Find x and y if the terms of an A. P. are x, 10, y and the terms of a G. P. are x, y, 5.

We are given that x, 10 and y are terms of an AP and x , y
and 5 are terms of a GP.

Now for an AP we know that twice
any term is the sum of the term before and after it.

So we
have x+ y = 2*10 = 20 ...(1)

Similarly for a GP we know
that the square of any term is the product of the terms before and after the
term.

So we have y^2 =
5x....(2)

From (1) and (2) we can frame y^2/ 5 + y
=20

=> y^2 + 5y =
100

=> y^2 + 5y -100 =
0

Therefore y = [-b + sqrt ( b^2 - 4ac)] / 2a and y = [-b -
sqrt ( b^2 - 4ac)] / 2a

Substituting we
get

y = [-5 + sqrt( 25 + 400) ]/ 2 and [-5 - sqrt( 25 +
400) ]/ 2

y = -5/2 + (sqrt 425)/2 adn y = -5/2 - (sqrt
425)/2

x= 20 -
y

Therefore for

y = -5/2 + (sqrt 425)/2, x=
22.5 - (sqrt 425)/2

and for y
= -5/2 - (sqrt 425)/2, x= 22.5 + (sqrt 425)/2.