com
Use our video course along with Beat The GMAT's free 60-Day Study Guide

Check out the online reviews of our course
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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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