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There is a set of 160 numbers, beginning at 6, with each subsequent term increasing
by an increment of 3. What is the average of this set of numbers?

I am not able to understand how to calculate the last digit?to calculate the average.

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by Brent@GMATPrepNow » Mon Oct 01, 2012 1:08 pm
[email protected] wrote:There is a set of 160 numbers, beginning at 6, with each subsequent term increasing
by an increment of 3. What is the average of this set of numbers?

I am not able to understand how to calculate the last digit?to calculate the average.

There's an important rule that says, "If the numbers in a set are equally spaced, then the mean and median of that set are equal"

So, for example, the mean and median of this set {1, 4, 7, 10, 13, 16} are equal since the numbers are evenly/equally spaced.

In this question, we have 160 numbers. They look like this: {6, 9, 12, 15, . . . }
We can find either the mean or the median, since they'll be equal here.

Let's find the median.
Since there's an even number of integers in the set, we'll need to find the mean of the two middle numbers. In other words, we'll need to find the mean of the 80th value and the 81st value in the set.

But what are the 80th and 81st values in the set?

To find out, let's rewrite the numbers in the set {6, 9, 12, 15, ..}

1st value = 6 = 3 + 3
2nd value = 9 = 3 + 2(3)
3rd value = 12 = 3 + 3(3)
4th value = 15 = 3 + 4(3)
5th value = 18 = 3 + 5(3)
.
.
.

From this we can see a pattern.
80th value = 3 + 80(3) = 243
81st value = 3 + 81(3) = 246

The median (aka the mean) of all 160 terms will equal the mean of 243 and 246.

Median = (243 + 246)/2 = 244.5

So the mean and median are both 244.5

Cheers,
Brent
Last edited by Brent@GMATPrepNow on Mon Oct 01, 2012 8:36 pm, edited 1 time in total.
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by truplayer256 » Mon Oct 01, 2012 8:14 pm
Great Job by Brent up there! Just one correction, I think it should be 244.5 and not 245.5.

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by Brent@GMATPrepNow » Mon Oct 01, 2012 8:36 pm
truplayer256 wrote:Great Job by Brent up there! Just one correction, I think it should be 244.5 and not 245.5.
Nice catch - I've corrected it.

Cheers,
Brent
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by Brent@GMATPrepNow » Mon Oct 01, 2012 9:19 pm
Brent@GMATPrepNow wrote: If the numbers in a set are equally spaced, then the mean and median of that set are equal
The above rule can save you a lot of time on some GMAT questions.

Here's an article on the topic: https://www.beatthegmat.com/mba/2012/05/ ... d-the-mean

And here are 4 related practice questions:
- https://www.beatthegmat.com/og-question- ... 16004.html
- https://www.beatthegmat.com/set-nds-t82611.html
- https://www.beatthegmat.com/og-question- ... 16004.html
- https://www.beatthegmat.com/0-is-added-t ... 92300.html

Cheers,
Brent
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by gmat6087 » Thu Oct 04, 2012 1:55 am
Brent@GMATPrepNow wrote:
[email protected] wrote:There is a set of 160 numbers, beginning at 6, with each subsequent term increasing
by an increment of 3. What is the average of this set of numbers?

I am not able to understand how to calculate the last digit?to calculate the average.

There's an important rule that says, "If the numbers in a set are equally spaced, then the mean and median of that set are equal"

So, for example, the mean and median of this set {1, 4, 7, 10, 13, 16} are equal since the numbers are evenly/equally spaced.

In this question, we have 160 numbers. They look like this: {6, 9, 12, 15, . . . }
We can find either the mean or the median, since they'll be equal here.

Let's find the median.
Since there's an even number of integers in the set, we'll need to find the mean of the two middle numbers. In other words, we'll need to find the mean of the 80th value and the 81st value in the set.

But what are the 80th and 81st values in the set?

To find out, let's rewrite the numbers in the set {6, 9, 12, 15, ..}

1st value = 6 = 3 + 3
2nd value = 9 = 3 + 2(3)
3rd value = 12 = 3 + 3(3)
4th value = 15 = 3 + 4(3)
5th value = 18 = 3 + 5(3)
.
.
.

