Question

In an A.P the fourth term exceed four times the 12th term by one and the third term exceeds twice than 10 th term by five find the ap​

Answer 1

Answer:

According to the question

a

4

=4a

12

+1

a

1

+3d=4(a

1

+11d)+1

a

1

+3d=4a

1

+44d+1

−3a

1

=41d+1

−a

1

=

3

41d+1

............(1)

a

3

=2a

10

+5

a

1

+2d=2(a

1

+9d)+5

a

1

+2d=2a

1

+18d+5

−a

1

=16d+5

From eq (1)

3

41d+1

=16d+5

41d+1=48d+15

−7d=14

d=−2

a

1

=

−3

41d+1

=

−3

41×−2+1

=21

∴ series in AP =27,25,23,21