BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
GMATLiveTeach Starts Oct 17
Chris Peckover, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Chris Peckover, 100th-Percentile GMAT Scorer

Oct 17 · Chris Peckover
Sat · 11:00 AM to 2:00 PM ET
Oct 20 · Chris Peckover
Tue, Thu · 8:00 to 10:00 PM ET
Oct 25 · Josh Braslow
Sun · 1:00 to 4:00 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemand 5 seats left Start anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

Que: If set N consists of odd numbers of consecutive integers, starting with 1, what is the difference between...

Expert replies
by Max@Math Revolution » Sat Dec 26, 2020 11:12 pm

Timer

00:00

Answers

A

B

C

D

E

Stats

Difficulty—

Que: If set N consists of odd numbers of consecutive integers, starting with 1, what is the difference between the average of the odd integers and the average of the even integers in set N?

(A) −1
(B) \(\frac{1}{2}\)
(C) 0
(D) 1
(E) 2
Join the discussion
Source: — Problem Solving |

Solution: Since the set starts with an odd number (1) and has an odd number of integers, the set would end with an odd number, too.

Let’s see the set. Set N: {1, 2, 3, 4, 5, ..., (2n + 1)}, where n is a positive integer.

=> Number of odd terms is one more than the number of even terms.

Thus, the number of odd terms = (n + 1) and we derive a new set from set N, that is {1,3,…, 2n+1}.

Then we get the average of the odd integers = \(\frac{\left[1\ +\ 3\ +\ ......\ +\ \left(2n\ +\ 1\right)\right]}{\left(n\ +\ 1\right)}\) and since \(1\ +\ 3\ +\ ......\ +\ \left(2n\ +\ 1\right)\ =\ \left(n\ +\ 1\right)^2\), we get \(\frac{\left[1\ +\ 3\ +\ ......\ +\ \left(2n\ +\ 1\right)\right]}{\left(n\ +\ 1\right)}=\frac{\left(n\ +1\right)^2}{\left(n\ +\ 1\right)}\ =\ n\ +\ 1\)

The number of even terms = n and we derive a new set from set N, that is {2,4,…, 2n}.

Then we get the average of the even integers = \(\frac{\left(2\ +\ 4\ +\ ....\ +\ 2n\right)}{n}=\frac{2\cdot\left(1\ +\ 2\ +\ ....\ +\ n\right)}{n}\) and since \(1\ +\ 2\ +\ ........\ +\ n\ =\ \frac{n\left(n\ +\ 1\right)}{2}\), we get \(\frac{\left(2\ +\ 4\ +\ .......\ +\ 2n\right)}{n}=\frac{2\cdot\left(1\ +\ 2\ +\ .......\ +\ n\right)}{n}=\frac{n\left(n\ +\ 1\right)}{n}\ =\ n\ +1\)

Therefore, we get the difference of the average of the odd integers and the average of the even integers in set N=(n + 1) - (n + 1) = 0

Say there are only three terms in the set

S = {1, 2, 3}

=> X = \(\frac{\left(1\ +\ 3\right)}{2}=\frac{4}{2}=2\)

=> Y = 2

=> X – Y = 0

Again, say there are five terms in the set

S = {1, 2, 3, 4, 5}

=> X = \(\frac{\left(1\ +\ 3\ +\ 5\right)}{3}=\frac{9}{3}=3\)

=> Y = \(\frac{\left(2\ +\ 4\right)}{2}=\frac{6}{2}=3\)

=> X – Y = 0

Therefore, C is the correct answer.

Answer C
Join the discussion

Max@Math Revolution wrote: ↑
Sat Dec 26, 2020 11:12 pm
Que: If set N consists of odd numbers of consecutive integers, starting with 1, what is the difference between the average of the odd integers and the average of the even integers in set N?

(A) −1
(B) \(\frac{1}{2}\)
(C) 0
(D) 1
(E) 2
Solution:

We can let set N be {1, 2, 3}. So the average of the odd integers is (1 + 3)/2 = 2, and the average of the even integers is 2. We see that the difference between the two averages is 0. Although we can say the correct choice must be B, let’s make sure it is the case by using another example.

If we let set N be {1, 2, 3, 4, 5}, the average of the odd integers is (1 + 3 + 5)/3 = 3 and the average of the even integers is (2 + 4)/2 = 3. Again we see that the difference between the two averages is 0.

Answer: C

Scott Woodbury-Stewart
Founder and CEO
[email protected]

Image

See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews

ImageImage
Join the discussion