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Consecutive Integers (m06q28)

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Consecutive Integers (m06q28) [#permalink] New post 14 Nov 2008, 14:02
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What is the value of k if the sum of odd, consecutive, positive integers from 1 to k equals 441?

(A) 47
(B) 41
(C) 37
(D) 33
(E) 29

[Reveal] Spoiler: OA
B

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I'm using the formulas:

1. Sum of S = n/2(2a+(n-1)d) and
2. Sum of S = n/2(a+l)

where a = first number
l= last number
d= difference between the numbers

And I'm not getting the right answer.
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Re: Consecutive Integers [#permalink] New post 14 Nov 2008, 15:57
bigfernhead wrote:
The sum of the odd positive integers from 1 to k equals to 441. What is the value of k?

47
41
37
33
29

I'm using the formulas:

1. Sum of S = n/2(2a+(n-1)d) and
2. Sum of S = n/2(a+l)

where a = first number
l= last number
d= difference between the numbers

And I'm not getting the right answer.


you will get the answer using the first formula (arithmetic progression)

s=n/2(2a+(n-1)d)

where, n= number of terms
a=1 is the first term
d=2 is the common difference

solving , 441=n/2(2+ (n-1)2)= n^2 ==> n= 21 ==> number of terms = 21 ie k =41 as there are 21 positive consecutive intergars starting 1 hence 20 even on the way to 41. Hence answer is 41
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Re: Consecutive Integers [#permalink] New post 15 Nov 2008, 12:28
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bigfernhead wrote:
The sum of the odd positive integers from 1 to k equals to 441. What is the value of k?

47
41
37
33
29

I'm using the formulas:

1. Sum of S = n/2(2a+(n-1)d) and
2. Sum of S = n/2(a+l)

where a = first number
l= last number
d= difference between the numbers

And I'm not getting the right answer.



Here is what I did. The number 441 was really odd and some how it occurred to me that it is square of 21. So we need 21 21's in this sequence to get the sum to 441. We are starting with 1. The last number possibly is 41 because 1 and 41 average is 21. So are (3,39) (5,37) and so on.

Not a great way but helpful when we have strange numbers that have some significance and lets us connect dots.
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Re: Consecutive Integers [#permalink] New post 17 Nov 2008, 22:57
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sum of consecutive positive odd integers between 1 and k is (1/4)*(k+1)^2

from the sum of arithmetic progression S= n/2(2a+(n-1)*d)
put n=(k+1)/2 and d=2

therefore, (k+1)^2=1764
k+1=42
k=41
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Re: Consecutive Integers (m06q28) [#permalink] New post 05 Aug 2010, 04:32
Math (PS)

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

(A) 47
(B) 41
(C) 37
(D) 33
(E) 29

Explanation:

The sum is given as 441 which means by using the sum to n terms of an AP we can arrive at the following

n*(2a+(n-1)*d)/2 = 441

here, a=1 (given)
d=2 (consecutive odd integers)
k= a +(n-1)*d = 1+(n-1)*2 =2n-1

put these values in the above eqn and you'll get n=21. So the total number of terms in the AP = 21.
the value of k will be = 1+(21-1)*2 = 41 which is option A.
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Re: Consecutive Integers (m06q28) [#permalink] New post 05 Aug 2010, 04:53
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based on the following formulae:
sum of the first consecutive ODDS is N^2
sum of the first consecutive EVENS is N*(N+1)
=> 441=N^2 => N=21 => k=21*2-1=41
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Re: Consecutive Integers (m06q28) [#permalink] New post 05 Aug 2010, 05:11
somewhat different approach

sum of all consecutive numbers from 1....n is \frac{n*n+1}{2}

sum upto what such n is approx 441*2

find out that n so that \frac{n*n+1}{2} = 441 *2

so n*n+1 = 1764

looking at the answer choices 41 and 47 are the closest, and 41 more so

Last edited by abhtiw on 05 Aug 2010, 07:45, edited 1 time in total.
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Re: Consecutive Integers (m06q28) [#permalink] New post 05 Aug 2010, 05:38
I used the answer choices and worked backwards knowing that since this is an evenly spaced set, the sum = midpoint or avg * number of items in the set.

