M06-28

Question

What is the value of  k  if the sum of consecutive odd integers from 1 to  k , inclusive equals 441?

A. 47
B. 41
C. 37
D. 33
E. 29

Bunuel · Accepted Answer

Official Solution:

What is the value of  k  if the sum of consecutive odd integers from 1 to  k , inclusive equals 441?

A. 47
B. 41
C. 37
D. 33
E. 29

Consecutive odd integers represent an evenly spaced set (aka arithmetic progression). Now, the sum of the terms in any evenly spaced set is the mean (average) multiplied by the number of terms, where the mean of the set is  \frac{	ext{first term} + 	ext{last term}}{2} .
 
•  	ext{average}=\frac{	ext{first term} +	ext{last term}}{2}=\frac{1+k}{2} ;
 
•  	ext{number of terms}=\frac{k-1}{2}+1=\frac{k+1}{2}  (number of terms in an evenly spaced set is  \frac{	ext{last term} - 	ext{first term}}{	ext{common difference}}+1 )
 
•  	ext{sum} = \frac{1+k}{2}*\frac{k+1}{2}=441 .
 
The above gives  (k+1)^2=4*441 , which simplifies to  k+1=2*21=42  giving  k=41 .
 
  Alternative Solution:  
 
Use the formula for the sum of the first  n  positive odd integers:  (\frac{x+1}{2})^2  where  x  is the last of these integers. Then  \frac{k+1}{2} = \sqrt{441} = 21 , and  k = 41 .

Answer: B

him1985 · Answer

Alternate approach:

Sum of n evenly spaced numbers = n/2

Where n is the number of odd no, a = 1st number and d= difference
No of integers - (last - first)/2 + 1

Number of odd no = k-1/2, a=1, d=2

441= k+1/4  
441= k+1/4  
441*4 = (K+1)^2 = (441 * 4)
k+1 = 21 * 2 = 42
k=41

Hence 41 is answer

buddyisraelgmat · Answer

60 seconds appraoch

Sum of First "n" +ve ODD Numbers = n^2

Quickly check... 
20^2 = 400
21^2 = 441 ... Bingo!

Hence (21*2) - 1 = 41 .... or you can even quickly check this manually ... will take only 30 seconds more
Hence k = 41

Just FYI.... 
Sum of First "n" natural nos = n(n+1)/2
Sum of First "n"   ODD   natural nos = n^2
Sum of First "n"   EVEN   natural nos = n (n+1)

amkoam · Answer

1+3 = 4 (# of terms)^2 = (2)^2
1+3+5 = 9 (# of terms)^2 = (3)^2 
1+3+5+7 = 16 (# of terms)^2 = (4)^2 
1+3+5+7+....+k=441

Therefore # of terms = square root of 441 = 21
21st consecutive odd term = 41.

sergge · Answer

I've follwed this logic while solving this question on CAT:
1. The odd numbers end with 1, 3, 5, 7, 9, and the sum of them is 25 for the 1st ten (e.g. 1, 3, 5, 7, 9), 75 for the 2nd ( e.g. 11, 13, 15, 17, 19), 125 for the 3rd etc. So the sum ends with number 5.
2. Following this logic, 47 will ends with 4 *5 + 1 + 3+ 5 + 7, {number of tenth in the 47 = 4} * {sum of each tenth ends with 5} + {sum of all other odds till end of 47} 
3. Obviously with the info from (1) and (2) only number which ends with 1 e.g. 41 will give a sum which ends with 1 e.g. 441

It took me ~10 seconds to evaluate the correct answer.

mvictor · Answer

I used the second method...
the sum of first x odd integers is x^2.
441 = 21^2.
this means that we are looking for the 21st odd integer.
39 is the 20th odd integer, and 41 is the 21st.

arhumsid · Answer

Got this question wrong at first because i couldn't make out that the K is not the number of terms in this series, its just the last term in it.

We already know the formula of the  sum of first N odd integers =  N^2  
So,  N^2  = 441
N = 21

Hence, the number of terms in this series is 21.

