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If n is a positive odd number and the sum of the even numbers from 1

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If n is a positive odd number and the sum of the even numbers from 1  [#permalink]

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New post 16 Oct 2019, 02:25
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A
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Difficulty:

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Question Stats:

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If n is a positive odd number and the sum of the even numbers from 1 to n is 79*80, n=?

(A) 79

(B) 81

(C) 157

(D) 159

(E) 161

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Re: If n is a positive odd number and the sum of the even numbers from 1  [#permalink]

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New post 18 Oct 2019, 11:36
Sum of even number from 1 to n =79*80
N =Odd positive no
So Sum of Even no will be equal to
2+4+........+(n-1)
2(1+2+.....+(n-1)/2) taking 2 common
So we get sum of first n naturals no equation
=2*(n^2-1)/8 Putting #n=(n-1)/2

So (n^2-1)/4=79*80
Which gives n=159

Answer D
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Re: If n is a positive odd number and the sum of the even numbers from 1  [#permalink]

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New post 18 Oct 2019, 14:26
rohan2345 wrote:
If n is a positive odd number and the sum of the even numbers from 1 to n is 79*80, n=?

(A) 79

(B) 81

(C) 157

(D) 159

(E) 161


because 2 and n-1 are even, mean has to be even also
so [2+(n-1)]/2=80→
n+1=160→
n=159
D
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Re: If n is a positive odd number and the sum of the even numbers from 1  [#permalink]

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New post 18 Oct 2019, 21:47
One can go through options as well, there must be 79 even terms in the sequence... Take 157 , 156 would be the highest even term in the sequence .. 156/2 will give you 78 even terms. Our number should have one more even term to make the total even terms in the sequence to 79.. next highest is 159, gives you 158 as highest even term, 158/2 = 79, this is exactly what we wanted.. hence n = 159..

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If n is a positive odd number and the sum of the even numbers from 1  [#permalink]

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New post 19 Oct 2019, 04:02
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rohan2345 wrote:
If n is a positive odd number and the sum of the even numbers from 1 to n is 79*80, n=?

(A) 79

(B) 81

(C) 157

(D) 159

(E) 161


Formula: Sum of n even numbers = n(n+1)

From the question stem, we are told that the sum of even numbers is 79*80.

It is safe to assume that n = 79, as 79(79+1) is the sum which is in the form of n(n+1)
Since we are talking about 79 even numbers, the last even number we are looking must
be 2*79(158)

We are told that the sum of even numbers from 1 to n is 79*80 and n is a positive odd number.
Therefore, n = 159(Option D) as sum of even numbers from 1 to 159(79 even numbers) is 79*80.
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If n is a positive odd number and the sum of the even numbers from 1  [#permalink]

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New post 19 Oct 2019, 11:14
Sum of n and 1 =n+1;
Thus, Average of 1st and last numbers =(n+1)/2;
Number of Even numbers =(n-1)/2
Now, (n-1)(n+1)/(2*2)=79*80=(159-1)(159+1)/(2*2)
-->n=159
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If n is a positive odd number and the sum of the even numbers from 1   [#permalink] 19 Oct 2019, 11:14
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