Given x,16,y as the terms of an A.P. and x,y,8 as the terms of a G.P. What are x and y?

Given x,  16 and y are in
AP.

Give x, y and 8 ar in
G.P.

To find x and .

Since x,
6 and y are in AP, the successive terms should have the same
difference.

Therefore 16-x =
y-16.

Therefore 32 = y+x.
Or

x+y =
32.................(1).

Since x, y and 8 are in GP, the
successive terms should have the same ratio:

y/x =
8/y.

We multply by yboth
sides:

y^2 = 8x. Or

x =
(y^2)/8..................(2)

We substitute  x = (y^2)/8  in
(1):

 (y^2/8) + y =
32

Multiply by 8.

y^2 + 8y=
32*8

y^2+8y -256 = 0.

y1 =
{-8+sqrt(8^2+4*1*256)}/2

y1 =
{-4+4sqrt7)

y2 =
(-4-4sqrt17).

if y1 = -4+4sqrt17, then x = 32-y = 32
-(-4+4sqrt17) = 36 -4sqrt17.

If y2 = -4-4sqrt17, then, x =
32- y1 = 32 - (-4 - 4sqrt17) = 36+4sqrt17.

Therefore (x ,
y)  =  (36-4sqrt17 ,  -4+4sqrt17)

 Or (x , y) = (36+4sqrt17
,  -4-4sqrt17).