Quote:
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
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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