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Average question! clear solution please!

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by cramya » Wed Dec 03, 2008 7:49 pm
There are different ways to solve this

One way

6*1 = 6

6*166 = 996

There are 166 multiples of 6 greater than 0 and less than 1000 (6,12,18...996)

It can be treated as an ARITHMETIC PROGRESSION

Sn = n/2(2a1+(n-1)d
Sn->sum of n terms
n->number of terms (166 in this case)
a1->first term in sequence (a1=6)
d->common difference(d=6)

WE get 83*1002 as the sum. This divided by number of terms 166 is

83*1002/166 = 501

C)
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by hwiya320 » Wed Dec 03, 2008 7:54 pm
I need to memorize this! thanks! thanksgivn.
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by logitech » Wed Dec 03, 2008 7:59 pm
6 + 12 + 18 + ..... + 696

6 ( 1+2+3+...166)

1+2+...+n = nx(n+1)/2

SO the MEAN of 166 numbers is :

= 6x166x167/2x166

= 167x3

Hence, C
LGTCH
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by niraj_a » Wed Dec 03, 2008 11:29 pm
i see

i was thinking in terms or evenly spaced consecutive numbers, where the mean would be equal to the median. i thought since 0 and 1000 and not included, there are 998 values. A and E sorta fit the bill on that idea.

could anyone advise how one would solve this question using that idea?
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by cramya » Wed Dec 03, 2008 11:31 pm
could anyone advise how one would solve this question using that idea?
Yes it can be done like that also(mean=median )

It will 166/2 = 83


a83+a84/2

a83 = 6*83
a84 would be 6*84

6*83+6*84/2

6(83+84)/2 = 3 (167) = 501
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by logitech » Wed Dec 03, 2008 11:55 pm
And I am sure you have already seen that the median and the mean of consecutive integers are the same:

6x1 = 6 and 6x166 = 996

So the answer is : 996+6 = 1002/2 = 501
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by niraj_a » Wed Dec 03, 2008 11:58 pm
supersonic !
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