Sum of Consecutive Positive Integers Making My Life **** :P

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BilalAnsar

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Let S be the sum of the elements of X, a set of 10 consecutive positive integers.
Quantity A: S/10 - 5
Quantity B: The smallest of the integers in X
Which of the two quantities is greater?

I solved the question by using numbers 1-10. The sum came out to be 55 which made Quantity A equal to 0.5 (55/10 - 5) and Quantity B equal to 1 (smallest integer in the set). 0.5 < 1 so Quantity B is greater.

However, I don't understand the method used in the book. The book says that the average value of the set is S/10, which I agree with. It then says that the average value can be found by averaging the 5th and 6th numbers in the set, which I agree with as well as the set is made of 10 consecutive integers. If the set was 50 integers then the average value would be found by averaging the 25th and 26th numbers.

But then the book says that this average value is "exactly 4.5 greater than the lowest number in the set". How do we know that? I understand that if I plug in numbers that turns out to be true, but that is a pattern that the solution in the book does not establish. How then can they say so confidently that the smallest number is exactly 4.5 less than the average? Is there a rule here that I'm missing?

(The book then goes on to write the smallest integer in the set as S/10 - 4.5 and compares this value to S/10 - 5. Since S/10 - 4.5 is greater, Quantity B is greater.)

I would appreciate any help very much. Thanks!
 
have a look at the general sum of any 10 positive integers where n represents the starting integer ...

[MATH]n + (n+1) + (n+2) + (n+3) + \, ... \, + (n+9) = 10n + 45[/MATH]
does that help?
 
BilalAnsar said:
Let S be the sum of the elements of X, a set of 10 consecutive positive integers.
Quantity A: S/10 - 5
Quantity B: The smallest of the integers in X
Which of the two quantities is greater?

I solved the question by using numbers 1-10. The sum came out to be 55 which made Quantity A equal to 0.5 (55/10 - 5) and Quantity B equal to 1 (smallest integer in the set). 0.5 < 1 so Quantity B is greater.

However, I don't understand the method used in the book. The book says that the average value of the set is S/10, which I agree with. It then says that the average value can be found by averaging the 5th and 6th numbers in the set, which I agree with as well as the set is made of 10 consecutive integers. If the set was 50 integers then the average value would be found by averaging the 25th and 26th numbers.
I would have said it is average of the largest and smallest numbers in the list. That is the same but the "largest" and "smallest" numbers, the "first" and "last" numbers, are easier to find. In case of 10 consecutive integers, taking the first number to be "n" the last number is 10+ 9 and the average is (n+ (n+ 9)/2= (2n+ 9)/2= n+ 9/2.

But then the book says that this average value is "exactly 4.5 greater than the lowest number in the set". How do we know that?
Click to expand...
Because 9/2= 4.5.

I understand that if I plug in numbers that turns out to be true, but that is a pattern that the solution in the book does not establish. How then can they say so confidently that the smallest number is exactly 4.5 less than the average? Is there a rule here that I'm missing?
Click to expand...
With n the smallest number we have n, n+1, n+2, n+3, n+4, n+5, n+6, n+7, n+8, and n+9. The two middle numbers (corresponding to the "25th and 26th numbers" in your 50 number example) are n+4 and n+ 5. Their average is (n+ 4+ n+ 5)/2= (2n+ 9)/2= n+ 4.5 again. Notice that if you were to add all 10 numbers then divide by 2 you would get (10n+ 1+2+3+4+5+6+7+8+9)/2= (10n+ 45)/10= n+ 4.5, the same as averaging the first and last numbers or the fifth and sixth numbers.

(The book then goes on to write the smallest integer in the set as S/10 - 4.5 and compares this value to S/10 - 5. Since S/10 - 4.5 is greater, Quantity B is greater.)

I would appreciate any help very much. Thanks!
Click to expand...
Click to expand...
Click to expand...
 
skeeter said:
have a look at the general sum of any 10 positive integers where n represents the starting integer ...

[MATH]n + (n+1) + (n+2) + (n+3) + \, ... \, + (n+9) = 10n + 45[/MATH]
does that help?
Click to expand...

Yes. So the average would be (n + 4 + n + 5)/2 = n + 4.5. Which means the average number will always be 4.5 more than the first number. Thanks a lot!
 
The fact that the (last number + first number)/2 = average of middle two numbers = sum of all numbers / no. of numbers is really helpful to know. Thanks for taking out the time!
 
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