SOLUTION: A sequence of positive integers with a_1 = 1 and a_9 + a_{10} = 646 is formed so that the first three terms are in geometric progression, the second, third, and fourth terms are in

Question 1176668: A sequence of positive integers with a_1 = 1 and a_9 + a_{10} = 646 is formed so that the first three terms are in geometric progression, the second, third, and fourth terms are in arithmetic progression, and, in general, for all i>=1, the terms a_{2i - 1}, a_{2i}, a_{2i + 1} are in geometric progression, and the terms a_{2i}, a_{2i + 1}, and a_{2i + 2} are in arithmetic progression. Find the greatest term in this sequence that is less than 1000.

Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

A formal algebraic solution would probably be very messy....

But since the sequence is all integers, we can find the sequence we are looking for simply by investigation.

Second term 2: 1, 2, 4, 6, 9, 12, 16, 20, 25, 30; sum 55
Second term 3: 1, 3, 9, 15, 25, 35, 49, 63, 81, 109; sum 190
Second term 4: 1, 4, 16, 28, 49, 70, 100, 130, 169, 208; sum 377
Second term 5: 1, 5, 25, 45, 81, 117, 169, 221, 289, 357; sum 646

That is the sequence we want; continue the sequence until the numbers exceed 1000.

1, 5, 25, 45, 81, 117, 169, 221, 289, 357, 441, 525, 625, 725, 841, 957, 1089

The largest number in the sequence less than 1000 is 957.

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