Problem Solving

Joy Shaha wrote:Q. 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 an+2 in terms of t?

A) 4t
B) t2
C) t3
D) t4
E) t8

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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) t²
C) t³
D) t�
E) t�

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) t²
C) t³
D) t�
E) t�

Let's examine some terms:
a1 = 3
a2 = 5
a3 = 15
a4 = 225 = 15²
a5 = 15�
a6 = 15�
a7 = 15¹�
a8 = 15³²
.
.
.

Notice that the exponent of each term is FOUR TIMES the exponent of the term that is two terms before it.

For example, a6 = 15� and a8 = 15³²
Notice that 15³² = (15�)�

Likewise, a5 = 15� and a7 = 15¹�
Notice that 15¹� = (15�)�

So, if an = t, then a(n+2) = t�
Answer: D

Cheers,
Brent

You may find a couple of different approaches here.

https://www.beatthegmat.com/og-17-a-sequ ... 93432.html

-Jay
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Joy Shaha wrote:Q. 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 an+2 in terms of t?
A) 4t B) t2 C) t3 D) t4 E) t8

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

Alternate solution:

Let's list the first five terms of this sequence:

a(1) = 3
a(2) = 5
a(3) = 3 x 5 = 15
a(4) = 3 x 5 x 3 x 5 = 3^2 x 5^2 = 15^2
a(5) = 3 x 5 x 3 x 5 x 3 x 5 x 3 x 5 = 3^4 x 5^4 = 15^4

Recall that a(n) = t and that n > 2. Our goal is to compare the value of a(n) to that of a(n + 2). Let's let n = 3. This means that we will need to compare the value of a(3) to that of a(5).

From the list above, we see that a(3) = 15 = t. Now note that a(5) = 15^4, which we see is also equal to t^4.

Answer: D