SOLUTION: The sum of the first dozen terms of an arithmetic sequence is 120. The sum of the second dozen terms is 408. Find the sum of the third dozen terms.

Question 1144731: The sum of the first dozen terms of an arithmetic sequence is 120. The sum of the second dozen terms is 408. Find the sum of the third dozen terms.
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

Here are two paths to the answer....

(1) Each of the terms in the second dozen is larger than the corresponding term in the first dozen by 12 times the common difference. Since the difference in the two sums is 408-120 = 288, the common difference in the sequence is 288/12 = 24.

So each of the terms in the third dozen is 24 more than the corresponding term in the second dozen; that means the sum of the third dozen is 12*24 = 288 more than the sum of the second dozen.

ANSWER: 408+288 = 696

(2) You might see, from the discussion of that first method, that there is a much faster path to the answer.

In any arithmetic sequence, if you take three successive groups of the same number of terms, the sum of the middle group will be the halfway between (i.e., the average of) the sums of the first and last group. So if S is the sum of the third dozen terms,

.

And if you are understanding this discussion and learning anything from it, the next time you see a problem like this, you will simply say
408-120 = 288; 408+288 = 696
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