Problem Solving

I found this question in the difficult list of the quant questions, but not able to figure out how this is to be solved. Looks very simple though:

The sum of the even numbers between 1 and n is 79*80, where n is an odd number, then n=?

If you mean consecutive even numbers between 1 and n, we can solve it the following way:

let us express the terms in the sum as 2X_i as they are all even. X_i are consecutive #s.

Sum(2X_i) for i=1 to n is 79*80.

This implies:

2 * Sum(X_i) (for i=1 to n) = 79*80.

Since X_i are consecutive, Sum(X_i)(for i=1 to n) = n(n=1)/2.

So, 2 * n(n+1)/2 = 79*80 => n(n+1) = 79*80 => n = 79.

indeed sum{2, 4, 6 ... 158} (where there are n=79 terms) = 79*80.

When you say sum is for numbers between 2 to 158, it is not possible that these numbers are between 1 and 79 as mentioned in the question itself.
Suppose n=9, numbers we are concerned with are
2, 4, 6, 8
But 2 + 4 + 6 + 8 NOT = 2(1+2+....+9) as you indicated.
It will be 2(1+2+3..+(n-1/2))
Correct me if I am wrong.

first even number=2
last even number=n-1
total number of even numbers= (n-1-2)/2 + 1= n/2-3/2+1
=(n-1)/2
sum of A.P =total number/2(first term + last term)
in our case
(n-1)/4 * (2+n-1)=79*80
(n-1)(n+1)=4*79*80= 158*160
n=159

my trial:

(2+n-1)*(n-1)/4=79*80
(n+1) * (n-1) = 158*160 ... therefore, n=159

Ah Yes, 158 is correct. I used n to denote the number of terms in the sequence. n is indeed the the last term of the sequence.