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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Here are a couple of ways to think about this problem:

OG 17 q201 Method 1:



OG 17 q201 Method 2:

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

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 = \(D\)

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

Bunuel - can I call the above mentioned problem as a recursive sequence ? I know this formula An= n1+(n-1)*d but could not apply it here.

Hello hazelnut - how did you figure out that A_n+2 is the fifth term and not sixth ?

let n = 3. \(a_3 = a_n = a_1*a_2 = t\)
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an = t = .......(an - 1)

(an + 1) = .......(an - 1) * an = t^2

(an + 2) = .....(an - 1) * an * (an + 1)

=) t^2 * t ^2 = t^4 = D the answer

thanks

Hello pushpitkc, hope you are having an awesome gmat weekend

i have a few questons regarding the above solution, so let me break it down in into following clauses

1.) it says "Say n = 3 [given n>2]. Hence we must find value of a5" my question what does \(n\) mean is it the last term ? based on which rule do we concude " If n = 3 [given n>2]. Hence we must find value of \(a_5\) i mean why if n = 3 then we need to find \(a_5\) and not for example \(a_6\) ?


2.) if \(a_3 = a_1*a_2 = 3*5 = 15\), then to find \(a_4\) we need to perform following \(a_1*a_2*a_3 = 3*5*15 = 225\)

and following this logic in order to find \(a_5\)we need to do this \(a_1*a_2*a_3*a_4 = 3*5*15*225 = 3375\)

i cant understand why in the solution above to find \(a_5\) we multiply by the same numbers \(a_5= 15*15*15*15 = t^4\), to find the next term shoudlnt we mupltiply by all previous terms

3.) Also based on this solution \(a_5= 15*15*15*15 = t^4\) why the correct answer is \(t^4\) because there are FOUR numbers 15 ? but there could be infinite NUMBERS of 15

And the last question is it geometric sequence question ?

many thanks for taking time to explain

enjoy the weekend

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 .

Hey EMPOWERMathExpert i have been waiting for your explanation for many years:-) what took you so long welcome to Club of forever students

Answer:
\(a_{n+2}\) = \(a_{n+1}\) * \(a_{n}\) * \(a_{n-1}\) * \(a_{n-2}\) _ _ _ _ \(a_{1}\)

now also we can write \(a_{n+1}\) = \(a_{n}\) * \(a_{n-1}\) * \(a_{n-2}\) _ _ _ _ \(a_{1}\), so substituting this in the above equation

\(a_{n+2}\) = (\(a_{n}\) * \(a_{n-1}\) * \(a_{n-2}\) _ _ _ _ \(a_{1}\)) ^ 2

again, \(a_{n}\) = \(a_{n-1}\) * \(a_{n-2}\) _ _ _ _ \(a_{1}\) , substituting above

\(a_{n+2}\) = ((\(a_{n}\) * \(a_{n}\))) ^ 2

we are given that \(a_n =t\), so substituting above

\(a_{n+2}\) = (t^2)^2 = t^4

So, option D is correct

I don't think plugging in numbers would be necessarily faster. Here is my solution, which in fairness should be much easier take on (if you are OK with handling series)

Forget about a1=3 and a2=5, since it mentions n>2 anyways.

\(a_3 = a_1.a_2\)
\(a_4 = a_1.a_2.a_3\)

Therefore, we can write
\(a_n = a_1.a_2.a_3......a_{n-1} = t\)

\(a_{n+2} = a_1.a_2.a_3....a_{n-1}.a_n.a_{n+1}\)

\(a_{n+2} = t.a_n.a_{n+1}\)

\(a_{n+2} = t.t.a_{n+1} = t^2.a_{n+1}\) ....................equation (1)

Now we need to find out what \(a_{n+1}\) is..
\(a_{n+1} = a_1.a_2.a_3......a_{n-1}.a_n\)
\(a_{n+1} = t.a_n\)
\(a_{n+1} = t.t = t^2\)

Substituting the value of \(a_{n+1}\) back in equation (1)
\(a_{n+2} = t^2.t^2 = t^4\)