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
Start free 5-day trial
Read 559 reviewsTarget Test Prep

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

tricky VIAC (Variables in the answer choices) question

Expert replies
Post new topic Post Reply
Previous Topic Next Topic

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members
Location: Vancouver, BC
Thanked: 5254 times
GMAT Score:770
by Brent@GMATPrepNow » Mon May 08, 2017 6:07 am
If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + 97 + 99, then 1² - 2² + 3² - 4² + 5² - 6² + . . . . . + 97² - 98² + 99² - 100² =

A) J² - K²
B) -50(J² - K²)
C) -K - J
D) K² - J²
E) (-J - K)²

Answer: C

Source: www.gmatprepnow.com
Difficulty level: 700
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members
Location: Vancouver, BC
Thanked: 5254 times
GMAT Score:770

by Brent@GMATPrepNow » Wed May 10, 2017 1:05 pm
Brent@GMATPrepNow wrote:If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + 97 + 99, then 1² - 2² + 3² - 4² + 5² - 6² + . . . . . + 97² - 98² + 99² - 100² =

A) J² - K²
B) -50(J² - K²)
C) -K - J
D) K² - J²
E) (-J - K)²
We have several differences of squares hiding in the expression 1² - 2² + 3² - 4² + 5² - 6² + . . . . . + 97² - 98² + 99² - 100²

1² - 2² + 3² - 4² + 5² - 6² + . . . . . + 97² - 98² + 99² - 100² = 1² - 2² + 3² - 4² + 5² - 6² + . . . . . + 97² - 98² + 99² - 100²
= (1 - 2)(1 + 2) + (3 - 4)(3 + 4) + (5 - 6)(5 + 6) + . . . . . + (97 - 98)(97 + 98) + (99 - 100)(99 + 100)
= (-1)(1 + 2) + (-1)(3 + 4) + (-1)(5 + 6) + . . . . . + (-1)(97 + 98) + (-1)(99 + 100)
= (-1)[(1 + 2) + (3 + 4) + (5 + 6) + . . . . . + (97 + 98) + (99 + 100)]
= (-1)(1 + 2 + 3 + 4 + . . . . . 97 + 98 + 99 + 100)

IMPORTANT: within the sum, 1 + 2 + 3 + 4 + . . . . . 97 + 98 + 99 + 100, we have all of the ODD integers from 1 to 99 inclusive, and we have all of the EVEN integers from 2 to 100 inclusive.

So, we can say that 1 + 2 + 3 + 4 + . . . . . 97 + 98 + 99 + 100 = K + J

So, we're replace 1 + 2 + 3 + 4 + . . . . . 97 + 98 + 99 + 100 with K + J.
We get: (-1)(1 + 2 + 3 + 4 + . . . . . 97 + 98 + 99 + 100) = (-1)(K + J)
= -K - J

Answer: C

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
GMAT Instructor
Posts: 2630
Joined: Wed Sep 12, 2012 3:32 pmFollowed by: 119 members
Location: East Bay all the way
Thanked: 625 times
GMAT Score:780

by Matt@VeritasPrep » Thu May 11, 2017 8:34 pm
I'm thinking ...

1² - 2² + 3² - 4² ± ... + 99² - 100² =>

(1² + 3² + 5² + ... + 99²) - (2² + 4² + ... + 100²) =>

(1² + 3² + 5² + ... + 99²) - ((1+1)² + (3+1)² + ... + (99+1)²) =>

(1² + 3² + 5² + ... + 99²) - (1² + 3² + ... + 99² + 2*1 + 2*3 + ... + 2*99 + 1 + 1 + ... + 1) =>

(1² + 3² + 5² + ... + 99²) - ((1² + 3² + 5² + ... + 99²) + 2*(1 + 3 + ... + 99) + 50) =>

-2*(1 + 3 + ... + 99) - 50 =>

-2*K - 50 =>

-K - K - 50

And since J = K + 50, we can write this as

-K - (K + 50)

or

-K - J
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
GMAT Instructor
Posts: 2630
Joined: Wed Sep 12, 2012 3:32 pmFollowed by: 119 members
Location: East Bay all the way
Thanked: 625 times
GMAT Score:780

by Matt@VeritasPrep » Thu May 11, 2017 8:37 pm
We could also cheat with a pattern:

n² - (n + 1)² =>

n² - (n² + 2n + 1) =>

-(2n + 1) =>

-(n + n + 1)

for any value of n.

Since we've got 1² - 2² + 3² - 4² ..., we've really got -(1 + 2) -(3 + 4) .... -(99 + 100), or -1 -2 -3 -4 .... - 99 - 100, or -(1 + 2 + 3 + ... + 100), or -(K + J), or -K - J.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 157
Joined: Sat Nov 19, 2016 5:34 amFollowed by: 4 members
Thanked: 2 times

by hazelnut01 » Thu May 11, 2017 9:44 pm
Matt@VeritasPrep wrote:I'm thinking ...

1² - 2² + 3² - 4² ± ... + 99² - 100² =>

(1² + 3² + 5² + ... + 99²) - (2² + 4² + ... + 100²) =>

(1² + 3² + 5² + ... + 99²) - ((1+1)² + (3+1)² + ... + (99+1)²) =>

(1² + 3² + 5² + ... + 99²) - (1² + 3² + ... + 99² + 2*1 + 2*3 + ... + 2*99 + 1 + 1 + ... + 1) =>

(1² + 3² + 5² + ... + 99²) - ((1² + 3² + 5² + ... + 99²) + 2*(1 + 3 + ... + 99) + 50) =>

-2*(1 + 3 + ... + 99) - 50 =>

-2*K - 50 =>

-K - K - 50

And since J = K + 50, we can write this as

-K - (K + 50)

or

-K - J
Hi Matt, How do we know J = K + 50?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1