Latest TCS Aptitude Question SOLUTION: consider the sequence 1,-2,3,-4,5,-6......what is the Average of the first 200 terms of the sequence?

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consider the sequence 1,-2,3,-4,5,-6......what is the Average of the first 200 terms of the sequence?

Asked In TCS Shyam Agrawal (8 years ago)

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Ganesh  (8 years ago)

(1-2)=-1
(3-4)=-1.....similarly (199-200)=-1
so total 100 set.
(-1)100/200 =-0.5
ans=-0.5

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aakash  (8 years ago)

1+3+5.....n=n^2 {here n=100} , -(2+4+6.....)=-[n(n+1)] {here n=100 }
total sum=n^2-n(n+1)=-n
average= sum/no.of terms
= -n/200
=-100/200
=-0.5

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Parvathi.R  (8 years ago)

7
odd positions--> 1 3 5 and hence 7

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aakash  (8 years ago)

no. of terms =200 means 100 odd and 100 negative even

sum of 1+3+5.........n = n^2 and sum of -(2+4+6......n) = -[n(n+1)]

total sum = n^2-n(n+1) = 100^2-[100(100+1)] = 10000-10100 = -100

average=Total sum/no. of total terms

= -100 / 200

=-0.5

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gautam kumar pandey  (8 years ago)

-100/200=-.5

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Aparna Takkellapati  (8 years ago)

-100/200=-.5

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