Question

The terms of an arithmetic sequence add to $715.$ The first term of the sequence is increased by $1,$ the second term is increased by $3,$ the third term is increased by $5,$ and in general, the $k$ th term is increased by the $k$ th odd positive integer. The terms of the new sequence add to $836.$ Find the sum of the first, last, and middle terms of the original sequence.

Answer

Solution:

If the sequence has $n$ terms, the amounts added are the first $n$ odd numbers, whose sum is $n^2.$ Thus $n^2 = 836 - 715 = 121,$ so $n = 11.$

The average of the $11$ terms is $\frac{715}{11} = 65,$ which equals the middle (sixth) term of the arithmetic sequence. The first and last terms also average to $65,$ so they add to $130.$

The requested sum is $65 + 130 = 195.$

2012 AIME I Problem 2

Solution:

Problem 2 in Other Years