Determine the AP whose third term is 16

Solution:

$a_{3}=16$

$a+(3-1) d=16$

$a+2 d=16$ ...(1)

$a_{7}-a_{5}=12$

$[a+(7-1) d]-[a+(5-1) d]=12$

$(a+6 d)-(a+4 d)=12$

$2 d=12$

d = 6

From equation (1), we obtain

a + 2 (6) = 16

a + 12 = 16

a = 4

Therefore, A.P. will be

$4,10,16,22, \ldots$