n(n+1)/2 = 2008 => n² + n - 4016 = 0

Find 'n' from the above equation and that is the answer.......

1 occurs last as the 1st term, 2 occurs last as the 3rd term, 3 occurs last as the 6th term, 4 occurs last as the 10th term

we observe a general pattern

n occurs last as the n(n+1)/2 th term

so n(n+1)/2 = 2008

here solving for n, and rejecting -ve values you get: n is approximately 62.8739 so.. the last time 62.8739th term occurs is 2008 the last time the 62nd term would occur would be before 2008, and last time the 63rd term would occur would be after 2008, so currently, we have the ongoing sequence of the 62nd termso n is 62

thanks

 Hi,

the series if looked upon like this is somewhat easier. Any no.'s last occurrence is at a place which is equal to the sum of n natural numbers till that term. In easy words (Hindi, if u understand it all right :)  ), " agar aap 2 ko lete ho to woh akhiri baar teesre number par ata hai, aur 1+2 bhi 3 hota hai, usi tarah agar ap 3 ko chk karo to woh las baar 6th poition par hai, aur 1+2+3 bhi 6 hi hota hai ", similarly for other numbers. Now the sum of first n natural nos. is n(n+1)/2, or roughly n^2/2. Hence equate 2008 to n^2/2 or n^2 = 4016, which is around 63. something something. Now that means the possibilities can be 62 x 63 or 63 x 64. For the first it is 3906 , that is 1953, hence 62 ends at the 1953rd term, and 63 starts from there. The second possibility has 4032 or 2016th last term that is 63 will continue till 2016th term, hence 2008th term is 63. Waise dekha jaye to itna nahi karna tha because if it gives '63+' as the answer it will be 63 x 64 only cause that can only be bigger than 63, but above is also ok, it clarifies a bit. Hope you get it. :)

Thanks and Cheers!!!