In an evenly spaced set, the avg and midpoint = first + last/2 using the answer choices, where k equals 1, 1 + 41 =42/2 = 21. 21 should be midpoint. Second step was sum divided by midpoint or avg which we just found to be 21. So, 441/21 = 21 to find the number of items in the set.

Since sum = avg*number....441 = 21*21.

B is the correct answer. I was lucky in that the correct answer was earlier in the answer choices.

I dont understand the sum progression formula. Can some please help me on this one?
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Re: Consecutive Integers (m06q28) [#permalink] New post 05 Aug 2010, 06:58
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An = A1 + (n-1)d

therefore
k = 1 + (n-1)2
n = (k+1)/2

Also
Sn = n/2 ( A1 + An)
441 = (K+1)/4 ( 1 + k )
k = 41
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Re: Consecutive Integers (m06q28) [#permalink] New post 05 Aug 2010, 07:08
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Sn = (n/2){2a + (n-1)*d}
tn = a + (n - 1 ) * d

Sn = Sum of terms in the series
tn = nth term of the series
a = First term in series = 1
n = Number of terms in the series = is Unknown
d = Difference between consecutive terms in the series = 2

Sn = 441

so 441 = (n/2) * { 2 * 1 + (n-1) * 2}
From this n = 21 (discarding the negative value)

Question , it is asked k , which is the n th term

tn = t21 = 1 + (21 - 1) * 2 = 41
Answer is 41 (B)
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Re: Consecutive Integers (m06q28) [#permalink] New post 06 Aug 2010, 06:50
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The sum of any evenly distributed numbers = (Total numbers in the distribution) * ((first number + lastnumber)/2)

Example1: The sum of first 5 integers (1, 2, 3, 4, 5) is:
(5)*((1+5)/2) = 5*3 = 15

Example2: The sum of first 5 odd integers (1 3 5 7 9) is:
5*((1+9)/2) = 5 * 5 = 25

Example3: The sum of first 6 odd integers ( 1 3 5 7 9 11) is: 6 * 6 = 36

========
What's given in the problem is the sum of 1 to k odd consecutive integers is: 441 (i.e: 7*7*3*3, i.e, 21*21)
===> 21 * (( 1+k)/2) = 21 * 21
===> (1+k)/2 = 21
===> k = 42 - 1 = 41


Cheers!
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Re: Consecutive Integers (m06q28) [#permalink] New post 06 Aug 2010, 15:23
nravi4 wrote:
The sum of any evenly distributed numbers = (Total numbers in the distribution) * ((first number + lastnumber)/2)

Example1: The sum of first 5 integers (1, 2, 3, 4, 5) is:
(5)*((1+5)/2) = 5*3 = 15

Example2: The sum of first 5 odd integers (1 3 5 7 9) is:
5*((1+9)/2) = 5 * 5 = 25

Example3: The sum of first 6 odd integers ( 1 3 5 7 9 11) is: 6 * 6 = 36

========
What's given in the problem is the sum of 1 to k odd consecutive integers is: 441 (i.e: 7*7*3*3, i.e, 21*21)
===> 21 * (( 1+k)/2) = 21 * 21
===> (1+k)/2 = 21
===> k = 42 - 1 = 41


Cheers!
Ravi


keep finding stuff i dont understand in explanations and for the first time i will try to point out something with the risk of looking stupi.....

