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In a certain sequence, the term A_n is given by the formula A_n= {16-(

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Re: In a certain sequence, the term A_n is given by the formula A_n= {16-( [#permalink]
sahil7389 wrote:
sharma123 wrote:
series upto seventh term is 7,12,12,16,15,15,7. mean is 12 and median is 12 , so B

Though I think answer is correct but series is wrong..
series should be 7,12,15,16,17,20,25
Median ans average would only be equal for integral value if we have the odd number of terms.
So i just checked for 7 and 9. But if u check for 6 and 8, you will come to know the avg would be either less or more than median just by seeing the terms.

Yes you are correct , series i write is wrong. Thanks for correcting
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Re: In a certain sequence, the term A_n is given by the formula A_n= {16-( [#permalink]
Bunuel wrote:
In a certain sequence, the term $$A_n$$ is given by the formula $$A_n= 16-\frac{(4- n)^3}{|n-4|}$$ for all positive integers n ≠ 4, while $$A_4 = 16$$. For which integer value of k greater than 2 is the mean of all values in the sequence from $$A_1$$ through $$A_k$$ equal to the median of those values?

(A) 6
(B) 7
(C) 8
(D) 9
(E) 10

Kudos for a correct solution.

MANHATTAN GMAT OFFICIAL SOLUTION:

As is the case with most sequence problems, your first step is to compute the first several terms. The formula looks ugly, but if you work from the inside out, respecting order of operations, you’ll be in good shape. Do the calculations carefully and clearly, not only so that you don’t make a mistake but also so that you can spot emerging patterns.

$$A_n=16-\frac{(4-n)^3}{|n – 4|}$$

If n = 1, then $$A_1=16-\frac{(4-1)^3}{|1 – 4|}$$

$$= 16-\frac{(3)^3}{|- 3|}$$

$$= 16-\frac{3^3}{3}$$

$$= 16-3^2$$

$$= 16 – 9$$

$$= 7$$

If n = 2, then $$A_2=16-\frac{(4-2)^3}{|2 – 4|}$$

$$= 16-\frac{(2)^3}{|- 2|}$$

$$= 16-\frac{2^3}{2}$$

$$= 16-2^2$$

$$= 16 – 4$$

$$= 12$$

If n = 3, then $$A_3=16-\frac{(4-3)^3}{|3 – 4|}$$

$$= 16-\frac{(1)^3}{|- 1|}$$

$$= 16-\frac{1^3}{1}$$

$$= 16-1^2$$

$$= 16 – 1$$

$$= 15$$

So far, the sequence seems to be converging on 16. The first term, $$A_1$$, is 16 minus 9; the second term is 16 minus 4; the third term is 16 minus 1. The separate definition of $$A_4$$ as 16 confirms our suspicion. (The purpose of the separate definition is to prevent division by 0 in that one case.)

What happens when n = 5?

If n = 5, then $$A_5=16-\frac{(4-5)^3}{|5 – 4|}$$

$$= 16-\frac{(-1^3)}{|1|}$$

$$= 16+\frac{1^3}{1}$$

$$= 16+1^2$$

$$= 16 + 1$$

$$= 17$$

So, on the other side of 16, the terms pick up by adding a square to 16.

Consider what we are asked. We are looking for how far we need to go past the second term to make the mean of all the values up to that point equal the median. For the mean to equal the median, the values must be symmetrically spaced around the mean. Our sequence is not evenly spaced, but at a certain point, it will be symmetrically spaced around some number.

From the special role of 16, we might suspect that 16 will wind up being the number around which the sequence will be symmetrically spaced. That suspicion is correct:

If n = 1, then $$A_1=16-3^2=16-9 =7$$
If n = 2, then $$A_2=16-2^2=16-4=12$$
If n = 3, then $$A_3=16-1^2 =16-1=15$$
If n = 4, then $$A_4=16$$
If n = 5, then $$A_5=16+1^2=16+1=17$$
If n = 6, then $$A_6=16+2^2=16+4=20$$
If n = 7, then $$A_7=16+3^2=16+9=25$$

At that point, the sequence is symmetrical around 16. For all higher values of n, the sequence is no longer symmetrical, so the mean and the median will not be equal.

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Re: In a certain sequence, the term A_n is given by the formula A_n= {16-( [#permalink]
This is a preety cheesy question as it requires us to get the values
Hmm textbook maths

A1=7
A2=12
A3=15
A4=16
A5=17
A6=20
I calculated upto A6 as its the minimum that the options mention.
Here median = 3rdterm +4thterm/2 => 15.5
Now we dont have to calculate the mean here actually
We can use the basic definition of mean which states that the sum of deviations around the mean has to be zero
here sum of deviations is not zero
lets get A7=> 25
median = 16 again (4th term )
now the deviations are => 16-7+16-12+16-16+16-16+16-17+16-20+16-25 => ZERO
BINGO
SMASH THAT B
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Re: In a certain sequence, the term A_n is given by the formula A_n= {16-( [#permalink]
Bunuel wrote:
In a certain sequence, the term $$A_n$$ is given by the formula $$A_n= 16-\frac{(4- n)^3}{|n-4|}$$ for all positive integers n ≠ 4, while $$A_4 = 16$$. For which integer value of k greater than 2 is the mean of all values in the sequence from $$A_1$$ through $$A_k$$ equal to the median of those values?

(A) 6
(B) 7
(C) 8
(D) 9
(E) 10

Kudos for a correct solution.

Since we only care about median and mean, we can ignore the 16 part of the terms, as it is fixed.

When we divide $$(4-n)^3$$ by $$|n-4|$$, see that the argument is the same, with the sign flipped. This means that we can divide one by the other, as long as we keep the sign of the dividend, so we can rewrite the expression like this:
$$A_n' = -Sign(4-n)*(4-n)^2$$

With the same mean and median, shifted 16.

Since we know that, we can conclude:
1) At n=4, $$A_n' = 0$$, this is the tipping point - as n furthers from 4, the difference between consecutive terms increase
2) It's a parabola with the sign of the left half flipped, therefore the moduli are symmetric around $$A_4'$$. (This means that $$A_3'=-A_5', A_2'=-A_6'...$$)
3) $$A_1' = -9$$

We want n such that $$|A_n| = |A_1| = 9$$
It's easy to see than that n = 7 gives that.
B
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Re: In a certain sequence, the term A_n is given by the formula A_n= {16-( [#permalink]
2
Kudos
Bunuel wrote:
In a certain sequence, the term $$A_n$$ is given by the formula $$A_n= 16-\frac{(4- n)^3}{|n-4|}$$ for all positive integers n ≠ 4, while $$A_4 = 16$$. For which integer value of k greater than 2 is the mean of all values in the sequence from $$A_1$$ through $$A_k$$ equal to the median of those values?

(A) 6
(B) 7
(C) 8
(D) 9
(E) 10

Kudos for a correct solution.

The question is not hard and the logic becomes apparent the moment we do what we are required to do in sequence questions - find out the first few values:

$$A_1= 16-3^2$$ (there is likely a pattern here so we should keep this as it is)
$$A_2= 16-2^2$$
$$A_3= 16-1^2$$
$$A_4= 16$$
$$A_5= 16+1^2$$ (and with this we already know that the answer must be 7. Watch why)
$$A_6= 16+2^2$$
$$A_7= 16+3^2$$

Note that when we add these 7 terms, the deviation from the mean 16 will become 0 because deficit = excess.
The median will be 16 too.

Hence we need the first 7 terms.