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

Question

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

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

IanStewart · Answer

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} ight)^2 &= L^2 - S^2 \
\left( \frac{1}{4} ight) (L + S)^2 &= (L + S)(L - S) \
\left( \frac{1}{4} ight) (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?

gracie · Answer

let x=first term; y=last term
 ^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

hellosanthosh2k2 · Answer

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)

Adit_ · Answer

Questions like these require arithmetic more than trial and error.
Simplify the statements:
(x+y)(x-y) = (x+y/2)^2
You will eventually get 4(x-y) = x+y
OR
4/2(x-y) = x+y/2 = Mean.
2(x-y) = Mean

Now is the right time to think.
What is mean for consecutive positive numbers?
Its the median.

If its the median, increase to the left = decrease to the right.
Hmm now mean is a 2(diff between the extremes). 
What is x-y, its precisely the distance between the two extremes and the mean sits right in between this distance!
Since increase and decrease are equal to the mean/median this value has to be divisible by 2 as well!

For example take Mean = 4.

Take the series 3,4,5.
Distance = 2  
Now our mean = 2*2 = 4 exactly what we need!
Why did I choose such an example? Because we now notice that the means are multiples of 4. Simple as that!

Take 8. On reversing the equation if 8 is the mean then the distance must be 4, which is 2 to the right and 2 to the left.
Sequence = 6,8,10
And it continues..

We have been given mean^2<1000
Which means mean<32
Which means all multiple of 4 < 32
= 4 to 28 = 7 multiples.

Answer:  Option E

___________________________

Nice enough this is the rephrasing of Pythagorean triplets with special properties.

Fictional26 · Answer

can you please explain how 4m<~32 came? I know it must be something basic, but I don't seem to understand.

Bunuel · Answer

It comes from taking the square root from  (4m)^2 < 1000 :

4m < \sqrt{1000}

Since  \sqrt{1000} ≈ 31.6 , they wrote  4m < ≈32 .

Adit_ · Answer

Second post here but just wanted to point out that problems like these are best solved when treated as a proper word problem rather than forcing matter using trial and error then there is almost not way to solve it correctly within a stipulated time.

In this case we get the last and first integer in that ratio of 5:3.
Once we get this, we can substitute into the equation that (a+b/2)^2<1000
Then we subsequently get b<24.
or b<= 23.
We can now see that b is a multiple of 3, thus highest value of b = 21, which is 3*7.
So multiples of 3 from 1 to 7 all form individual sequences.

Thus  7 .

Fictional26 · Answer

Thank you  Bunuel  for the clarification.

Kinshook · Answer

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?

Let the least and greatest terms be L & G respectively

(L+G)^2/4 = G^2 - L^2 = (G+L)(G-L)
(G+L) = 4(G-L)
3G = 5L
L = 3G/5

L=3; G=5
L=6; G=10
L=9; G=15

(G+L)^2/4 < 1000
(G+L)^2 < 4000
G+L < 63
G + 3G/5 < 63
G < 39

L=3; G=5
L=6; G=10
L=9; G=15
L=12;G=20
L=15;G=25
L=18;G=30
L=21;G=35

IMO E

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

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