Question

If the 5th, 9th terms of an A.P is 19& 35 then find 12th term of the sequency? ​

Answer 1

Given: 5th and 9th terms of an AP are 19 and 35 respectively.

To find: 12th term of the sequence.

Solution:

We know any term of an arithmetic progression (AP) is given by,

an = a + (n - 1)d

Therefore,

5th term of the AP is given by,

• a5 = a + 4d ______(1.)

And also,

9th term of the AP is given by,

• a9 = a + 8d _______(2.)

Now, subtract equation 1 from equation 2

=> a9 - a5 = (a + 8d) - (a + 4d)

=> 35 - 19 = a + 8d - a - 4d

=> 16 = 4d

=> 16/4 = d

=> 4 = d

Therefore, the common difference of AP is 4.

Now substitute this value in equation (1).

=> a5 = a + 4d

=> 19 = a + 4(4)

=> 19 = a + 16

=> 19 - 16 = a

=> 3 = a

So the first term of AP is 3.

Now, the required 12th term of AP is given by,

=> a12 = a + 11d

=> a12 = 3 + 11(4)

=> a12 = 3 + 44

=> a12 = 47

Hence the required answer is 47.

Answer 2

Given :   5th, 9th terms of an A.P is 19& 35

To Find : 12th term

Solution:

nth term = a  + (n - 1) d

5th term = a + 4d  = 19

9th term = a + 8d  = 35

=> 4d = 16

=> d = 4

12th term = a  + 11d

= a  + 8d  + 3d

= 35 + 3(4)

= 35 + 12

= 47

Hence 12th term is 47

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