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?

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giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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If x,16,y are terms of an AP, then the middle terms is the arithmetical average of the neighbor terms:

16 = (x+y)/2

We'll cross multiply and we'll get:

x + y = 32 

y = 32 - x (1)

If x,y and 8 are the terms of a GP, then the middle terms is the geometric mean of the neighbor terms:

x*8 = y^2

y = sqrt (8x) (2)

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

32 - x = sqrt (8x)

We'll raise to square both sides:

(32-x)^2 = 8x

1024 - 64x + x^2 - 8x = 0

We'll combine like terms:

x^2 - 72x + 1024 = 0

Ww'll apply the quadratic formula:

x1 = [72+sqrt(5184 - 4096)]/2

x1 = (72+8sqrt17)/2

x1 = 36 + 4sqrt17

x2 = 36 - 4sqrt17

y = sqrt[8*(36+4sqrt17)]

y1 = 20.48

y2 = 12.49

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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).

 

 

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