From this we can see a pattern.
80th value = 3 + 80(3) = 243
81st value = 3 + 81(3) = 246

The median (aka the mean) of all 160 terms will equal the mean of 243 and 246.

Median = (243 + 246)/2 = 244.5

So the mean and median are both 244.5

Cheers,
Brent

Hi Brent,

I may sound silly, but I just need a clarification. The question says the number begins with 6 and subsequent number is incremented by 3.

So the pattern I assume Should be:

1st value = 6 = 6 + 0
2nd value = 9 = 6 + 1(3)
3rd value = 12 = 6 + 2(3)
4th value = 15 = 6 + 3(3)
5th value = 18 = 6 + 4(3)
.
.
.

From this we can see a pattern.
80th value = 6 + 80(3) = 246
81st value = 6 + 81(3) = 249

Please correct me if I an wrong.

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by das.ashmita » Thu Oct 04, 2012 2:45 am
Hi nishat

Another approach:

The given series is an AP with
(1st term) a = 6,
(no of terms) n = 160,
(common diff) d = 3
(last term) l = a+(n-1)d = 6+(159)3 = 483

Avg = (a+l)/2 = (6+483)/2 = 244.5

Hope it helps :)

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by Brent@GMATPrepNow » Thu Oct 04, 2012 7:17 am
gmat6087 wrote: Hi Brent,

I may sound silly, but I just need a clarification. The question says the number begins with 6 and subsequent number is incremented by 3.

So the pattern I assume Should be:

1st value = 6 = 6 + 0
2nd value = 9 = 6 + 1(3)
3rd value = 12 = 6 + 2(3)
4th value = 15 = 6 + 3(3)
5th value = 18 = 6 + 4(3)
.
.
.

From this we can see a pattern.
80th value = 6 + 80(3) = 246
81st value = 6 + 81(3) = 249


Please correct me if I an wrong.
Your way works perfectly fine as well.
In my approach, I arranged it so that the numbers match up.
For example, 5th value = 18 = 3 + 5(3) . . . and 80th value = 3 + 80(3) = 243

In your approach, we need to be careful and recognize that the numbers don't match up. That is they are 1 off. Example: 5th value = 18 = 6 + 4(3).
So, continuing along . . . 80th value = 6 + 79(3) = 243
(you have 80th value = 6 + 80(3) = 246) . . . and 81st value = 6 + 80[/blue](3) = 246

Cheers,
Brent
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by Brent@GMATPrepNow » Thu Oct 04, 2012 7:19 am
gmat6087 wrote: Hi Brent,

I may sound silly, but I just need a clarification. The question says the number begins with 6 and subsequent number is incremented by 3.

So the pattern I assume Should be:

1st value = 6 = 6 + 0
2nd value = 9 = 6 + 1(3)
3rd value = 12 = 6 + 2(3)
4th value = 15 = 6 + 3(3)
5th value = 18 = 6 + 4(3)
.
.
.

From this we can see a pattern.
80th value = 6 + 80(3) = 246
81st value = 6 + 81(3) = 249


Please correct me if I an wrong.
Your way works perfectly fine as well.
In my approach, I arranged it so that the numbers match up.
For example, 5th value = 18 = 3 + 5(3) . . . and 80th value = 3 + 80(3) = 243

In your approach, we need to be careful and recognize that the numbers don't match up. That is they are 1 off. Example: 5th value = 18 = 6 + 4(3).
So, continuing along . . . 80th value = 6 + 79(3) = 243
(you have 80th value = 6 + 80(3) = 246) . . . and 81st value = 6 + 80(3) = 246

Cheers,
Brent
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by LalaB » Thu Oct 04, 2012 7:56 am
one more solution-

average= (last term+first term)/2=(A160th+A1)/2

note that we have the following set-
6 9 12 15 etc

it means we have multiple of 3 plus 3 . a1=1*3+3=6 a2=2*3+3=9 etc

so, A160=160*3+3=483

(483+6)/2=244.5
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