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The square of the arithmetic mean of a sequence of consecutive positiv

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The square of the arithmetic mean of a sequence of consecutive positiv  [#permalink]

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New post 20 Nov 2017, 15:52
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The square of the arithmetic mean of a sequence of consecutive positive integers is equal to the difference between the squares of the least and greatest terms. If the square of the mean is <1000, how many such sequences are there?

A. 3
B. 4
C. 5
D. 6
E. 7
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Re: The square of the arithmetic mean of a sequence of consecutive positiv  [#permalink]

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New post 20 Nov 2017, 16:30
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Say S is the smallest of our consecutive integers, and L is the largest. In any equally spaced list, the mean is always equal to the average of the smallest and largest terms, so the mean of these consecutive integers must be (L + S)/2. We are told the square of the mean is equal to the difference L^2 - S^2. Notice that's a difference of squares, so is equal to (L+S)(L-S). So we have

\(\begin{align}
\left( \frac{L+S}{2} \right)^2 &= L^2 - S^2 \\
\left( \frac{1}{4} \right) (L + S)^2 &= (L + S)(L - S) \\
\left( \frac{1}{4} \right) (L + S) &= L - S \\
L + S &= 4L - 4S \\
5S &= 3L \\
L &= \frac{5S}{3}
\end{align}\)

So we just need L to be 5/3 of S. Notice, since our letters represent integers, that this means S will be divisible by 3, and L will be a multiple of 5. But S can be any multiple of 3 at all, so we have, for example, these sequences:

3, 4, 5
6, 7, 8, 9, 10
9, 10, 11, ..., 14, 15

Finally, we need the mean (which is equal to the median) of the sequence to be less than √1000, which is slightly less than 32. That will clearly be true of this sequence:

18, 19, ..., 29, 30

which would be the sixth smallest sequence we could make. If we look at this sequence:

21, 22, ..., 34, 35

the mean is 28, so again this sequence works. But if we look at this sequence:

24, 25, ..., 39, 40

the mean is 32, and 32^2 = 2^10 = 1024 is too large. So there are seven such sequences.

There are some interesting things about the question, but it also involves about four more steps than any real GMAT question. Where is it from?
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The square of the arithmetic mean of a sequence of consecutive positiv  [#permalink]

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New post 21 Nov 2017, 11:26
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gracie wrote:
The square of the arithmetic mean of a sequence of consecutive positive integers is equal to the difference between the squares of the least and greatest terms. If the square of the mean is <1000, how many such sequences are there?

A. 3
B. 4
C. 5
D. 6
E. 7


let x=first term; y=last term
[(y+x)/2]^2=(y+x)(y-x)➡
y+x=4y-4x➡
5x=3y
thus, first three sequences are:
3,4,5;
6,7,8,9,10;
9,10,11,12,13,14,15
means--4,8,12--are consecutive multiples of 4
let m=number of means
(4m)^2<1000
4m<≈32
4m=28 (next lowest multiple of 4)
m=7
E
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GMAT 1: 700 Q50 V34
The square of the arithmetic mean of a sequence of consecutive positiv  [#permalink]

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New post 15 Dec 2017, 14:35
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First thing to note => since the square of the arithmetic mean = difference between the squares of the least and greatest terms (which is an integer) => square of arithmetic mean is integer => mean of consecutive positive integers is integer => number of integers is odd

Let number of terms in the sequence be \(2 * n + 1\) (as odd number can be expressed of this form)

Let the mean of the sequence be x,
so sequence will read as

x - n, x - (n-1), .................., x - 1, x , x + 1, x + 2, ...... , x + n

The square of the arithmetic mean = difference between the squares of the least and greatest terms
=> \(x^2 = (x + n)^2 - (x - n)^2\)
=> \(x^2 = x^2 + n^2 + 2xn - x^2 - n^2 + 2xn\)
=> \(x^2 = 4xn\)
=> \(x = 0\) or \(x = 4n\) (x (middle term) cannot be zero, as it is sequence of positive integers)
=> \(x = 4n\)

given : \(x^2\) < 1000
=> \(16n^2 < 1000\) => \(n^2 < 62.5\) => \(n <= 7\)
if n = 1, total terms (2*n+1) = 3
if n = 2, total terms = 5
.
.
n = 7

so we can 7 such sequences => (E)
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The square of the arithmetic mean of a sequence of consecutive positiv &nbs [#permalink] 15 Dec 2017, 14:35
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