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 $4, 10, 16, 22, ….$