At this point i realised that N and K are different here, for eg:
1,3,5,7 => N=4; K=7
1,3,5,7,9,11,13 => N=7; K=13

We can see that,  K = 2N-1 
since we have N=21 for this series, we can find out K
 K=2*21 - 1 
 K=41

Side note (not quite reqd here but could be helpful in similar questions): "Sum of odd number of odd numbers is odd and Sum of even number of odd numbers is even "

rishovchicago · Answer

Sum of first n odd integers is n^2.

So sum of first K odd integers in K^2.

K^2=441, K= 21.

Can anyone explain this?

Bunuel · Answer

The sum of first n odd integers is n^2 means that, for example, the sum of first 4 consecutive odd integers, 1, 3, 5, and 7 is 4^2 = 16.

The question does not say that the sum of first k consecutive integers is 441. It says that the sum of consecutive odd integers  from 1 to k  equals 441. After solving we get that the sum of odd integers, from 1 to 41, so 1 + 3 + 5 + ... + 41 = 441. The number of terms we add is indeed 21 but that's not the value of k, the value of k is 41.

TheNightKing · Answer

Shorter approach.

Since 441 is 21^2 and this is an AP series, The sum of numbers at two extremes has to average to 21.

(1+41)/2=21

Hence B.

AndrewN · Answer

Hello, everyone. For the less mathematically inclined, you can also pair numbers and keep track of the sum by the number of pairs. To illustrate, say you started with 33:

1 + 3 + 5 + ...

Since we know that the last number will be 33, the  next to last  number will be 31, then 29, and so on.

1 + 3 + 5 + ... + 29 + 31 + 33

By pairing the outermost numbers and working our way in, we can appreciate that each individual sum will be 34.

(1 + 33) + (3 + 31) + (5 + 29)

The question is, do we have enough pairs to work up to 441? Well, clearly we do not, since the last pair, 15/19, only gets us to eight total pairs (plus an unmatched 17), and

34 * 8 = well below 400

We can comfortably eliminate (D) and (E). Now it would make sense to choose (B), 41, the number in the middle of the three remaining answers. The sum of each pair will be higher this time, but the process should be faster this time around, since we are already keen to the process.

1 + 3 + 5 + ... + 37 + 39 + 41

(1 + 41) + (3 + 39) + (5 + 37)

How many pairs of 42 will we have? You can count them on your fingers if you want. The answer will be ten, ten pairs, and of course, 21 cannot be doubled, so it is the odd number out that we have to add back in.

(42 * 10) + 21 = 441

To be clear, I am  not  advocating this method over the more formulaic ways of solving the question given above, but number sense can come in handy in a pinch.

- Andrew

loryanna · Answer

Hi Bunuel,
so the sum of odd consecutive integers: Sum_odd =( n+1/2)^2 = n^2 right?
Is there a similar formula for calculating sum of even integers in a set, and for natural integers?

Bunuel · Answer

the sum of the first n positive odd integers: (\frac{x+1}{2})^2 where x is the last of these integers. For example, the sum of the first five positive odd integers (1, 3, 5, 7, 9) is (\frac{9 + 1}{2})^2=25 . For more, check here: https://gmatclub.com/forum/math-number- ... 88376.html Hope it helps.

theGisforGorilla · Answer

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

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.

Arpitamitra · Answer

My 2 cents on this:

Sn=n/2{2a+(n-1)d}
or 441*2=n{2+2(n-1)}, where  
882={2n+2n(n-1)}
882=2n+2n^2-2n
882=2n^2
441=n^2
n=41

vn049636 · Answer

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

ruchirk · Answer

n^2=441
n=21
so 21 st term is a+(n-1)2=1+20*2=41

Geminiggl · Answer

I did it differently. Wanted to go for testing the answer choices, started from E. Realised it's going to be 15*15. Then D gave me 17*17.

441 = 49 * 9 = 7 * 3 * 7 * 3 = 21 * 21

Hence, B.

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