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

Question

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

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

stne · Accepted Answer

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

anjanita · Answer

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*263

Therefore, the Arithmetic Mean of the squares of these 9 integers is = (3*263)/9 = 263/3

chetan2u · Answer

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

samarpan.g28 · Answer

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

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