## num props

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### num props

by vinviper1 » Thu Jun 19, 2008 10:06 am
Whats the formula for these types of problems?

1+2+3+4+...+99=

A 4700
B 4750
C 4850
D 4900
E 4950

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by netigen » Thu Jun 19, 2008 10:36 am
When you see this think sets

This is a consecutive number set from 1 to 99 so the number of terms is 99

the mean (avg) of the terms is (1+99)/2 = 50

so the sum = mean x number of terms = 50 x 99 = 4950

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by wawatan » Thu Jun 19, 2008 12:18 pm
you could also use this formula

[n(n+1)]/2

1+2+3+4+...+99

so n=99

(99*100)/2 =4950

this is a relatively easy problem if you know the formula. also be aware that this formula only works for consecutive positive integers and when the sequence starts from 1 to n.

PS (this should be a 500-600 level math problem IMO)

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