Aptitude - Numbers - Discussion

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"Act well your part; there all honor lies."

- Alexander Pope

57.

1 - 1 + 1 - 2 + 1 - 3 + ... up to n terms = ?

n n n

[A].

1	n
2

[B].

1	(n - 1)
2

[C].

1	n(n - 1)
2

[D].

None of these

Answer: Option D

Explanation:

Given sum

= (1 + 1 + 1 + ... to n terms) - 1 + 2 + 3 + ... to n terms

n n n

= n - n 1 + 1 [ Ref: nth terms = (n/n) = 1]

2 n

= n - n + 1

2

= 1 (n - 1)

2

Richa said: (Wed, Dec 1, 2010 02:31:06 AM)

Not getting it. Please explain more.

Venumadhav said: (Sun, Jan 9, 2011 04:22:43 AM)

Hai richa. In the given question substitute n=2 (for example). You get (1-1/2) =0. 5. Now cross check in the given options. 1/2(n-1) gives you the answer.

Vamsi Krishna said: (Tue, Sep 27, 2011 12:53:15 PM)

Sum of (1/n + 2/n + 3/n + 4/n + ........ n terms) is Sum of all the terms which are in A.P is Sn = n/2 (2a+(n-1)d)) Using the above formula and substitute as n=n,a=1/n and d=1/n. We will get as n/2(1+ 1/n). If we subtract the above from 'n', we will get the result as (n-1)/2.

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Aptitude - Numbers - Discussion

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