Question

determine the the ap whose third term is 16 and the 7th term exceeds the 5th term by 12.

Answer 1

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...

Answer 2

Solution:

Given:

$\sf{\implies a_{3}=16}$

=> 7th term exceeds 5th term by 12

To Find:

=> A.P

Formula used:

$\sf{\implies a_{n}=a+(n-1)d}$

So, it is given that 7th term exceeds 5th term by 12

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

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

=> 2d = 12

=> d = 12/2

=> d = 6

Now, it is given that 3rd term is 16. So,

=> a + 2d = 16

Now, put the value of d in above equation

=> a + 2d = 16

=> a + 2 × 6 = 16

=> a + 12 = 16

=> a = 16 - 12

=> a = 4

∴ Therefore A.P is 4, 10, 16, 22........