Question

determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12

Answer 1

a3=16
a+2d=16.....(1)
a7=a5+12
a+6d=a+4d+12
a-a+6d-4d=12
2d=12
d=12/2=6
Put d=6 in (1)
a+2d=16
a+2*6=16
a=16-12=4
AP is 4,10,16,22....

Answer 2

Answer

a3 = 16

=> a + 2d = 16

=> a = 16 - 2d

a7 = a5 + 12

=> a + 6d = a + 4d + 12

=> 2d= 12

=> d = 6

a = 16 - 2d

=> a = 16 - 12

=> a = 4

AP = 4, 10, 16...