Live GMAT classes with Target Test Prep start Nov 21 and Nov 22. First class free.

See classes
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.

Newsweek Readers’ Choice Awards 2026 winner: Target Test Prep, #1 Test Prep

Target Test Prep was voted #1 Best Test Prep in the 2026 Newsweek Readers’ Choice™ Awards.

★★★★★5.0563 reviews
GMATLiveTeachStarts Nov 22
Josh Braslow, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Josh Braslow, 20+ years in test prep

Nov 22 · Josh Braslow
Sun · 9:30 AM to 12:30 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
EALiveTeachOnDemandStart anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

EXECUTIVE ASSESSMENT

with

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
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
GMATOnDemandStart 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

sum of consecutive terms

Expert replies
by singhsa » Mon Sep 13, 2010 10:25 pm
The sum of n consecutive positive integers is 45. What is the value of n?
(1) n is odd
(2) n >= 9
Join the discussion
Source: — Problem Solving |

by hi.itz.mani » Mon Sep 13, 2010 10:51 pm
I guess this a data sufficiency question and not PS

sum of n consecutive numbers starting with A as first number = ( (2A + n-1 ) / 2 ) n

( Sum of AP series ( ( 2a + (n-1)d )/ 2 )n here since its consecutive therefore d = 1 )

its given that the sum is 45

hence (2a + N - 1 )N = 90

factor of 90 = 2 * 5 * 9

hence n can be either 2 or 5 or 9

for n is odd we can have n = 5 or n = 9

for n >=9 we will have just n =9

so the second choice is sufficient.

Thanks!
-Mani
Join the discussion

by rb90 » Mon Sep 20, 2010 11:36 pm
Is B the correct answer?because i am getting C as the answer.Here's how:

n(n+1)/2=45
= n^2 + n = 90
= n^2 + n - 90 = 0
= (n+10)(n-9) = 0
which gives us n=9 (since n cant be -ve)

Is this correct?
Please let me know
Thanks
Join the discussion

by singhsa » Tue Sep 21, 2010 2:31 am
rb90 wrote:Is B the correct answer?because i am getting C as the answer.Here's how:

n(n+1)/2=45
= n^2 + n = 90
= n^2 + n - 90 = 0
= (n+10)(n-9) = 0
which gives us n=9 (since n cant be -ve)

Is this correct?
Please let me know
Thanks
I think n(n+1)/2 is a formula used for the FIRST n nos. OA is definitely correct.
Join the discussion

by Brian@VeritasPrep » Tue Sep 21, 2010 9:11 am
Hey everyone:

The OA is correct here and it's a pretty good question. Ultimately this one is more of a divisibility problem than anything else.

We need the sum of the digits to be 45, and that sum is going to be made of up of x + (x+1) + (x+2).... for n number of values. x in this case will be the first value, and each one goes up by one from there.

In the first statement, we know that we need an odd number of terms, so we could have:

x + x+1 + x+2 = 45

or

x + x+1 + x+2 + x+3 + x+4 = 45

The key on these is to determine whether, once we combine like terms, we'll get integer values of x in each case. For the first, we'd have:

3x + 3 = 45
3x = 42
because 42 is divisible by 3, we'll end up with an integer value of x, so we CAN have 3 consecutive integers.

For the second, we'd have:

5x + 10 = 45
5x = 35
because 35 is divisible by 5, we'll have an integer value of x, so we CAN have 5 consecutive integers and therefore statement 1 is not sufficient - we can have either 3 or 5 values.


For Statement 2, we know that the minimum number of values is 9, which would give us:

9x + (1+2+3+4+5+6+7+8) = 45
9x + 36 = 45
9x = 9
x = 1
So we can have 9 values...but can we have 10 or more?

For 10 values, the setup would be:

10x + 36 (from before) + 9 (for that last new integer) = 45
10x + 45 = 45
10x = 0
x = 0, which is not possible because we need POSITIVE integers. Therefore, the only number of values we can have is 9, because anything larger would give us either 0 or a negative number. Because statement 2 is sufficient, the correct answer is B.
Brian Galvin
GMAT Instructor
Chief Academic Officer
Veritas Prep

Looking for GMAT practice questions? Try out the Veritas Prep Question Bank. Learn More.
Join the discussion

by singhsa » Wed Sep 22, 2010 4:55 pm
Brian@VeritasPrep wrote:Hey everyone:

The OA is correct here and it's a pretty good question. Ultimately this one is more of a divisibility problem than anything else.

We need the sum of the digits to be 45, and that sum is going to be made of up of x + (x+1) + (x+2).... for n number of values. x in this case will be the first value, and each one goes up by one from there.

In the first statement, we know that we need an odd number of terms, so we could have:

x + x+1 + x+2 = 45

or

x + x+1 + x+2 + x+3 + x+4 = 45

The key on these is to determine whether, once we combine like terms, we'll get integer values of x in each case. For the first, we'd have:

3x + 3 = 45
3x = 42
because 42 is divisible by 3, we'll end up with an integer value of x, so we CAN have 3 consecutive integers.

For the second, we'd have:

5x + 10 = 45
5x = 35
because 35 is divisible by 5, we'll have an integer value of x, so we CAN have 5 consecutive integers and therefore statement 1 is not sufficient - we can have either 3 or 5 values.


For Statement 2, we know that the minimum number of values is 9, which would give us:

9x + (1+2+3+4+5+6+7+8) = 45
9x + 36 = 45
9x = 9
x = 1
So we can have 9 values...but can we have 10 or more?

For 10 values, the setup would be:

10x + 36 (from before) + 9 (for that last new integer) = 45
10x + 45 = 45
10x = 0
x = 0, which is not possible because we need POSITIVE integers. Therefore, the only number of values we can have is 9, because anything larger would give us either 0 or a negative number. Because statement 2 is sufficient, the correct answer is B.
Like always, wonderful explanation Brian!!!
Join the discussion