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k_abhi19 · Tue Sep 23, 2008 1:20 pm

n=159 should be the answer... I used a very simple approach.... by taking few cases...... for eg n=3........ in this case the sum of all even nos is 2.. or 1*2 .... similarly for n=5.. sum of all even nos between 1 & 5 will be 6 or 2*3 .. n=2+3=5.... for n=7... sum of all even nos in between will be 12 or 3*4 where 3+4=7=n.... thus when sum of all even nos bettwen 1 & n where n is odd... is 79*80.. then n=79+80 or 159....

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davy420 · Tue Sep 23, 2008 2:49 pm

80 + 79 = 159

Can this problem be solved this way?

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Xins · Fri Oct 31, 2008 11:56 am

Answer n = 159.

I did scratch my head for a few minutes.

Heres what I found

Lets take n = 3 Sum of all even number from 1-3 = 2 = 2x1
Lets take n = 5 Sum of all even number from 1-3 = 6 = 2X3
Lets take n = 7 Sum of all even number from 1-3 = 12 = 4X3

I used the X pattern since the question stem says 79*80 i.e product of 2 consequtive numbers

All the above product needs to be represented in terms on n to get their n

2X1 = [ 2 -(2-1)/2] [ 2- (2+1)/2 ]
3x2 = [ 5 -(5-1)/2] [ 5- (5+1)/2 ]
4X2 = [ 7 -(7-1)/2] [ 7- (7+1)/2 ]

Hence 80 X 79 = [n- (n-1)/2] [n -(n+1)/2]

If you solve for n ; n = 159

Hope this helps

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Xins · Fri Oct 31, 2008 11:59 am

Pardon the typo above. It should be 4X3

4X3 = [ 7 -(7-1)/2] [ 7- (7+1)/2 ]

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Xins · Fri Oct 31, 2008 2:27 pm

I am on a roll of typos today ......

2X1 = [ 2 -(2-1)/2] [ 2- (2+1)/2 ] should be
2X1 = [ 3 -(3-1)/2] [ 3- (3+1)/2 ]

Sorry again.......But I hope you get the point.........

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earth@work · Sun Nov 02, 2008 7:53 pm

nice question 800guy

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piyushdabomb · Sun Nov 02, 2008 9:50 pm

What difficulty level is this sort of question? approximation of 650?

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gmatbedamned · Tue Nov 18, 2008 5:31 pm

Love the list of tougher questions!!

I think we have a confusion of variables by the folks giving 79 as the answer.

I assume they are using the equation Sum(n terms)= n/2 * (2a + (n-1)d) where a is the first term (2) and d is the common difference (2). The equation yields "79" as the answer. Which is right, there are 79 TERMS.

The confusion stems from the fact the original question refers to n as the last term wihich is odd, NOT the number of terms. After you get the n=79 terms from the equation, to find the largest even term = #of terms * common differnece: 79*2 = 158. But n is odd and must be greater than 158 and less than 160. So n =159.

I personally like the equation Sum(X Terms) = X(First+Last)/2. The first is 2 and the last is n-1 and the number of terms (X) is (n-1)/2. It is a messier caculation but more fun!

Anyway, my two cents...

Sorry for reposting a dead topic but I was confused and I thought others would be helped by the clarification.

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jeffm · Sat Sep 26, 2009 8:51 am

Strictly speaking the problem could be looked at as "the sum of all consecutive even integers from 2 to (n-1), inclusive".

first even number= 2
last even number = n-1

Average of all the even #s from first to last =
[(n-1)+2]*1/2 =(n+1)/2
(i.e (first + last)/2)

Total number of even integers from first to last =
1/2*[(n-1) - 2] + 1
= (n-1)/2
(i.e (last -first)/2 + 1)

Sum = (Average) * (Total number of even integers from first to last)

=>
(n+1)/2 * (n-1)/2 =79*80
=> n ~ SR(4*79*80)
n ~SR(4*80*80)
n~160

This takes a minute to do.

Well, we all know n=158.993711...

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liciniudragos · Mon Jan 25, 2010 5:50 am

After some quick search in wikipedia This is what I found:

Sum of all even numbers up to n = N(N+1) where N is the number of consecutive even numbers

So: 79*80=N(N+1) therefore N=79 the last term of the sequence

the nth number value is expressed as n(N)=a+(N-1)*d where a = 1st term here 2 and d = increment here 2

n(79)=2+(79-1)*2 so n(79)=158 so the next odd number is 159

link: http://wiki.answers.com/Q/What_is_the_sum_of_the_first_113_consecutive_even_numbers

q.e.d Smile

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