A sequence of odd integers begins at 61 and ends at 119. What is the s

Question

A sequence of odd integers begins at 61 and ends at 119. What is the sum of all numbers in this sequence?

A) 2610
B) 2700
C) 2790
D) 5400
E) 6000

Can someone tell me a quick way to find the number of odd terms between 61 to 119.

noTh1ng · Accepted Answer

Another way of finding out how many terms are there:
Last Term = First Term + Common difference * (n-1)

119 = 61 + 2*(n-1) 
 119 = 61 + 2n - 2 
 58 = 2n - 2 
 60 = 2n 
 n = 30

Multiply 30 with the average from (61+119)/2 = 90 => 90*30 = 2700

EMPOWERgmatRichC · Answer

Hi rohankant12,

There are a couple of different ways to find the total number of terms involved, but here's a relatively simple one:

How many odd integers are there from 61 to 69, inclusive?  5 
Shouldn't that be the SAME number of odd integers from 71 to 79, inclusive?  YES 
...
How about from 111 to 119, inclusive?  Again, 5

How many 'sets' of odd integers are there in this group?  6 sets of 5 integers each = 30 odd integers

GMAT assassins aren't born, they're made,
Rich

bhaskar438 · Answer

An odd number,N, can be represented as N=2q+1, where q is a non-negative integer.
61=2(30)+1 and 119=2(59)+1
The number of terms is equal to 59-30+1, which is 30.

bhaskar438 · Answer

Let's say you wanted to find all the terms between 61 to 119 inclusive which is 5 more than a multiple of 3.
You can do it the following way:
1. 61<=3q+5<=119 
2. 56<=3q<=114 
3. 18.something<=q<=28 
q can take values from 19 to 28 inclusive, which means there are 10 terms.

mejia401 · Answer

(\frac{n+1}{2})^2-(\frac{m-1}{2})^2 , where  m  and  n  are the even integer boundaries.
 (60)^{2}-(30)^{2}=3600-900=2700

This is possible because the difference of the boundaries automatically include the first and last odd digits. Plus, in this way, knowing that odd (or even) sums are half the total number of integers eliminates three or four mental calculations right off the bat.

Kr, 
Mejia

ersheet · Answer

or you can simply subtract the sum of even integers from the total sum of all the numbers that lie between 61 and 119:
-for sum of all the numbers between 61 and 119: 
n=119-61+1=59
sum=59(61+119)/2=5310
-for sum of the even numbers between 61 and 119:
these would be from 62 to 118
we can write the sum of these numbers as 2(31+32+33+....+57+58+59)
=2{29(31+59)/2}=2610
in this case n=59-31+1=29

So the required sum=5310-2610=2700

nishantdoshi · Answer

can someone please post a detailed solution

KVISHUINFO · Answer

Here is a quick solution:

1. Find the total numbers of odd integers

(119-61)/2 +1 = 30

2. Find the average

(119+61)/2 = 90

3. Sum of the odd integers

90*30 = 2700

Ans B

Posted from my mobile device

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A sequence of odd integers begins at 61 and ends at 119. What is the s

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A sequence of odd integers begins at 61 and ends at 119. What is the s

Prep Toolkit

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