The sum of the first in positive integers is given by n(n + 1)/2. What

Question

The sum of the first in positive integers is given by n(n + 1)/2. What is the sum of the first 100 positive odd integers? A. 10,100 B. 10,000 C. 9,950 D. 9,900 E. 5,050 2024-01-30_01-18-49.png

Bunuel · Accepted Answer

There are many ways to solve this question. However, let's try solving it using the formula in the stem to flex our number properties, logic, and algebra skills.

The sum of the first 100 positive odd integers is 1 + 3 + ... + 199.

The sum of the integers from 1 to 200, 1 + 2 + ... + 200, equals:

n(n + 1)/2 = 200(200 + 1)/2 = 100 * 201.

The sum of the even integers from 2 to 200, 2 + 4 + ... + 200 = 2(1 + 2 + ... + 100), equals:

2 * n(n + 1)/2 = 2 * 100(100 + 1)/2 = 100*101.

Therefore, the sum of the first 100 positive odd integers is the first sum minus the second sum:

100 * 201 - 100 * 101 =

= 100(201 - 101) =

= 10,000.

Answer: B.

gmatophobia · Answer

The sum of the first  n  positive odd integers is given by  n^2

Here   n = 100

The sum of the first 100 positive odd integers =  100^2 = 10^4

Option B

Archit3110 · Answer

sum will be
n/2 *( 2a+(n-1)*d)
n is 100 ; a is 1 d is 2
solve we get 
10^4
option B

trepsixore · Answer

1+3+5+7+....193+195+197+199 Notice the pattern 1+199=200; 3+197=200; 5+195=200 so on, so forth. Hence there are 50 pairs of 200 which is equal to 10,000

GMATinsight · Answer

A few sequences that one must know for GMAT are mentioned here.

We can discuss three methods here

M-1:

Using the formula given in the attached picture i.e. sum of first n odd integers = n^2

i.e. Sum of first 100 odd integers = 100^2 =10,000

M-2:

We could use the formula given here which is for the sum of n consecutive integers

Sum of first 100 odd integers = Sum of first 200 consecutive numbers - sum of first 100 even numbers
Sum of first 100 odd integers =   -  =   - 2*  = 10,000

M3:   
Use Arithmetic progression sum formula which is {first term+Last term)*number of terms / 2

i.e. 1+3+5+-----+199 = (1+199)*100/2 =10,000

Answer: Option B

gladgaming · Answer

Sum of n (200) numbers:( 100 odd + 100 even ) 200x201/2 = 20,100
-
Sum of n(100) even numbers: 100(101) = 10,100

= 10,000

Usernamevisible · Answer

you can also simply use

n/2(2a+(n-1)d)

The sum of the first in positive integers is given by n(n + 1)/2. What

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

The sum of the first in positive integers is given by n(n + 1)/2. What

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards