Question

Determine the AP whose third term is \(16\) and the \(7^{th}\) term exceeds the \(5^{th} \) term by \(12\).

Answer 1

\(a_3 = 16\)
\(a + (3 – 1) d = 16\)
\(a + 2d = 16  \)     \(……….(1)\)
\(a_7 − a_5 = 12\)
\(a_7 − a_5 = 12 [a+ (7 − 1) d] − [a + (5 − 1) d] = 12\)
\((a + 6d) − (a + 4d) = 12\)
\(2d = 12\)
\(d = 6\)
From equation (1), we obtain
\(a + 2 (6) = 16\)
\(a + 12 = 16\)
\(a = 4\)

Therefore, A.P. will be \(4, 10, 16, 22, ….\)