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The average of 9 consecutive numbers is 9. What is the

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Manager
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The average of 9 consecutive numbers is 9. What is the [#permalink] New post 13 Feb 2004, 01:45
The average of 9 consecutive numbers is 9. What is the average of there squares.
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 [#permalink] New post 13 Feb 2004, 02:15
First is it numbers or integers? I assumed integers.x+(x+1)...+(x+8)/9=9=> 9X+36=81=>X=5. Now calculate sum of squares=789/9
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 [#permalink] New post 13 Feb 2004, 06:51
Hmmm, usually when the numbers are consecutive, I think it implies that they are integers. BG's technique is good. Another way to look at it is to realize that when the average of the numbers is 9 and there are 9 numbers(odd amount of numbers), it means that the median number is just 9 and there are 4 consecutive numbers on each side of the median. In this case:
5 - 6 - 7 - 8 - 9 - 10 - 11 - 12 - 13
You then square each term and add them up to get 789/9
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 [#permalink] New post 13 Feb 2004, 07:13
I liked Paul's scheme. I will remember that.
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 [#permalink] New post 15 Feb 2004, 02:40
Well what if i said that the average of 20 consecutive integers is 20. What is the average of there squares. Would u do the same way?????
Is there any better ways to do it????
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 [#permalink] New post 15 Feb 2004, 08:58
Same way except that it is VERY unlikely that such a lenghty problem will be asked on the actual test
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 [#permalink] New post 16 Feb 2004, 02:50
Well i don't know if it is so lengthy...but here is a better way to do it.......
I will go by ur approach only.
Sorry I am changing my question from 20 to 21.

11...........................21.....................................32
Place 21 in between and sub. (21-1)/2 and add the same.It will give u the extremes.
They are 11 and 32.
We need to find sqr(11) + sqr(12) +.............sqr(32)
Use the formula of the sum of squares of first N natural numbers.
N(N+1)(2N + 1)/6.
Place N = 32 we get Sum = 11440
Place N = 10 we get Sum = 385

So req. average = (11440 - 385)/21
or 3685/7.
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 [#permalink] New post 16 Feb 2004, 04:20
Quote:
Use the formula of the sum of squares of first N natural numbers.
N(N+1)(2N + 1)/6.

Nice one mdfrahim
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 [#permalink] New post 16 Feb 2004, 07:46
Quote:
Sorry I am changing my question from 20 to 21.

11...........................21.....................................32
Place 21 in between and sub. (21-1)/2 and add the same.It will give u the extremes.
They are 11 and 32.


Be careful..

the way to find the extremes described here is not accurate. In order to find the extrems for consecutive integers, you must know the average and the number of integers. (Both of which were given in the original problem you posted).

If the number of consecutive integers is even then the average will not be an integer. It would be a number with decimal.

For example: average of 2,3,4, 5 = (2+5) / 2 = 3.5

If the number of consecutive integers is oddthen only the average will be an integer.

for eaxmple, the average of 2, ,3 ,4 is = middle number 3

In the above calculation you meant that 21 is average and 11 and 32 are extremes. However, the average of consecutive integers from 11 to 32 is (11+32)/2 = 43/2 = 21.5

So the point is the method to find the extreme is not accurate. We must know the number of consecutive integers we are talking about.
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 [#permalink] New post 16 Feb 2004, 22:35
Agreed to Gmatblast........we need to be more carefull when dealing with the even numbers.......
In the example which i gave the extremes will be 11 and 31 (not 32)..it was a mistake. Thanks for the correction gmatblast.
  [#permalink] 16 Feb 2004, 22:35
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