What is the average of the following numbers?

We know, Average of a series = (Sum of all Numbers) / (Total Quantity of Numbers)

As sum of 1st n square numbers = {n(n+1)(2n+1)}/6

Hence average = {n(n+1)(2n+1)}/6*n ; where n = quantity of numbers

As n=10 ; therefore avg = {10*(10+1)(2*10 + 1)} / (6 * 10)

= 10*11*21 / 60

= 38.5 ; Hence C

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Thank you for the great solutions and formulas!

Is there a formula if the first consecutive number is not 1?
What if (for example) the question says:

What is the average of the following numbers?
11^2, 12^2, 13^2...20^2.

---------
What I would do is calculate from 1^2 to 10^2 and from 1^2 to 20^2. and then substract


Do you think the GMAT can ask this or this type of questions always start with 1 as the first digit?
Thank you!!

A. 5.5
B. 25
C. 38.5
D. 55
E. 385

As the answer choices are quite spreaded, we can use approximation.

(1 + 4) + (9 + 16) + (25 + 36) + 49 + (64 + 81) + 100

We can add them roughly

5+25+60+50+150+100

390/10 = 38.5 approx.

(C)

My biggest problem is that sometime I dont see the easy solutions and end up wasting my time on prolonged calculations. Shortcuts are always useful. And I would also say that GMAT usually provides reasonably different answer choices so that shortcuts tend to work fine.

n is the last term

In this case, n=10

So \(\frac{10 * 11 * 21}{6} = 385\)

Average \(= \frac{385}{10} = 38.5\)

Yes, even if you do not know the formula, a bit of logic and number properties can help you solve the question. The numbers will be

1, 4, 9, 16, 25, 36, 49, 64, 81, 100
Note that as you go to higher numbers, the square spreads out more. the difference between 8^2 and 9^2 will be much more than the difference between 6^2 and 7^2.

Only 3 numbers are greater than 50 and that too not all very far from it but we do have a lot of very small numbers (1, 4, 9) so the average is certainly less than 50.

In increasing order, 30 lies about mid way (between 25 and 36) but deviation of the numbers to the right of it is much more than the deviation to the left of it. Hence, the average should certainly be greater than 25.

Answer (C)

Great alternative approach. Very insightful. Thank you.

The average is:

(1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100)/10

385/10 = 38.5

Answer: C

Asked: What is the average of the following numbers?
1^2, 2^2, 3^2...10^2.
Sum of {1^2, 2^2, 3^2...10^2} = n(n+1)(2n+1)/6 = 10*11*21/6 = 5*11*7 = 385
Average of {1^2, 2^2, 3^2...10^2} = 385/10 = 38.5

IMO C

Hi Vips,

The strategy applied by you will work if the answer choices are not similar for example
A. 39.5
B. 38.5

Are such questions common in Gmat?

P.S: My apporach to solution was just like yours

Thanks,
Pritish

Hi Pritish,

Truely speaking, from what I've seen and what I've heard 'experts' say is that GMAT doesnt test number crunching but the application of knowledge. For the given problem we were able to solve easily using number properties. if the numbers in options are closer as u mentioned, usually calculation would still be minimal on actual gmat problem.

Hi Craky,
I had the same problem too. Reading one of gmat blogs I understood that gmat will not test your calculation ability (how fast we can add or multiply numbers) rather it tests if we can apply any learnt math priniciple here.

In case you come across such questions, please feel free to share in the forum.

Some instances :
1. A problem where applying (a+b)(a-b) rule gave the answer quickly
2. Take out the common terms and break the numbers

Regards,
Pritish

Hi.. Can anyone tell me, what does n mean in the equations? n = total number????? or first term????

sum of square of number from 1,2,3...n = {n(n+1)(2n+1)}/6
average = {n(n+1)(2n+1)}/6 * 1/n = {(n+1)(2n+1)}/6

putting n =10 in the above expression we get
{(10+1)(2*10+1)}/6
= 11*21/6
= 38.5

Option C

Sum of the squares of first n natural numbers = \(\frac{n(n+1)(2n+1)}{6}\)

So, Sum of the squares of first 10 natural numbers = \(\frac{10*11*21}{6}\) = \(385\)

Average of these 10 numbers is 38.50, Answer must be (C)

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