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Well, I'm going to start by running a scenario with the first 3 odd integers. Those would be 1, 3, and 5 for a total of 9. The first 3 non-negative integers would be 0, 2, and 4 for a total of 6.

It seems that each even integer is 1 less than its odd counterpart. So my sum is 3 less for the evens because I had 3 terms. So if I use 5 terms, it should be 5 less. Let's test that.

1 + 3 + 5 + 7 + 9 = 15 0 + 2 + 4 + 6 + 8 = 20

Sure enough, each even integer is 1 less than its counterpart. So the answer must be k-30. I'll pick answer choice (B).
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Sum of First n +ve odd integers is n^2

Sum of First n +ve even integers is n(n+1)

If we take the scenario for first 5 even/odd numbers

First 5 +ve odd integers would be 25 which is K here

First 5 +ve even integers would be 5(6) = 30 (Here "0" is not counted)

"0" is considered as a non-positive and non-negative even integer. and hence will go with K-N i.e. (B)
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Sum of First n +ve odd integers is n^2

Sum of First n +ve even integers is n(n+1)

If we take the scenario for first 5 even/odd numbers

First 5 +ve odd integers would be 25 which is K here

First 5 +ve even integers would be 5(6) = 30 (Here "0" is not counted)

"0" is considered as a non-positive and non-negative even integer. and hence will go with K-N i.e. (B)

Odd sum is 25 --> This means k = 25 Even sum is 30. --> This is 5 more than k. n = 5 (No. of elements) --> This means Sum for even is k+5 --> k+n

If the sum of the first 30 positive odd inte [#permalink]

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26 May 2017, 11:06

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We should know that in an evenly spaced set (Arithmetic Progression), Mean or average is always = Median Also, for a series having EVEN number of terms, Median = average of two middle terms

First 30 positive odd integers: 1,3,5,7,...59 This is an evenly spaced set hence mean = median = average of two middle terms = average of 15th and 16th terms = (29+31)/2 = 30 Mean=30, So sum = 30*30 (Sum = mean*n) Sum = 900, this is given as K.

Now, first 30 non negative even integers: 0,2,4,6,8,...58 This is again an evenly spaced set hence mean = median = average of 15th and 16th terms = (28+30)/2 = 29 Mean=29, So Sum = 29*30 Sum = 870

If K=900, we can clearly see that 870 = k-30

Hence B answer

Last edited by amanvermagmat on 26 May 2017, 21:10, edited 1 time in total.

Re: If the sum of the first 30 positive odd inte [#permalink]

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26 May 2017, 14:00

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Since the sum of n positive integers(odd) can be got by simple formula Sum(Odd positive numbers) = n^2, which is equal to k. We can deduce that k = 30^2 as we are asked the sum of 30 odd integers.

Coming to the second part of the question, we have a formula Sum(Even positive numbers) = n*n-1 Also, the sum of the first 30 non-negative even numbers is 30*29

Since we need to find it in form of k & we already know that k = 30^2, we can use k-30 to get the value k-30 = 30^2 - 30 = 30(30-1) = 30*29 Hence, Option B is the correct answer
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Re: If the sum of the first 30 positive odd inte [#permalink]

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11 Oct 2017, 19:14

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Can someone tell me if this is another way of doing these kind of problems?

I saw pushpitkc use n(n-1) and was wondering if that was just a shortcut from what I did on 2 below.

N^2 = sum of odd numbers N(N+1) = sum of even numbers. To find n = (First Even + Last Even)/2

1. The sum of the first 30 positive odd integers (0 is not included).... N^2=k=30^2=900

2. The sum of first 30 non-negative even integers (Non-Negatives include 0)... 0,2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32,34,36,38,40,42,44,46,48,50,52,54,56,58... (58 + 0)/2 = 29 = n 29(29+1)= 870