Question

there are four numbers in ap ,the sum of two extremes is 8 and the product of the middle two is 15 find the numbers

Answer 1

Let the first term be a and the C.D be d.
The series is a,a+d,a+2d,a+3d 
Accordingly, 
a + a + 3d = 8 ......................................... (1)
(a+d)(a+2d)= 15......................................... (2)
from (1),
a= 8-3d/2
Now in (2),
(8-3d/2 + d)(8-3d/2 + 2d) = 15
=> (8-3d+2d)/2*(8-3d+4d)/2 = 15
=> (8-d)(8+d) = 15*4
=> 64-d^2=60
=> d^2 = 64-60 =4
=> d = +2 or -2
a = 8-3.2/2
= 8-6/2 
=1
a = 8-3.(-2)/2
= 8+6/2
=14/2=7
The series is : (1,3,5,7) / (7,5,3,1)

Answer 2

Answer:

1,3,5,7

Step-by-step explanation:

Let the first term be a and the C.D be d.

The series is a,a+d,a+2d,a+3d 

Accordingly, 

a + a + 3d = 8 ......................................... (1)

(a+d)(a+2d)= 15......................................... (2)

from (1),

a= 8-3d/2

Now in (2),

(8-3d/2 + d)(8-3d/2 + 2d) = 15

=> (8-3d+2d)/2*(8-3d+4d)/2 = 15

=> (8-d)(8+d) = 15*4

=> 64-d^2=60

=> d^2 = 64-60 =4

=> d = +2 or -2

a = 8-3.2/2

= 8-6/2 

=1

a = 8-3.(-2)/2

= 8+6/2

=14/2=7

The series is : (1,3,5,7) / (7,5,3,1)