Elegant solution to prickly arithmetic progression problem?

This one really bothers me a great deal:

An AP with first term a and common difference d has value of 121 for the sum to n terms. The sum of the first three terms of this series is 9, and the sum of the last three terms of this series is 57. Find n.

I managed to get the answer, however my workings were both clumsy and lengthy. Is there any elegant solution to this problem?

Student X

There is a special property linking any three consecutive terms of an AP, ie the middle term is always given by the sum of these three terms divided by three aka its arithmetic mean (it isn't all that difficult to prove, try to work this out). Hence, the second term of the overall AP we are examining would be 9/3=3, and the second last term of this same overall AP would be 57/3=19.

With these information at hand, we can see that the first and last terms of this AP are simply 3-d and 19+d respectively. Since the total sum is equivalent to121, then (n/2)* (first term +last term)=121 ======>(n/2)*(3-d+19+d)=121. Solving this yields n=11.

Hope this helps. Peace.

Best Regards,

Mr Koh