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April 10th, 2018, 07:48 PM   #1
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What is the average of all the odd natural numbers upto 51?

Find the the average of all the odd natural numbers upto 51.

Sum of odd numbers is $\displaystyle n^2$. Here n is 51.

51 stands in 26th position in odd natural numbers.

Average is $\displaystyle \frac{n^2}{26}$ = 100.03

But answer is 26.

Please help me.

Last edited by skipjack; April 11th, 2018 at 12:31 PM.
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April 10th, 2018, 07:57 PM   #2
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Quote:
Originally Posted by Ganesh Ujwal View Post
Find the the average of all the odd natural numbers upto 51.

Sum of odd numbers is $\displaystyle n^2$. Here n is 51.

51 stands in 26th position in odd natural numbers.

Average is $\displaystyle \frac{n^2}{26}$ = 100.03

But answer is 26.

Please help me.
You discovered a great problem-solving tactic earlier. Work hard to solve the problem yourself; since you can't depend on us slackers and mathematical posers. You should challenge yourself to solve this.
Thanks from SDK

Last edited by skipjack; April 11th, 2018 at 12:32 PM.
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April 10th, 2018, 08:03 PM   #3
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Quote:
Originally Posted by Ganesh Ujwal View Post
Find the the average of all the odd natural numbers upto 51.

Sum of odd numbers is $\displaystyle n^2$. Here n is 51.

51 stands in 26th position in odd natural numbers.

Average is $\displaystyle \frac{n^2}{26}$ = 100.03

But answer is 26.

Please help me.
1+3+5+..+25+26+...+47 + 49 + 51. Now write the sum as (1+51) + (3+49) + (25+26) = 52*13. So the average will be (52*13)/26 = 26.
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Last edited by skipjack; April 11th, 2018 at 12:32 PM.
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April 10th, 2018, 08:24 PM   #4
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$n$ is not 51. Rethink what the formula $n^2$ is counting. Start by looking at small examples.
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April 10th, 2018, 08:28 PM   #5
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Originally Posted by Maschke View Post
You discovered a great problem-solving tactic earlier. Work hard to solve the problem yourself; since you can't depend on us slackers and mathematical posers. You should challenge yourself to solve this.
Ok, now can you tell me the procedure how to solve this problem?

Last edited by skipjack; April 11th, 2018 at 12:33 PM.
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April 10th, 2018, 09:35 PM   #6
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Ok, now can you tell me the procedure how to solve this problem?
1+3+5+..+25+26+...+47 + 49 +51. Now write the sum as (1+51) + (3+49) + ...+(25+26) = 52*13. So the average will be (52*13)/26 = 26.
Thanks from topsquark

Last edited by skipjack; April 11th, 2018 at 12:34 PM.
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April 10th, 2018, 09:41 PM   #7
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Now write the sum as (1+51) + (3+49) + ...+(25+26) = 52*13. So the average will be (52*13)/26 = 26.
Why did you write (1+51) + (49 +3)..... etc?

Last edited by skipjack; April 11th, 2018 at 12:35 PM.
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April 10th, 2018, 10:21 PM   #8
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Quote:
Originally Posted by Ganesh Ujwal View Post
Why did you write (1+51) + (49 +3)..... etc?
That's basically what Gauss did to invent the arithmetic summation formula.
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Last edited by skipjack; April 11th, 2018 at 12:36 PM.
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April 10th, 2018, 11:02 PM   #9
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Originally Posted by SDK View Post
$n$ is not 51. Rethink what the formula $n^2$ is counting. Start by looking at small examples.
If n is not 51. Then n is 26. Can you tell me why n is not 51?

Last edited by skipjack; April 11th, 2018 at 12:35 PM.
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April 11th, 2018, 11:14 AM   #10
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Why did you write (1+51) + (49 +3)..... etc?
In order to get the same sum in each parenthesis. If you keep adding 52 to itself say 11 times, the sum will be 52*11.

Last edited by skipjack; April 11th, 2018 at 12:36 PM.
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