Question

Determine the AP whose 3rd term is 16 and 7th term exceeds 5th term by 12

Answer 1

HEYA!!!!

Here is your answer :
__________________________

Given,

7th term exceeds 5th term by 12

a + 6d = a + 4d + 12

6d - 4d = 12

2d = 12

d = 6,,


From,

3rd term = 16

a + 2d = 16

a + 2 (6) = 16

a + 12 = 16

a = 4

Therefore,

AP = 4 , 10 , 16.........




HOPE THIS HELPS U. . .

Answer 2

The answer is given below :

Let us consider that the first term of the AP is a and the common difference is d.

Now,

3rd term = 16

=> a + (3 - 1)d = 16

=> a + 2d = 16 .....(i)

7th term = a + (7 - 1)d = a + 6d

and

5th term = a + (5 - 1)d = a + 4d

Given that,

7th term = 5th term + 12

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

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

=> 2d = 12

=> d = 6

So, common difference, d = 6.

Putting d = 6 in (i), we get

a + (2 × 6) = 16

=> a + 12 = 16

=> a = 4

So, first term of the AP is 4.

The AP is an infinite one.

Therefore, the required AP is

4, 4 + 6, 4 + 6 + 6, 4 + 6 + 6 + 6, ...

i.e., 4, 10, 16, 22, 28, 34, 40, ...

Thank you for your question.