Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
4, 10, 16, 22, ...
Let the AP be \(a, a + d, a + 2d, a + 3d, \ldots\), where \(a\) is the first term and \(d\) is the common difference.
The 3rd term of the AP is given as 16.
3rd term: \(a_3 = a + 2d\).
So,
\[a + 2d = 16 \quad ...(1)\]
We are told that the 7th term exceeds the 5th term by 12.
7th term: \(a_7 = a + 6d\).
5th term: \(a_5 = a + 4d\).
Condition: \(a_7 - a_5 = 12\).
Substitute:
\[(a + 6d) - (a + 4d) = 12\]
\[a + 6d - a - 4d = 12\]
\[2d = 12\]
So,
\[d = 6\]
Substitute \(d = 6\) into equation (1):
\[a + 2d = 16\]
\[a + 2(6) = 16\]
\[a + 12 = 16\]
\[a = 16 - 12 = 4\]
First term: \(a = 4\).
Common difference: \(d = 6\).
So the AP is:
\[4, 4 + 6, 4 + 2 \cdot 6, 4 + 3 \cdot 6, \ldots = 4, 10, 16, 22, \ldots\]
Conclusion: The required AP is 4, 10, 16, 22, ...