Latest TCS Aptitude Question SOLUTION: The sum of first n terms in a sequence is always 1/n.Find the product of the first 2014 terms (n! denotes the factorial of n,that is the product of the numbers(1,2……n)) a)-1/(

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The sum of first n terms in a sequence is always 1/n.Find the product of the first 2014 terms (n! denotes the factorial of n,that is the product of the numbers(1,2……n))

a)-1/(2013!2013!)

b) 1/(2013!2013!)

c) 1/(2013!2014!)

d)- 1/(2013!2014!)

Asked In TCS milan jena (8 years ago)

Unsolved Read Solution (4)

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Read Solution (4) :

Shashank Sourabh Singh  (8 years ago)

we solve it by generalized example
like we have a series like: 1 -1/2 -1/6 -1/12 we take 4 term
so it is clear that if term is even then ans is negative
multiply= -1/144 that means -1/ 3!*4!
means answer d)- 1/(2013!2014!) is correct

Is this solution Helpfull? Yes (14) | No (6)  

Sayan De  (8 years ago)

as per the ques sum of 1 term is 1/1=1,2 terms=1/2,3 terms=1/3 and so on. now from here we can find the 2nd term ,3rd term which are -1/2, -1/6 and so on. Now as Shashank has correctly stated the product thus becomes (-1)^n-1/n!(n-1)! so the correct answer is d.

Is this solution Helpfull? Yes (8) | No (3)  

minu sen  (8 years ago)

cannot understand plz explain....

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SRIJAN SEAL  (6 years ago)

Sn = 1/n
Let, Ti denote the ith term, Si denote the sum upto ith term.
S1=T1=1/1=1.

S2=1/2
hence, T2=S2-S1=1/2 - 1 = -1/2

S3=1/3
hence, T3=S3-S2= 1/3 - 1/2 = -1/6

similarly, T4= -1/12 and so on.

hence, for 4 terms, product = (1) (-1/2) (-1/6) (-1/12) = (-1/144) = -1/(3! * 4!)

hence by observing, option (d) -1/(2013! * 2014!) is correct.

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