Question

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

Answer 1

3rd term =
a+2d = 16

5th term =
a+4d

7th term =
a+6d

According to question,
7th term - 5th term = 12
a+6d - (a+4d) = 12
a+6d - a-4d =12
2d = 12
d = 6

Common difference (d) is 6

Now, putting the value of d in 3rd term equation.

a+2d = 16
a+2*6 = 16
a = 16-12
a = 4

First term of AP is 4

AP is 4, 10, 16, 22, 28 ....

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...