Question 260663
Both of these sequences are arithmetic sequences. The general formula for the terms of an arithmetic sequence is:
{{{a[n] = a[1] + (n-1)d}}}
For the first sequence, since the first term, {{{a[1]}}}, is 1 and the common difference, d, is 3 we get:
{{{a[n] = 1 + (n-1)3}}}
For the second sequence we get:
{{{a[n] = 99 + (n-1)(-4)}}}
The positive difference in these terms, which we'll call D, is:
{{{D = abs((1 + (n-1)3) - (99 + (n-1)(-4)))}}}
(Note how we use absolute value to ensure a positive difference.) Simplifying this we get:
{{{D = abs((1 + 3n-3) - (99 + -4n + 4))}}}
{{{D = abs((3n-2) - (103 + -4n))}}}
{{{D = abs(7n - 105)}}}
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!