SOLUTION: A sequence begins with 1, 4, 7, � Another sequence begins with 99, 95, 91, � At which position is the positive difference between the respective terms of the two sequences the mi

Question 260663: A sequence begins with 1, 4, 7, � Another sequence begins with
99, 95, 91, � At which position is the positive difference between
the respective terms of the two sequences the minimum?

Answer by jsmallt9(3759) (Show Source): You can put this solution on YOUR website!
Both of these sequences are arithmetic sequences. The general formula for the terms of an arithmetic sequence is:

For the first sequence, since the first term, , is 1 and the common difference, d, is 3 we get:

For the second sequence we get:

The positive difference in these terms, which we'll call D, is:

(Note how we use absolute value to ensure a positive difference.) Simplifying this we get:

So the question is: What value of n makes 7n - 105 closest to zero? After a little effort we find that if n = 15, then 7n - 105 is zero! In other words, the 15th term of both sequences is the same!