whilst your examples are very good, why do you multiply by 21? I get that 441=21*21 but how do you know that
(Total numbers in the distribution) = 21?
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Re: Consecutive Integers (m06q28) [#permalink] New post 06 Aug 2010, 15:26
abhtiw wrote:
somewhat different approach

sum of all consecutive numbers from 1....n is \frac{n*n+1}{2}

sum upto what such n is approx 441*2

find out that n so that \frac{n*n+1}{2} = 441 *2

so n*n+1 = 1764

looking at the answer choices 41 and 47 are the closest, and 41 more so


again, using the example formula, \frac{n*n+1}{2} = 441 and not =441*2
!!! y do you multiply by 2 the right hand side?
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Re: Consecutive Integers (m06q28) [#permalink] New post 06 Aug 2010, 23:59
why you try to make it more complicated.

Variant #1
sum of the first consecutive odd integers is N^2. that's it.

Variant #2
If you want to make it complex: sum of consecutive integers is N*(a1+an)/2=N*(2*a1+D*(N-1))/2
where a1=1, D=2 =>
N*(2*1+2*(N-1))/2=N^2=441 and further as in the first variant.
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Re: Consecutive Integers (m06q28) [#permalink] New post 07 Aug 2010, 00:23
bigfernhead wrote:
What is the value of k if the sum of odd, consecutive, positive integers from 1 to k equals 441?

(A) 47
(B) 41
(C) 37
(D) 33
(E) 29

[Reveal] Spoiler: OA
B

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Sum of 1+3+5+7+9 = 25

Now every next 5 odd number will be greater by 50

for eg....11+13+15+17+19 = 75
21+23+25+27+29 = 125 and so on...


SO reaching 441 is like this...

25+75+125+175 = 400 ( from 1 to 40)

now adding 41 gives 441 so the K is 41....

No need to use formula......heh heh...
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GMAT 08/08 question of the day [#permalink] New post 08 Aug 2011, 05:12
Here is a non-algebraic solution

Put given solutions in ascending order:

29
33
37
41
47

TEST 37 (in this way we can eliminate 2 choices)

S=1+3+5+....+37
#of terms: (37-1)/2+1=19
mean: (37+1)/2=19
Sum of terms: 19*18= 342 < 441 (so we eliminate 29,33,37)

We can either try 41 and 47, or using common sense:

361+ 39 + 41 = 441

Answer is B. Any comments welcome

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Re: Consecutive Integers (m06q28) [#permalink] New post 08 Aug 2011, 05:18
Here is a non-algebraic solution

Put given solutions in ascending order:

29
33
37
41
47

TEST 37 (in this way we can eliminate 2 choices)

S=1+3+5+....+37
#of terms: (37-1)/2+1=19
mean: (37+1)/2=19
Sum of terms: 19*18= 342 < 441 (so we eliminate 29,33,37)

We can either try 41 and 47, or using common sense:

361+ 39 + 41 = 441

Answer is B. Any comments welcome

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Re: Consecutive Integers (m06q28) [#permalink] New post 08 Aug 2011, 05:31
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bigfernhead wrote:
What is the value of k if the sum of odd, consecutive, positive integers from 1 to k equals 441?

(A) 47
(B) 41
(C) 37
(D) 33
(E) 29



Average of elements*Number of Elements=Sum of elements

\frac{k+1}{2}*(\frac{k-1}{2}+1)=441

k=41

Ans: "B"
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Re: Consecutive Integers (m06q28) [#permalink] New post 08 Aug 2011, 09:09
I got the answer but with a longer solution.

(k-1)/2 +1 =N

a) => N= (k+1)/2

b) Sum/N=Average

c) Average in this evenly-spaced sets is (K+1)/2

a,b,c) 441/(k+1)/2 = (K+1)/2

Solving for K, K=41
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Re: Consecutive Integers (m06q28) [#permalink] New post 08 Aug 2011, 10:02
answer is B....n^2/4 or n+1 ^2/4 depend on whether n is odd or even
Re: Consecutive Integers (m06q28)   [#permalink] 08 Aug 2011, 10:02
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