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The average of 7 consecutive integers is 7. Find the average of the sq

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Bunuel
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The average of 7 consecutive integers is 7. Find the average of the sq [#permalink] New post   29 Jan 2019, 04:31
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The average of 7 consecutive integers is 7. Find the average of the squares of these integers.

A. 7
B. 49
C. 53
D. 57
E. 371
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C
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The average of 7 consecutive integers is 7. Find the average of the sq [#permalink] New post   29 Jan 2019, 04:36
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Bunuel wrote:
The average of 7 consecutive integers is 7. Find the average of the squares of these integers.

A. 7
B. 49
C. 53
D. 57
E. 371


For terms in Arithmetic Progression (Such as consecutive terms),
Mean = Median = (First term+last term)/2

Mean of 7 consecutive terms = median of those terms = 7

i.e. terms are _ _ _ 7 _ _ _

Filling the spaces, backwards gives us, 4, 5, 6, 7, 8, 9, 10

i.e. the Sum of squares of the terms = \(4^2 + 5^2+ 6^2 + 7^2 + 8^2 + 9^2 + 10^2 = 16+25+36+49+64+81+100 = 371\)

For sum of the square one may also apply the property, \(1^2 + 2^2 + 3^2 + ... + n^2 = (1/6)*(n)*(n+1)*(2n+1)\)

Answer: Option E
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The average of 7 consecutive integers is 7. Find the average of the sq [#permalink] New post   Updated on: 29 Jan 2019, 10:11
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Bunuel wrote:
The average of 7 consecutive integers is 7. Find the average of the squares of these integers.

A. 7
B. 49
C. 53
D. 57
E. 371



mean = median for consecutive terms
so
4,5,6,7,8,9,10
squaring
16,25,36,49,64,81,100
add all = 371
avg = 371/7 = 53 IMO D

Originally posted by Archit3110 on 29 Jan 2019, 04:46.
Last edited by Archit3110 on 29 Jan 2019, 10:11, edited 1 time in total.
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Re: The average of 7 consecutive integers is 7. Find the average of the sq [#permalink] New post   29 Jan 2019, 10:03
The question is asking for the average of the squares of the integers, not the sum.

[ x + (x+1) + (x+2) + ... + (x+6) ] / 7 = 7
7x + 21 = 49
x = 4 <-- represents the first integer in the sequence

[ 4^2 + 5^2 + 6^2 + ... + 10^2 ] / 7 = 371 / 7 = 53

Answer: C
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Re: The average of 7 consecutive integers is 7. Find the average of the sq [#permalink] New post   29 Jan 2019, 16:17
Bunuel wrote:
The average of 7 consecutive integers is 7. Find the average of the squares of these integers.

A. 7
B. 49
C. 53
D. 57
E. 371


sequence=4,5,6,7,8,9,10
the average of the squares of the seven integers
equals the sum of the squares of the second and sixth integers divided by 2
(25+81)/2=53
C
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Re: The average of 7 consecutive integers is 7. Find the average of the sq [#permalink] New post   01 Feb 2019, 08:13
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Bunuel wrote:
The average of 7 consecutive integers is 7. Find the average of the squares of these integers.

A. 7
B. 49
C. 53
D. 57
E. 371


Since the average of 7 consecutive integers is 7, the integers are:

4, 5, 6, 7, 8, 9, 10

The average of the squares of those integers is:

(16 + 25 + 36 + 49 + 64 + 81 + 100)/7

371/7 = 53

Answer: C
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The average of 7 consecutive integers is 7. Find the average of the sq [#permalink] New post   01 Feb 2019, 08:22
Bunuel wrote:
The average of 7 consecutive integers is 7. Find the average of the squares of these integers.

A. 7
B. 49
C. 53
D. 57
E. 371


Let the average of the numbers be x + x + 1 + x + 2 + x + 3 + x + 4 + x + 5 + x + 6 = 49

You can get the value of x as 4.

Now you need to square them and take their average = \(\frac{16 + 25 + 36 + 49 + 64 + 81 + 100}{7}\)

Now you need not add the expression and then divide the value by 7

just calculate the unit digit of the above sum and you will notice it is satisfied by

C
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Re: The average of 7 consecutive integers is 7. Find the average of the sq [#permalink] New post   21 Jan 2021, 04:41
Bunuel I understand how we reached 53 as the answer, but my doubt is isn't the median and average of a set of odd no of integers the same?So why cant we apply this here and reach 49 as the answer? Or is there some detail I am missing?

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The average of 7 consecutive integers is 7. Find the average of the sq [#permalink] New post   23 Jan 2021, 06:17
Rather easier way is to assume consecutive numbers as a-3, a-2, a-1, a, a+1, a+2, a+3.
So the average of numbers is a, meaning a=7.
Now average of squares is taken as ((a-3)^2+(a-2)^2+(a-1)^2+a^2+(a+1)^2+(a+2)^2+(a+3)^2)/7
= (a^2+(a-1)^2+(a+1)^2+(a-2)^2+(a+2)^2+(a-3)^2+(a+3)^2)/7
= (a^2+2(a^2+1)+2(a^2+4)+2(a^2+9))/7
= (7a^2+28)/7
= a^2+4
=7^2+4 = 53
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