Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
The required AP is 4, 10, 16, 22, 28, . . .
Explanation
This question is similar to Example 5 in the textbook, where we find an AP given specific terms. The key insight is that when the 7th term exceeds the 5th term by 12, we can directly find the common difference since the difference between these terms equals 2d. This is a straightforward application of the nth term formula.
Solution Steps
Step 1: Let the first term be a and common difference be d. We know that the nth term of an AP is given by aₙ = a + (n-1)d
Step 2: Given that the third term is 16, so a₃ = a + 2d = 16 ... (1)
Step 3: Given that the 7th term exceeds the 5th term by 12. We have a₇ - a₅ = 12. This means (a + 6d) - (a + 4d) = 12, which gives 2d = 12, so d = 6
Step 4: Substituting d = 6 in equation (1): a + 2(6) = 16, which gives a + 12 = 16, so a = 4
Step 5: Therefore, the first term a = 4 and common difference d = 6. The required AP is 4, 10, 16, 22, 28, . . .