Question

determine the AP whose third term is 16 and 17 exceeds the 5th term by 12

Answer 1

a+2d=16. ----1

a+16d=a+4d+12
16d-4d=12
12d=12
d=1.

substituting d=1 in eq 1
a+2d=16
a+2(1)=16
a+2=16
a=14

Finding AP

a,a+d,a+2d....
14,14+1,14+2(1)...

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...