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by shashank.ism » Tue Feb 09, 2010 1:10 pm
Consider the set P = –5, –3, –1, 1, 3, 5…. consisting of 1998 numbers. If ‘a’ be the average of the elements in P and ‘b’ be twice the average of the first 1998 natural numbers, then which of the following is equal to (a – b)?

Correct Answer: 4
–6
5
–8
–7
–9
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by harsh.champ » Tue Feb 09, 2010 2:00 pm
shashank.ism wrote:Consider the set P = –5, –3, –1, 1, 3, 5…. consisting of 1998 numbers. If ‘a’ be the average of the elements in P and ‘b’ be twice the average of the first 1998 natural numbers, then which of the following is equal to (a – b)?

Correct Answer: 4
–6
5
–8
–7
–9
In the 1st series we are missing the even no.s adding them we have a complete series.
then dividing the even no.s by 2 we get the average of 1st 1998 no.s.

And if we consider 1998 x 2 nos. we get the average of 1st 3996 no.s .thus we can subtrat from it the average of 1st 1998 no.s to get the answer.

Hence , the answer would be -7.(this can be got by using the sum of the n natural no.s = n(n+1)/2 )
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