Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

In a certain sequence of numbers, a1, a2, a3, ..., an, the [#permalink]
06 Oct 2013, 09:30

2

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

55% (hard)

Question Stats:

58% (02:41) correct
42% (01:46) wrong based on 138 sessions

In a certain sequence of numbers, a1, a2, a3, ..., an, the average (arithmetic mean) of the first m consecutive terms starting with a1 is m, for any positive integer m. If a1=1, what is a10?

Re: In a certain sequence of numbers, a1,a2,a3,...,an, the avera [#permalink]
06 Oct 2013, 09:45

3

This post received KUDOS

nave81 wrote:

In a certain sequence of numbers, a1,a2,a3,...,an, the average (arithmetic mean) of the first m consecutive terms starting with a1 is m, for any positive integer m. If a1=1, what is a10?

A. 100 B. 55 C. 21 D. 19 E. 1

I got D: 19.

As stated above, (a1+a2+a3......+a10)/10=10 therefore a1+a2+a3.......a10=100 (1) using the same logic, we got a1+a2+a3..........+a9=81 (2) (2)-(1) we got a10=19

Re: In a certain sequence of numbers, a1,a2,a3,...,an, the avera [#permalink]
06 Oct 2013, 09:49

6

This post received KUDOS

Expert's post

nave81 wrote:

In a certain sequence of numbers, a1,a2,a3,...,an, the average (arithmetic mean) of the first m consecutive terms starting with a1 is m, for any positive integer m. If a1=1, what is a10?

A. 100 B. 55 C. 21 D. 19 E. 1

The average of the first 9 consecutive terms starting with a1 is 9 --> \(\frac{a_1+a_2+a_3+...+a_{9}}{9}=9\) --> \(a_1+a_2+a_3+...+a_{9}=81\).

The average of the first 10 consecutive terms starting with a1 is 10 --> \(\frac{a_1+a_2+a_3+...+a_{10}}{10}=10\) --> \(a_1+a_2+a_3+...+a_{10}=100\).

Subtract the first equation from the second: \(a_{10}=100-81=19\).

Re: In a certain sequence of numbers, a1,a2,a3,...,an, the avera [#permalink]
06 Oct 2013, 21:42

1

This post received KUDOS

nave81 wrote:

In a certain sequence of numbers, a1,a2,a3,...,an, the average (arithmetic mean) of the first m consecutive terms starting with a1 is m, for any positive integer m. If a1=1, what is a10?

A. 100 B. 55 C. 21 D. 19 E. 1

The average of 1st 10 numbers is 10

Hence, The average of 5th and 6th term is 10 and therefore 5th term is 9 and 6th term is 11 Extrapolating this trend we get 10th term = 19 i.e. 9,11,13,15,17,19

Re: In a certain sequence of numbers, a1,a2,a3,...,an, the avera [#permalink]
06 Oct 2013, 22:17

4

This post received KUDOS

Expert's post

nave81 wrote:

In a certain sequence of numbers, a1,a2,a3,...,an, the average (arithmetic mean) of the first m consecutive terms starting with a1 is m, for any positive integer m. If a1=1, what is a10?

A. 100 B. 55 C. 21 D. 19 E. 1

A slight different approach. Sum of first m numbers is: \([(First term\)+\(Last term)\)\(/\)\(2]\)*\(m\). Here it will be \((a1+a10)/2\)*\(m\). Also, since the average is given as m, therefore the sum would be \(m^2\).

On solving, \(a1+a10=2m\) \(a1+a10=20\) or \(a10=20-1=19\) _________________

Re: In a certain sequence of numbers, a1,a2,a3,...,an, the avera [#permalink]
07 Oct 2013, 23:33

Is it appropriate to do some good guess work out by looking at the options? For example if a1=1, and if a2, a3 ... a10 are in sequence then there are 3 possiblities for the answer of a10. 1st is that a1.. a10 are the first consecutive 10 nos. in that case the ans should be 10 which is not one of the options. 2nd is that a1... a10 could even consecutive nos. in that case nth term equals to 2n, so in our case 10 th term should be 2*10= 20. not one of the options. 3rd is that a1...a10 could be odd consecutive nos. in that case nth term equals to 2n-1, so in our case 10th term should be 2*10-1=19. Option D our ans.

Please correct me if my thought process is wrong.

Bunuel wrote:

nave81 wrote:

In a certain sequence of numbers, a1,a2,a3,...,an, the average (arithmetic mean) of the first m consecutive terms starting with a1 is m, for any positive integer m. If a1=1, what is a10?

A. 100 B. 55 C. 21 D. 19 E. 1

The average of the first 9 consecutive terms starting with a1 is 9 --> \(\frac{a_1+a_2+a_3+...+a_{9}}{9}=9\) --> \(a_1+a_2+a_3+...+a_{9}=81\).

