The sum of the first n consecutive positive odd integers is

Question

The sum of the first  n  consecutive positive odd integers is equal to  n^2 . What is the sum of all odd integers between 13 and 39, inclusive ?

A. 351
B. 364
C. 410
D. 424
E. 450

M20-31

Bunuel · Accepted Answer

Official Solution:

The sum of the first  n  consecutive positive odd integers is equal to  n^2 . What is the sum of all odd integers between 13 and 39, inclusive ?

A. 351
B. 364
C. 410
D. 424
E. 450

The sum of all odd integers between 13 and 39, inclusive is equal to  the sum of all odd integers from 1 to 39, inclusive   minus   the sum of all odd integers from 1 to 11, inclusive .
 
There are 20 odd integers from 1 to 39, inclusive, so the sum of all odd integers from 1 to 39, inclusive is  20^2 .
 
There are 6 odd integers from 1 to 11, inclusive, so the sum of all odd integers from 1 to 11, inclusive is  6^2 .
 
Thus, the required sum is  20^2 - 6^2 = 364 .

Answer: B

EMPOWERgmatRichC · Answer

Hi All,

These types of "sum of a sequence" questions can be approached in a number of different ways - you can use various formulas or you can approach this prompt by pattern-matching and minimizing the math that you have to do. For this question, you can use "bunching"....

We're dealing with a sequence of CONSECUTIVE ODD INTEGERS: 13 to 39, inclusive. We're asked for the SUM of this group.

1) Start with the sum of the smallest and the biggest: 13 + 39 = 52
2) Now look at the 'next smallest' and the 'next biggest': 15 + 37 = 52

From this, you can see that you're just going to be adding up a bunch of 52s. We DO have to check to see if there's a "middle" term in this sequence that doesn't get "bunched" though. To determine if that middle term exists, we just have find the last few 52s in the group....

21 and 31
23 and 29
25 and 27

Now we have proof that there is no middle term. We have 7 bunches of 52.

7(52) = 364

Final Answer:  B

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

Futuristic · Answer

I would suggest using the formula, for the simple reason that instead of an AP the problem might refer to some fancy formula for an unknown sequence of numbers. 

Anyway here we go...

For odd integers till 39, number of integers = 20
Sum = 20^2 = 400
For odd integers till (but not including 13), number of integers = 6
Sum = 6^2 = 36
Difference = 364 = B

Shahalikhan · Answer

13 & 39 inclusive

sum must be sum upto 39-sum upto 11

11 is (11+1)/2=6th term

39 is 20th term

20^2-11^2=364

Rohstar750 · Answer

Hi Bunnel,

I am struggling hard to understand why we cant treat this as a function question and take N^2 as a function to calulate 14^2 (14 are total odd integers between 13 to 39 and inclusive).

Bunuel · Answer

Becasue we are told that "The Sum of  first  N consecutive odd integers is N^2", not that the sum of any N consecutive odd integers is N^2.

siddhantvarma · Answer

To determine the sum of all odd integers between 13 and 39, inclusive, we first need to identify these integers. The odd integers in this range are:

13,15,17,19, .... ,39

These integers form an arithmetic sequence where the first term a=13 and the common difference d=2.

We can find the number of terms (n) in this sequence using the formula for the n-th term of an arithmetic sequence: A(n)=a+(n−1)d

Setting A(n)=39, the last term, we can solve for n:

39=13+(n−1)⋅2 => n=14There are 14 odd integers in the range from 13 to 39.

The sum of an arithmetic sequence is given by the formula: S(n)=(n*(a+l))/2

Substituting the values we found:

S(14)=(14*(13+39))/2 = 364Therefore, the sum of all odd integers between 13 and 39, inclusive, is 364 (B) .

sajibmannan · Answer

Sum of all odd integers between 13 and 39 is an example of an evenly spaced set.

Concept

Sum of evenly spaced set  =mean*n = \frac{first term + last term}{2}*n

13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39 = n = 14

Here, first term = 13, and last term = 39

Sum of evenly spaced set  = \frac{13+39}{2}*14

\frac{52}{2}*14

Answer:   364

Bhavya30301 · Answer

Hi. Why are we not taking 1 to 12 inclusive. Why have we ignored 12 here. Its an even number but why is it being ignored while calculating 6 as the number of terms?.

Bunuel · Answer

Because we are summing odd integers only, 12 is even, so it is not included.

When counting odd numbers from 1 to 11 inclusive, we get: 1, 3, 5, 7, 9, 11. That gives 6 terms.

12 is not an odd number, so it does not count toward the total.

Bhavya30301 · Answer

Ok got it. So it is like if we are calculating consecutive odd numbers in a set then we should take odd to odd inclusive. And for consecutive even numbers, we should take even to even inclusive.

essetempore · Answer

I have an easier method to approach this question: 
Odd consecutive numbers from 13 to 39 are: 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, and 39 (14 numbers)
Since they are evenly spaces- we can do= Median x No. of terms
Here, Median would be: (25 + 27)/ 2 = 26 
and then 26 x 14 (no of terms)= 364

Answer B

Is there any flaw?

shivani1351 · Answer

Let's first visually represent 1, 13 and 39:

1-------13----------------39

Now, we are given that the sum of the first n consecutive positive odd integers is equal to n^2.

Note:  Here, n refers to the number of the odd terms (and not that particular odd term as such)

2n-1 = 39 (thus, n = 20)
The sum of the first n consecutive positive odd integers till 39 = 400 (n^2 = 20^2 = 400)

Now, if you look at the above number line, we are required to find out the sum of all the consecutive odd numbers between 13 and 39, inclusive. This can also be thought of as the (sum of all consecutive odd numbers from 1 to 39) - (sum of all consecutive odd numbers from 1 to 11).

2n-1 = 11 (thus, here n = 6)
The sum of the first n consecutive positive odd numbers till 11 = 36 (n^2 = 6^2 = 36)

Therefore, the required sum = 400-36 = 364 (Option B) (Answer)

icandoit4u · Answer

for sum of odd integers from 13 to 39 inclusive, find the no of gaps = (last term-first term)/value of gap = (39-13)/2= 26/2= 13
and no of terms = no of gap+1 = 13+1= 14 so, there are total 14 terms from 13 to 39 inclusive.
sum of terms = n/2*(2a+(n-1)d)
= 14/2*(2*13+(14-1)2) 
=7*(26+26) 
=7*52
=364

theretheree4 · Answer

Know that to calculate the last term of the nth term is TN =a+(n-1)d here 39=13(n-1)2 since odd so diff =2 from here we get n=14 putting in sum formula s=n /2(first+last) 14/2*(52)=364

The sum of the first n consecutive positive odd integers is

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