Question

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

Answer 1

T3 =16

=> a + 2d = 16 --------(1)

T7 - T5 = 12

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

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

=> 2d = 12

=> d = 6

On Substituting the value of d in equation 1, we get

a + 2× 6 = 16

=> a + 12 = 16

=> a = 16 - 12

=> a = 4


Required AP is 4, 10, 16.........

Answer 2

A3=16
A7-A5=12
A+6d-A-4d=12
2d=12
D=6
A+2(6)=12
A+12=12
A=0
A2=a+d
0+6=6