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 7 seats left
Chris Peckover
NEXT LIVE COHORT

Oct 13 to Jan 7, 2027

with Chris Peckover

Schedule
Tue, Thu · 8:00 to 10:00 PM ET
Included
40 live hours + 6 months of GMAT OnDemand
  • Live instruction and real-time questions
  • Class recordings and assigned practice
View class & enroll
Limited cohort · enrollment openTarget Test Prep
EALiveTeach 5 seats left
Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
Live EA class + 6 months of EA OnDemand
  • Expert-led weekly online sessions
  • EA Masterclass access between classes
View EA class & enroll
Limited cohort · enrollment openTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

130-point 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.

Medium level - sum of integers

Expert replies
by Brent@GMATPrepNow » Thu Mar 05, 2009 1:28 pm
If the sum of the integers from 1 to 100 is 5050, then the sum of the odd integers from 1 to 99 is
(A) 2500
(B) 2525
(C) 2550
(D) 2575
(E) 2600
Brent Hanneson - Creator of GMATPrepNow.com
Image
Join the discussion
Source: — Problem Solving |

by Brent@GMATPrepNow » Thu Mar 05, 2009 4:42 pm
It's dinner time, so I better post my solution before I go.
The answer is A.

Image
Brent Hanneson - Creator of GMATPrepNow.com
Image
Join the discussion

by pakaskwa » Thu Mar 05, 2009 7:25 pm
For any arithmetic progression, the sum is

S = n(a1+an)/2

Where n is the number of sequence. For the question, there are 50 numbers from 1,3,5,... to 99, half of 100.
a1=1, a50=99,

So S=50(1+99)/2=2500

But the interesting part is that, why the sum is not half of 5050?
Join the discussion

by satish.nagdev » Fri Mar 06, 2009 12:38 am
pakaskwa wrote: But the interesting part is that, why the sum is not half of 5050?
because every even integer falls after odd and its value is +1 of the odd number, so 50 even numbers in set [1 to 100] result in sum of even numbers = sum of odd numbers +50

its like
1 2
3 4
....
.....
99 100

and thanks pakaskwa your formula for finding sum was quick one :)
Join the discussion

by pakaskwa » Fri Mar 06, 2009 3:08 pm
That makes sense! I didn't think enough.

By the way, I think Brent's idea is to let people know that, if they don't remember the formula, how they can get the correct numbers.
Join the discussion

Re: Medium level - sum of integers

by logitech » Sun Mar 08, 2009 3:39 pm
Brent Hanneson wrote:If the sum of the integers from 1 to 100 is 5050, then the sum of the odd integers from 1 to 99 is
(A) 2500
(B) 2525
(C) 2550
(D) 2575
(E) 2600
1+2+...+n = (n)x(n+1)/2

1+2+3+4+5+6+7+8+9+10 = (10)(10+1)/2 = 55

1+3+5+7+9 = 25

2+4+6+8+10 = 2( 1+2+3+4+5) = 2x(5)x(5+1)/2 = 30

I used this simple set to explain you the logic behind the calculation.

So:

2+4+6+..100 = 2( 1+2+3...+50) = 2x(50)(51)/2 = 50x51

So the sum of odd numbers is:

5050 - ( SUM OF EVEN ) = (101)x(50) - (50) x (51 ) = 50x50 = 2500
LGTCH
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"
Join the discussion

by deepoe » Thu Apr 16, 2009 2:59 am
I thought this


1+3+5+7+9 = 25


First digit = 10 numbers
Second digit = 10 numbers

So 10x10x25 = 2500

But is this a good way to calculate it?
Join the discussion