Question

determine the ap whose third term is 16 and the 7th term exceeds the fifth term by 12

Answer 1

Let a be the first term and d be common difference.
T3=16
a+2d=16--(i)
T7-T5=12
a+6d-(a-4d)=12
Or,2d=12
Or, d=6-(ii)
Putting d=6 in (i),
a+12=16
Or, a=16-12
Or, a=4
Therefore the ap is
a, a+d, a+2d,a+3d,...
=>4,10,16,22,....is the AP... (ans)

Answer 2

a + 2d = 16 || 3rd term || eq.i
a + 6d - a - 4d = 12 ||eq.ii
2d = 12
d = 6 ||eq.iii
Put eq.iii in eq.i
a + 2(6) = 16
a = 4
AP = 4,10,16,24...
HOPE IT HELPS YOU :)