The average of the first 10 consecutive terms starting with a1 is 10 --> \(\frac{a_1+a_2+a_3+...+a_{10}}{10}=10\) --> \(a_1+a_2+a_3+...+a_{10}=100\).

Subtract the first equation from the second: \(a_{10}=100-81=19\).

Re: In a certain sequence of numbers, a1,a2,a3,...,an, the avera [#permalink]
24 Oct 2013, 21:00

Expert's post

nave81 wrote:

In a certain sequence of numbers, a1,a2,a3,...,an, the average (arithmetic mean) of the first m consecutive terms starting with a1 is m, for any positive integer m. If a1=1, what is a10?

A. 100 B. 55 C. 21 D. 19 E. 1

Or use pattern recognition to solve it. a1 = 1 a2 = ? the mean of a1 and a2 must be 2. Since a1 is 1, a2 must be 3 a2 = 3 a3 = ? the mean of a1, a2, a3 = 3. Since a1 and a2 are 1 and 3, a3 must be 5 to give a mean of 3 a3 = 5 a4 = ? the mean of a1, a2, a3 and a4 is 4. Since we have 1, 3, 5 and we need a mean of 4, a4 must be 7 (use deviation from mean to figure out each of these in a couple of secs) a4 = 7

We see the pattern: 1, 3, 5, 7 ... nth term is given by 2n - 1. a10 is 2*10 - 1 = 19 _________________

Re: In a certain sequence of numbers, a1, a2, a3, ..., an, the [#permalink]
18 Dec 2013, 16:42

nave81 wrote:

In a certain sequence of numbers, a1, a2, a3, ..., an, the average (arithmetic mean) of the first m consecutive terms starting with a1 is m, for any positive integer m. If a1=1, what is a10?

A. 100 B. 55 C. 21 D. 19 E. 1

Sum of 'm' terms / m = Average of m

Now, average of m is also = (L + 1)/2 * (m) Where, L is the last number of the sequence and 'm' is the number of terms So then we have that L = 2m-1

Re: In a certain sequence of numbers, a1, a2, a3, ..., an, the [#permalink]
20 Feb 2014, 13:29

Instead of consecutive terms I would say evenly spaced set in the question. Anyway, from the example we know that they are talking about consecutive odd numbers (n^2 is the sum of the first n positive odd numbers) therefore, the largest number of the first 10 odd integers is 19.

Re: In a certain sequence of numbers, a1, a2, a3, ..., an, the [#permalink]
20 Feb 2014, 14:57

Expert's post

2 ways algebra focused and 1 way emphasizing pattern recognition:

1) by given formula:the mean of a1 is 1 the mean of a1+ a2 is 2 the mean of a1+a2+ a3 is 3 and so on. likewise for mean of numbers up toaX it is X.

if mean of numbers up to a10 is 10 then sum of numbers for a10 is 10*10. likewise, the sum of numbers for a9 is 9*9.

100 -81 is the final term: a10 =19

Formula for sum of number (( first + last)/2 )*m number of terms.

2) more elegant way:

we already know that a1 = 1 and formula for sum of numbers. ((a1 +a10)/2)*m

since average of numbers = m then sum of number = m*m

so we have an algebraic equivalency. we know a1= 1 , m= 10 thus only 1 variable to determine is a10.

3) pattern recognition

a1 = 1 a2 = ? since the mean of a1 and a2 is 2. Since a1 is 1, a2 must be 3 a2 = 3 a3 = ? since the mean of a1, a2, a3 = 3. Since a1 and a2 are 1 and 3, a3 must be 5 to give a mean of 3 a3 = 5

In a certain sequence of numbers, a1,a2,a3,...,an, average [#permalink]
20 Jul 2014, 03:28

In a certain sequence of numbers, a1,a2,a3,...,an, the average (arithmetic mean) of the first m consecutive terms starting with a1 is m, for any positive integer m. If a1=1, what is a10? A. 100 B. 55 C. 21 D. 19 E. 1

Re: In a certain sequence of numbers, a1, a2, a3, ..., an, the [#permalink]
20 Jul 2014, 03:30

Expert's post

goodyear2013 wrote:

In a certain sequence of numbers, a1,a2,a3,...,an, the average (arithmetic mean) of the first m consecutive terms starting with a1 is m, for any positive integer m. If a1=1, what is a10? A. 100 B. 55 C. 21 D. 19 E. 1

Merging topics.

Please refer to the discussion above. _________________

How the growth of emerging markets will strain global finance : Emerging economies need access to capital (i.e., finance) in order to fund the projects necessary for...

One question I get a lot from prospective students is what to do in the summer before the MBA program. Like a lot of folks from non traditional backgrounds...