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.
A7 = A5 + 12
A + 6d = A + 4d = 12
2d = 12
d = 6.
Now
Substituing the value of d= 6 in
A + 2 (6) = 16
A = 4
Thus,
A.P = 4,4+6,4+6+6,4+6+6+6
= 4,10,16,22,......,infinity

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...