Sequence S is the sequence of numbers a1, a2, a3, ... , an. For each

Question

Sequence S is the sequence of numbers  a_1, a_2, a_3,  ... ,  a_n . For each positive integer n, the  n_{th}  number  a_n  is defined by  a_n=\frac{n+1}{3n} . What is the product of the first 53 numbers in sequence S?

A.  \frac{2}{3^{53}}

B.  \frac{2}{3^{50}}

C.  \frac{2}{3^{49}}

D.  \frac{3}{2^{50}}

E.  \frac{2}{3^{25}}

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PyjamaScientist · Accepted Answer

Using,  a_n=\frac{n+1}{3n} ,
  a_1 = \frac{2}{1*3}, a_2 = \frac{3}{2*3}, a_3 = \frac{4}{3*3} .. so on
Now we need a product of first  53  numbers.
Now we can see a pattern here.
Numerators start at  2   for the first term and increase by  +1  so on till  53rd  term. So, the  53rd   terms numerator would be  54 . That would give product of all numerators =  54! .
Similarly, we can see the denominators are continuous  3 's ( 3^{53} ) with correspondingly increasing coefficients that start from  1  for first term and end at  53  for the last  53rd  term (thus  53! ).
Thus, total product becomes:  \frac{54!}{(53!*3^{53})} .
Upon solving you'd get: (B) :  \frac{2}{3^{50}}

Kinshook · Answer

Given: Sequence S is the sequence of numbers  a_1, a_2, a_3,  ... ,  a_n . For each positive integer n, the  n_{th}  number  a_n  is defined by  a_n=\frac{n+1}{3n} .

Asked: What is the product of the first 53 numbers in sequence S?

a_1 = \frac{2}{3} 
 a_2 = \frac{3}{6} 
 a_3 = \frac{4}{9} 
..
 a_n = \frac{n+1}{3n} 
..
 a_53 = \frac{54}{3*53}

Product of first 53 terms =  \frac{54!}{3^{53}*53!} = \frac{54}{3^{53}} = \frac{3^3*2}{3^{53}} = \frac{2}{3^{50}}

IMO B

BrushMyQuant · Answer

Given that  a_n=\frac{n+1}{3n}  and we need to find the product of the first 53 numbers in sequence S

a_1=\frac{1+1}{3*1}  =  \frac{2}{3*1} 
 a_2=\frac{2+1}{3*2}  =  \frac{3}{3*2} 
 a_3=\frac{3+1}{3*3}  =  \frac{4}{3*3} 
 a_4=\frac{4+1}{3*4}  =  \frac{5}{3*4}

So, Product of first 53 terms will be  \frac{2}{3*1}* \frac{3}{3*2} * \frac{4}{3*3}*....*\frac{53}{3*52} * \frac{54}{3*53} 
So, All numbers in numerator will cancel out apart from 54 and all numbers in denominator which are getting multiplied by 3 will cancel out

=> We will be left with  \frac{54}{3*3*....*3}  (3 in denominator will be 53 times one for each of the 53 terms)

=  \frac{54}{3^53}  =  \frac{27*2}{3^53}  =  \frac{3^3*2}{3^53}  =  \frac{2}{3^50}

So,  Answer will be B 
Hope it helps!

Watch the following video to learn How to Sequence problems

https://www.youtube.com/watch?v=EfNT2JoE5CM

SaquibHGMATWhiz · Answer

Solution:

We know  a_n=\frac{n+1}{3n}

We need the value of  a_1	imes a_2	imes a_3	imes ....	imes a_{51}	imes a_{52}	imes a_{53}

⇒\frac{2}{3}	imes \frac{3}{6}	imes \frac{4}{9}......	imes \frac{52}{153}	imes \frac{53}{156}	imes\frac{ 54}{159}

⇒\frac{54!}{3	imes 6	imes 9....	imes 153	imes 156	imes 159}

⇒\frac{54!}{(3	imes 3	imes ...)(2	imes 3....	imes 51	imes 52	imes 53)}

⇒\frac{54!}{3^{53}	imes 53!}

⇒\frac{54	imes 53!}{3^{53}	imes 53!}

⇒\frac{54}{3^{53}}

⇒\frac{54}{3^3	imes 3^{50}}

⇒\frac{2}{3^{50}}

Hence the right answer is    Option B

bumpbot · Answer

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Sequence S is the sequence of numbers a1, a2, a3, ... , an. For each

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