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BSchool Forum Moderator V
Joined: 29 Jan 2015
Posts: 1451
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WE: General Management (Non-Profit and Government)
If n is a positive odd number and the sum of the even numbers from 1  [#permalink]

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4 00:00

Difficulty:   65% (hard)

Question Stats: 53% (02:09) correct 47% (01:33) wrong based on 43 sessions

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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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Intern  B
Joined: 14 Jan 2018
Posts: 49
Location: India
Concentration: General Management, Entrepreneurship
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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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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

VP  P
Joined: 07 Dec 2014
Posts: 1221
Re: If n is a positive odd number and the sum of the even numbers from 1  [#permalink]

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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

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
Manager  S
Joined: 19 Jan 2019
Posts: 110
Re: If n is a positive odd number and the sum of the even numbers from 1  [#permalink]

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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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Senior PS Moderator V
Joined: 26 Feb 2016
Posts: 3307
Location: India
GPA: 3.12
If n is a positive odd number and the sum of the even numbers from 1  [#permalink]

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1
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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Joined: 07 Oct 2019
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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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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 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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