The sum of the first 50 positive even integers is 2,550. What is the

Question

The sum of the first 50 positive even integers is 2,550. What is the sum of the first 50 odd integers?

(A) 2,000
(B) 2,497
(C) 2,500
(D) 2,501
(E) 2,549

FacelessMan · Accepted Answer

2 ways to solve this problem :

Sol1 :
Sum of first n even numbers = n(n+1) = 2550
Sum of first n odd numbers = n^2 = 50*50 = 2500

Sol2 :
Now the sum of first 50 even numbers can be represented as 2 + 4 + 6 + .... 100 = 2550
Now the sum of first 50 odd numbers can be represented as 1 + 3 + 5 + ... 99 = (2-1) + (4-1) + (6-1) + .... (100 -1)
= (2+4+6+...100) -1 -1 -1 ... 50 times = 2550 - 50 = 2500

Hence C.

BrentGMATPrepNow · Answer

A different approach:

Let's observe some sums of consecutive odd integers and  look for a pattern

Sum of first 1 odd integers: 1 = 1
Sum of first 2 odd integers: 1 + 3 = 4
Sum of first 3 odd integers: 1 + 3 + 5 = 9
Sum of first 4 odd integers: 1 + 3 + 5 + 7 = 16
Sum of first 5 odd integers: 1 + 3 + 5 + 7 + 9 = 25
Sum of first 6 odd integers: 1 + 3 + 5 + 7 + 9 + 11 = 36
.
.
.
See the pattern?
.
.
.
  Sum of first n odd integers: n²

So, the sum of the first 50 odd integers = 50² = 2500

Answer:  C

Cheers,
Brent

fantaisie · Answer

I chose the solution similar to one presented by FacelessMan:

Completely ignore the information about the sum of the first 50 even integers since there is a quicker way to solve the problem:

Sum of the odd numbers of the first consecutive integers n is  n^2 , thus sum of odd integers =  50^2  = 2500

Answer:  C

Abhishek009 · Answer

The sum of the first n odd integers equals  n^2

= 50^2

= 2500

Thus, answer will be (C)

roadrunner · Answer

Sum of first 50 odd integers = sum of arithmetic series = average of the series * n

average = (first + last)/2 = (1+99)/2 =50
n = 50

==> Sum of 50 odd integers = 50*50 = 2500

generis · Answer

Sum of first 50 positive even integers is  \frac{(2 + 100)}{2}*(50) , which =

(51) (50) = 2,550

Sum of first 50 positive odd integers is  \frac{(1 + 99)}{2}*(50) , which =

(50) (50). There's  one extra 50  in the even integer sum.* 2,550 - 50 = 2500.

Answer (C).

*(Sum of first 50 positive integers) - (Sum of first 50 odd integers) =

(51)(50) - (50)(50) =

(50)(51-50) =

(50)(1) = 50 more in sum of even integers than in sum of odd integers. (2550 - 50) = 2500.

BrentGMATPrepNow · Answer

The sum of the first 50 positive EVEN integers is 2,550  
So,  2  +  4  +  6  + . . . +  98  +  100  =  2550

What is the sum of the first 50 odd integers?  
So, we want the value of 1 + 3 + 5 + . . . + 97 + 99

IMPORTANT: Notice that:
1 =  2  - 1
3 =  4  - 1
5 =  6  - 1
.
.
.
97 =  98  - 1
99 =  100  - 1

So, 1 + 3 + 5 + . . . + 97 + 99 = ( 2  - 1) + ( 4  - 1) + ( 6  - 1) + . . . + ( 98  - 1) + ( 100  - 1) 
=  2  +  4  +  6  + . . . +  98  +  100  ] - 1 - 1 - 1 - 1 . . . . -1 - 1
=  2550  ] - 1 - 1 - 1 - 1 . . . . -1 - 1

ASIDE: How many 1s are we subtracting from  2550  ?
Well, there are 50 even integers from 2 to 100, so we are subtracting fifty 1's
So, we get: 
1 + 3 + 5 + . . . + 97 + 99 =  2550  ] - 1 - 1 - 1 - 1 . . . . -1 - 1
=  2550  ] - 50
= 2500

Answer: C

Cheers,
Brent

Pritishd · Answer

We are asked to find the sum of first 50 odd integers i.e.  1 + 3 + 5 + 7 + .... + n

1. We will first find out  n 
2. We will use the derived form of  mean = \frac{sum-of-elements}{number-of-elements}  = mean * number of elements to find the sum

n   =  first term + (number of terms - 1) * common difference  =   1 + (50 - 1) * 2 = 99

mean = \frac{first-term + last-term}{2}  (for an evenly spaced set of numbers)  = \frac{1 + 99}{2} = 50

sum of elements  = 50 * 50 = 2,500

Ans.  C

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

The sum of the first 50 positive even integers is 2,550. What is the

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 50 positive even integers is 2,550. What is the

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