Two APs have the same common difference. The first term of one AP is \(2\) and that of the other is \(7\). Show that the difference between their 10th terms equals the difference between their 21st terms (indeed, between any two corresponding terms). Why?
The difference is constant and equals \(2-7=-5\) for every corresponding term.
Step 1: Recall the formula for the \(n\)th term of an AP:
\(a_n = a + (n-1)d\), where \(a\) is the first term and \(d\) is the common difference.
Step 2: Write the general term for both APs.
Step 3: Find the difference between the two corresponding terms.
\(T_n - T'_n = [2 + (n-1)d] - [7 + (n-1)d]\)
Step 4: Simplify:
\(T_n - T'_n = 2 - 7 = -5\)
Step 5: Notice that \((n-1)d\) cancels out, so the difference does not depend on \(n\). That means the difference will always be \(-5\), whether \(n=10\), \(n=21\), or any other \(n\).
Final Result: The difference between the 10th terms is also \(-5\), the difference between the 21st terms is also \(-5\), and in fact the difference between any corresponding terms is always \(-5\).