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Re: Consecutive Integers [#permalink]
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

Re: Consecutive Integers [#permalink]
15 Nov 2008, 12:28

4

This post received KUDOS

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.

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.

Re: Consecutive Integers (m06q28) [#permalink]
05 Aug 2010, 04:53

2

This post received KUDOS

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

Re: Consecutive Integers (m06q28) [#permalink]
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?

Re: Consecutive Integers (m06q28) [#permalink]
05 Aug 2010, 07:08

3

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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)

Re: Consecutive Integers (m06q28) [#permalink]
06 Aug 2010, 06:50

4

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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

Re: Consecutive Integers (m06q28) [#permalink]
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?

Re: Consecutive Integers (m06q28) [#permalink]
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.