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

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The average of 7 consecutive integers is 7. Find the average of the sq  [#permalink]

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New post 29 Jan 2019, 04:31
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A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

51% (02:03) correct 49% (01:41) wrong based on 48 sessions

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The average of 7 consecutive integers is 7. Find the average of the sq  [#permalink]

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New post 29 Jan 2019, 04:36
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]

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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]

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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]

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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]

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New post 01 Feb 2019, 08:13
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]

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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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The average of 7 consecutive integers is 7. Find the average of the sq   [#permalink] 01 Feb 2019, 08:22
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