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Re: PS: first 50 positive even integers [#permalink]

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06 Mar 2009, 13:23

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The first 50 even integers will be 2,4,6,8....100. Sum of the first 50 even integers = n(+1) = 50 * 51 = 2550. (Infact this fix is not needed to be explicitly specified. Already implied in the statement sum of the first 50 even integers)

Sum of even integers between 102 to 200 inclusive will be

102 + 104 + 108 + ....200 (there will be 50 terms)

Re: PS: first 50 positive even integers [#permalink]

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27 Feb 2011, 21:02

Although it defeats the purpose of the information presented in question, it is much simpler way to find out the sum.

Sum = (Ave) (number of terms)

Ave of an evenly spaced set = (First Term + Last Term) / 2 So, Ave = (102 + 200)/2 = 151

number of terms (inclusive) between an evenly spaced set = ((Last Term - First Term)/Difference between in each term) + 1 So, Number of terms N = ((200 - 98)/2)+1 = 50 (The diff between each of the two even integers is 2. In this particular case, you didn't even need to go thru this calculation since it is obvious that there are 50 even and 50 odd numbers in each 100. But the formula helps for other trickier questions)

Therefore, Sum = 151 * 50 = 7550

hope that helps!
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Re: PS: first 50 positive even integers [#permalink]

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27 Feb 2011, 21:26

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We are already given the sum of first fifty even numbers and now need to find the sum of next 50. If you notice, each number in this case (sum of next 50) is 100 more than the old series (sum of first 50) and there are 50 such numbers, so the sum will be just the sum of first 50 numbers + 50*100 = 2550+5000 = 7550

Re: PS: first 50 positive even integers [#permalink]

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19 Jul 2011, 09:37

The first piece of information is redundant here. We need the sum of even integers from 102 to 200(inclusive). 1. To find the sum in such questions, use the formula:

AVG or AM = Sum/no. of terms

2. The set is an evenly spaced set. i.e., each consecutive number has a constant difference with the prev. (const. diff here is 2, since even integers)

3. For an evenly spaced set, Properties: 1. AM or AVG = (first number + last number)/2 i.e., AM = Mid-term 2. AM = Median Now, we can find Avg. avg. = (102+200)/2 = 302/2 = 151 To find number of terms, Use this formula: No. of terms in a evenly spaced set

= [(Last term - First term)/constant difference] + 1

Re: PS: first 50 positive even integers [#permalink]

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20 Jul 2011, 03:11

nikhilsamuel89 wrote:

The first piece of information is redundant here. We need the sum of even integers from 102 to 200(inclusive). 1. To find the sum in such questions, use the formula:

AVG or AM = Sum/no. of terms

2. The set is an evenly spaced set. i.e., each consecutive number has a constant difference with the prev. (const. diff here is 2, since even integers)

3. For an evenly spaced set, Properties: 1. AM or AVG = (first number + last number)/2 i.e., AM = Mid-term 2. AM = Median Now, we can find Avg. avg. = (102+200)/2 = 302/2 = 151 To find number of terms, Use this formula: No. of terms in a evenly spaced set

= [(Last term - First term)/constant difference] + 1

no. of terms = (200-102)/2 +1 = 50

Therefore,

Sum = Avg * no. of terms = 151*50 = 7550

Should this even integer start with 0 ? is 0 an even integer ???? 0 + 2 + 4 .... 98? = 2550?

Re: PS: first 50 positive even integers [#permalink]

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20 Jul 2011, 03:15

siddhans wrote:

nikhilsamuel89 wrote:

The first piece of information is redundant here. We need the sum of even integers from 102 to 200(inclusive). 1. To find the sum in such questions, use the formula:

AVG or AM = Sum/no. of terms

2. The set is an evenly spaced set. i.e., each consecutive number has a constant difference with the prev. (const. diff here is 2, since even integers)

3. For an evenly spaced set, Properties: 1. AM or AVG = (first number + last number)/2 i.e., AM = Mid-term 2. AM = Median Now, we can find Avg. avg. = (102+200)/2 = 302/2 = 151 To find number of terms, Use this formula: No. of terms in a evenly spaced set

= [(Last term - First term)/constant difference] + 1

no. of terms = (200-102)/2 +1 = 50

Therefore,

Sum = Avg * no. of terms = 151*50 = 7550

Should this even integer start with 0 ? is 0 an even integer ???? 0 + 2 + 4 .... 98? = 2550?

Yes Zero is even integer

gmatclubot

Re: PS: first 50 positive even integers
[#permalink]
20 Jul 2011, 03:15

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