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by agatka16 » Wed May 28, 2008 8:12 am
Maybe you could look at it that way:
a=2+4+6+...+20 ==> you have 10 even integers
b=1+3+4+...+19==> you have 10 odd integers

a-b=(2-1)+(4-3)+(6-5)+...+(20-19)==> each term is 1 and you have 10 terms. Answer 10.
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by aatech » Wed May 28, 2008 8:21 am
Unfortunately, I am not able to follow your method... so can't comment..

I will suggest this would be easier...

a = 2+4+6.... +20
b= 1+3+5+....+19

a-b = (2+4+6.... +20)-(1+3+5+....+19) = 1+1+1.. (10 ones) = 10

You can also using Arithmetic Progression formula...
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by khurram » Wed May 28, 2008 8:52 am
why wont the short cut work

(largest +smallest)/2 for both sequences, to get avg of the sequence

largest -smallest +1 to get number of terms

times the two, avg times the number of terms to get the sum

eg
1, 2,3,4,5
1+5=6/2=3
3*5=15 same as 1+2+3+4+5=15

number of terms= 5-1+1=5.



Thanks
Khurram
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by khurram » Wed May 28, 2008 8:55 am
aatech
I like your thinking also. diff between each term is 1. so in toal 10 diff between the two.
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by khurram » Wed May 28, 2008 8:58 am
got it

was thinking consecutive not even set and odd set

thanks
khurram so my method cannot be used as that is for consecutive set only
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by mim3 » Thu May 29, 2008 2:41 pm
khurram wrote:got it

was thinking consecutive not even set and odd set

thanks
khurram so my method cannot be used as that is for consecutive set only
No, the shortcut can still be used as it's still a consecutive set:

sum of set= median x number of terms

2-20(e):
median= (20+2)/2 = 11
# terms= 10
sum= 110

1-19(o):
median= (19+1)/2= 10
# terms= 10
sum= 100

110-100= 10.
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