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The average (arithmetic mean) of 9 consecutive integers is 9. What is

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Bunuel
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The average (arithmetic mean) of 9 consecutive integers is 9. What is [#permalink] New post   04 Jan 2021, 02:13
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The average (arithmetic mean) of 9 consecutive integers is 9. What is the average (arithmetic mean) of the squares of these 9 integers?

A. 689/9
B. 263/3
C. 889/9
D. 989/9
E. 789

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The average of 9 consecutive numbers is 9. What is the average of there squares.
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B
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Re: The average (arithmetic mean) of 9 consecutive integers is 9. What is [#permalink] New post   04 Jan 2021, 02:58
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The answer is B.

Let us assume the 9 consecutive numbers are (x-4), (x-3), (x-2), (x-1), x, (x+1), (x+2), (x+3), (x+4)

The Arithmetic Mean of these 9 consecutive numbers is 9, therefore x=9

Now (x-4)^2 + (x+4)^2 = 2(x^2 + 16)
(x-3)^2 + (x+3)^2 = 2(x^2 + 9)
(x-2)^2 + (x+2)^2 = 2(x^2 + 4)
(x-1)^2 + (x+1)^2 = 2(x^2 + 1)

Therefore the sum of the square terms of 9 consecutive integers are = 2(x^2 + 16) + 2(x^2 + 9) + 2(x^2 + 4) + 2(x^2 + 1) + x^2 = 9*(x^2) + 2*(16+9+4+1) = 9*(x^2) + 2*30 = 9*9*9 + 3*20 = 3*[(3*81) + 20] = 3*263

Therefore, the Arithmetic Mean of the squares of these 9 integers is = (3*263)/9 = 263/3
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The average (arithmetic mean) of 9 consecutive integers is 9. What is [#permalink] New post   11 Jan 2021, 06:56
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Bunuel wrote:
The average (arithmetic mean) of 9 consecutive integers is 9. What is the average (arithmetic mean) of the squares of these 9 integers?

A. 689/9
B. 263/3
C. 889/9
D. 989/9
E. 789

Show Spoiler
The average of 9 consecutive numbers is 9. What is the average of there squares.


So the total of the 9 terms is 81.
We can find the first and last terms using \(\frac{9}{2}( x+x+8)=81\) ( Formula for sum of n consecutive terms )

Where \(x\) is the first term and \(x+8\) is the last term.

We get \(x= 5 \)

Hence the terms are \(5,6,7.......13\).
We need \(5^2 + 6^2+7^2.......13^2\)

We can use formula\( \frac{n(n+1)(2n+1)}{6}\) for the sum of squares for first n terms.

Sum of squares of first 13 terms - sum of squares for first 4 terms = sum of squares from 5 to 13

Sum of the required terms comes to 789 hence Average \(\frac{789}{9}\)= \(\frac{263}{3}\)

Ans-B

Hope it helps
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Re: The average (arithmetic mean) of 9 consecutive integers is 9. What is [#permalink] New post   20 Oct 2023, 22:38
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Bunuel wrote:
The average (arithmetic mean) of 9 consecutive integers is 9. What is the average (arithmetic mean) of the squares of these 9 integers?

A. 689/9
B. 263/3
C. 889/9
D. 989/9
E. 789

Show Spoiler
The average of 9 consecutive numbers is 9. What is the average of there squares.


Consecutive integers with average 9 means 9 is the middle number and there are FOUR numbers on each side of 9. => 5,6,7,8,9,10,11,12,13

The sum of square of these integers
\(5^2+6^2+7^2+8^2+9^2+10^2+11^2+12^2+13^2\)
It is all about easing the calculations now
\((9-4)^2+(9-3)^2+(9-2)^2+(9-1)^2+9^2+(9+1)^2+(9+2)^2+(9+3)^2+(9+4)^2\)
Take the pair \((9-4)^2+(9+4)^2=9^2+4^2-2*4*9+9^2+4^2+2*4*9=2(9^2+4^2)\)
Similarly for all other pairs
\(2(9^2+4^2)+ 2(9^2+3^2)+ 2(9^2+2^2)+ 2(9^2+1^2)+9^2\)
\(9*9^2+2(1^2+2^2+3^2+4^2)=9*81+60=789\)

Average = \(\frac{789}{9}\)

B
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Re: The average (arithmetic mean) of 9 consecutive integers is 9. What is [#permalink] New post   07 Apr 2024, 14:19
Bunuel wrote:
The average (arithmetic mean) of 9 consecutive integers is 9. What is the average (arithmetic mean) of the squares of these 9 integers?

A. 689/9
B. 263/3
C. 889/9
D. 989/9
E. 789

Show Spoiler
The average of 9 consecutive numbers is 9. What is the average of there squares.

­Let the first number be x. Then the consecutive numbers are x+1,x+2,...,x+8. The sum of all is 9x+36 and the average of this is 9x+36/9=x+4.
Given, x+4=9 => x=5. This means the nine consecutive numbers are 5,6,7,8,9,10,11,12,13.
We know, the sum of squares of the first n numbers are n(n+1)(2n+1) / 6.
Let's take 1,2,...,13. So the sum of squares of these 13 numbers is 13*14*27/6=819.
We have to exclude the sum of squares of 1,2,3,4 from the above sum, which is 4*5*9/6=30.
Therefore, the sum of squares of 5 to 13 is 819-30=789. The average will be 789/9=263/3. Option (B) is correct.
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Re: The average (arithmetic mean) of 9 consecutive integers is 9. What is [#permalink]
07 Apr 2024, 14:19
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