For the AP \(-3,-7,-11,\ldots\), can we find \(a_{30}-a_{20}\) directly without actually finding \(a_{30}\) and \(a_{20}\)? Explain.
Yes. \(a_{30}-a_{20}=(30-20)\,d=10(-4)=-40\).
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: To find the difference between two terms of an AP, say \(a_n - a_m\), we can subtract: \(a_n - a_m = [a + (n-1)d] - [a + (m-1)d]\).
Step 3: Simplify the subtraction: \(a_n - a_m = (n-1)d - (m-1)d = (n-m)d\).
Step 4: Apply this to our problem: \(a_{30} - a_{20} = (30 - 20) d\).
Step 5: The common difference of the given AP is \(d = -7 - (-3) = -4\).
Step 6: Substitute the values: \(a_{30} - a_{20} = (10)(-4) = -40\).
So, we do not need to calculate \(a_{30}\) and \(a_{20}\) separately. The answer is -40.