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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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 = \(D\)

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

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

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

Say n = 3 [given n>2]. Hence we must find value of a5

a1 = 3,
a2 = 5,
a3 = 5*3 = 15 = t [an = t given]
a4 = 15*15
a5= 15*15*15*15 = t^4
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\(a_3 = a_n = a_1*a_2 = t\)
\(a_4 = a_{n + 1} = a_1*a_2*a_3 = t*a_3 = t*t = t^2\)
\(a_5 = a_{n + 2} = a_1*a_2*a_3*a_4 = t*t*(t^2) = t^4\)
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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
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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.

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 .

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

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

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

GMAT Assassins aren’t born, they’re made,
Rich
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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
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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
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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.

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
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