M20-31

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

Bunuel · 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

priya1987 · Answer

HI.
 we can restate the problem as : what is the summation of evenly spaced series (arithmatic progression) 13,15,17...,39
no. of terms of the series= (39-13)/2 +1=14 and avg of the series =(1st term+last term)/2=52/2=26. so summation =26*14=364

spetznaz · Answer

Great explanations above. This is how I solved it .. Sum of all odd numbers from 13 to 39 is ((14/2) * (13+39)) This comes to 364. In other words, we are concerned about the sum of all numbers in an AP with a CD of 2 and the first term as 13. The answer is option B

patto · Answer

Hi  Bunuel

I don't get how you arrive to the conclusion that "It says that the sum of all odd integers between 13 and 39, inclusive equals to the sum of all odd integers from 1 to 39, inclusive minus the sum of all odd integers from 1 to 11, inclusive" from the question... would you pls further explain?

Bunuel · Answer

Take a simple example: the sum of all odd integers between 5 and 9, inclusive ( 5 + 7 + 9 ) equals to the sum of all odd integers from 1 to 9, inclusive ( 1 + 3  +  5 + 7 + 9 ) minus the sum of all odd integers from 1 to 3, inclusive ( 1 + 3 ).

Aadi01 · Answer

Hi  Bunuel ,
I am not able to understand this  The sum of all odd  i ntegers between 13 and 39, inclusive equals to  the sum of all odd integers from 1 to 39, inclusive   minus   the sum of all odd integers from 1 to 11, inclusive 
Where does this come from.Is there any basis, 11 is specific for this case or for every case

Bunuel · Answer

It's simple logic:

1 + 3 + 5 +7 + 9 + 11  +  13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29 + 31 + 33 + 35 + 37 + 39

RED (the sum of all odd integers between 13 and 39, inclusive) equals to TOTAL (the sum of all odd integers from 1 to 39, inclusive) minus BLUE (the sum of all odd integers from 1 to 11, inclusive).

Does this make sense?

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

Gemmie · Answer

=   -

1 = 0 * 2 + 1 (=> 1st positive odd integer)
...
11 = 5 * 2 + 1 (=> 6th)
...
39 = 19 * 2 + 1 (=> 20th)

Sum from 1 to 11 = Sum of first 6th positive odd integers =  6^2

Sum from 1 to 39 = Sum of first 20th positive odd integers =  20^2

Sum of all odd integers between 13 and 39, inclusive
 = 20^2 - 6^2 
= (20-6) * (20+6)
= 14 * 26
= 364

vn049636 · Answer

I think this is a  high-quality  question and I agree with explanation.

phmahi1997 · Answer

I like the solution - it’s helpful.

helloanand · Answer

I like the solution - it’s helpful.

arnabdas26 · Answer

This is a great question that’s helpful for learning.

RexLenny · Answer

I like the solution - it’s helpful.

obedear · Answer

Easiest for me to solve this by recognizing that it is asking for sum of a consecutive set, which I know the formula for is:

Sum = Avg. * Number

Avg of consecutive set = (First + Last)/2
Number = (Last-first)/Interval + 1

Avg = 26
Number = 14

26*14 = 364

Fisayofalana · Answer

I like the solution - it’s helpful. Sum = Avg x Quant

Avg = 39+13/(2) = 26
Quant = 39-13/2 (+1) = 14

26 x 14 = 364

komatsu · Answer

I like the solution - it’s helpful.

abhikc3004 · Answer

I solved this as a arithmetic progression, where a = 13, d=2, l = 39, and found the sum.

I am curious though about the given method of solving : the fact that sum of first 20 odd integers is 20^2 - is this a property for any n consecutive odd numbers? Is there something similar for consecutive even numbers too?

Bunuel · Answer

The property applies only to the  first   n   positive odd  integers:

1 + 3 + 5 + ... + (2n - 1) = n^2

For example, the sum of the first four positive odd integers is:

1 + 3 + 5 + 7 = 16 = 4^2

It does not apply to just any  n  consecutive odd integers.

Similarly, the sum of the first  n   positive even  integers is:

2 + 4 + 6 + ... + 2n = n(n + 1)

For example, the sum of the first five positive even integers is:

2 + 4 + 6 + 8 + 10 = 30 = 5(5 + 1)

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