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.

neela | High School Teacher | (Level 3) Valedictorian

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Given the information that x , 10 and y are in A.P. And x, y and 5 are in GP. We have to determine x and y.

x, 10 and y are in AP. Therefore the successive terms has the same common difference.

10-x = y-10

20 = x+y.

x+y = 20.............(1)

By second condition, x,y and 5 are in GP. Therefore the successive terms are in the same ratio.

y/x = 5/y

y^2/5 = x............(2)

We substitute x = y^2/5 in (1):

y^2/5 +y = 20

Multiply by 5:

y^2+5y = 100.

Subtract 100:

y^2+5y -100 = 0.

y1 = {-5+sqrt(5^2-4*1*-100)}/2

y1 = {-5+5sqrt17}/2

y2 ={-5-5sqrt17}/2.

Therefore x +y = 20

x = 20-y

x = 20-(-5+sqrt517)/2

x = (45-5sqrt17)/2

Therefore x = (45-5sqrrt17)/2 ,  y=  (-5+5sqrt17)/2 .

x = (45 +5sqrt17)/2 ,  y= -(5+5sqrt17)/2.

william1941 | College Teacher | (Level 3) Valedictorian

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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.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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If x,10,y are terms of an AP, then the middle terms is the arithmetical average of the neighbor terms:

10 = (x+y)/2

We'll cross multiply and we'll get:

x + y = 20

y = 20 - x (1)

If x,y and 5 are the terms of a GP, then the middle terms is the geometric mean of the neighbor terms:

x*5 = y^2

y = sqrt (5x) (2)

We'll substitute (1) in (2):

20 - x = sqrt (5x)

We'll raise to square both sides:

(20-x)^2 = 5x

400 - 40x + x^2 - 5x = 0

We'll combine like terms:

x^2 - 45x + 400 = 0

x1 = [45+sqrt(2025 - 1600)]/2

x1 = (45+5sqrt17)/2

y1 = sqrt[5*(45+5sqrt17)/2]

y1 = 12.80

x2 = (45-5sqrt17)/2

y2 = 60.975