Sum of even numbers:-

[samal]

Got this question from a compilation.
Q. The sum of the even numbers between 1 and n is 79*80, where n is an odd number, then n=?

My approach has been:-
The set of numbers give is (1,2,3,4,5,6,....n-1,n)

So the num of even numbers in that set is
sum = 2 + 4 + 6 + 8 ...(n-1)
=2*(1+2+3+4...+(n-1)/2)
=2*(((n-1)/2)*((n-1)/2)+1))/2 (sum of ni consecutive numbers)
= ((n-1)*(n+1))/4 = 79*80

Solving gives n = 159.

But the answer given is 79. Can somebody confirm my answer?

[Retake_GMAT]

First term a=2, common difference d=2 since even number

therefore sum to first n numbers of Arithmetic progression would be

n/2(2a+(n-1)d)

= n/2(2*2+(n-1)*2)=n(n+1) and this is equal to 79*80

therefore n=79 which is odd

[GmatG]

Originally Posted by Retake_GMAT

First term a=2, common difference d=2 since even number

therefore sum to first n numbers of Arithmetic progression would be

n/2(2a+(n-1)d)

= n/2(2*2+(n-1)*2)=n(n+1) and this is equal to 79*80

therefore n=79 which is odd

Completely agree with retake_gmat. PLease post the OA and source of this question

[CallMeDaOne]

Here we have calculated n which gives the number of terms of our A.P i.e.
the number of even terms between 1 and n. So IMO, the total number of terms between 1 and n should be 79*2 + 1 = 159. Thus IMO, n comes out to be 159 and not 79 which is just the number of even terms between 1 and n.

Could someone throw more light on this please.

Thanks and Regards.

Originally Posted by Retake_GMAT

First term a=2, common difference d=2 since even number

therefore sum to first n numbers of Arithmetic progression would be

n/2(2a+(n-1)d)

= n/2(2*2+(n-1)*2)=n(n+1) and this is equal to 79*80

therefore n=79 which is odd

[Shooter]

Originally Posted by Retake_GMAT

First term a=2, common difference d=2 since even number

therefore sum to first n numbers of Arithmetic progression would be

n/2(2a+(n-1)d)

= n/2(2*2+(n-1)*2)=n(n+1) and this is equal to 79*80

therefore n=79 which is odd

Retake_GMAT, your approach is perfect but I guess you missed out using (n-1) in the place of n:

As it has been given that last number is odd, which is 'n'.
Our immediate predecessor even number(last number in our series) should be (n-1).

Hence 2+4+6+.........+(n-1) = 79*80


substituting in the formulae n/2(2a+(n-1)d)

Here n=(n-1), a=2(first number), d=2(as it's even digits)

=> (n-1)/2(2*2+(n-1-1)2) = 79*80
=> n(n-1) = 79*80

=> n =79.

HTH

[sanjeev_sharma]

Good solution retake

[samal]

I just verified, the answer is 159 and not 79.
Consider this
sum1 = 2 + 4 + 6 + 8 ....+ 76 + 78 (even numbers less than 79)
= 2(1 + 2 + 3 + ...+38 + 39)
= 2*39*(40/2)
= 39.40

Now consider this
sum2 = 2 + 4 + 6..+ 156 + 158 (even numbers less than 159)
= 2(1 + 2 + 3 + 4.....+ 78 + 79)
= 2 * 79 * (80/2)
= 79 * 80

So n is 159 and not 79.

The mistake done in the sumation of series was in substituing the value of n (how many terms are there in the series), it is not n and not n-1 but rather n/2 and then the answer would come as 159.

[raunekk]

i had read d solution for this previously also..

Take numbers 1 to 7
Even numbers = 2+4+6=12= 3*4

Take number s 1 to 10
Even numbers = 2+4+6+8= 20 = 4*5

Take 1 to 15
even numbers = 2+4+6+8+10+12+14= 56 = 7*8

Thus if sum is 79*80 ( n (n+1))

Ans is 79

[bose]

i agree n=159

[Anson800]

n=159