Math question?

Question

This is HL Math:

"The first three terms of geometric sequence are 1,x,y and the first three terms of an arithmetic sequence are 1,x,-y. Prove that x^2+2x-1=0 and hence find “y” given that “x” is positive."

can someone help me with this question?

Answer 1

1f ist 2 terms of GP are 1,x,y then

y = x^2 as common ratio is x

now if 1 x and -y are in AP

x-1 = -y - x (common difference)
or y +2x -1 = 0

or x^2+2x-1 =0 as y = x^2
proved

Answer 2

Let the increase in the geometric sequence be g and the arithmetic sequence be a.

Then: x = g and y = g^2. ie. y = x^2.
Also, x = 1 + a and -y = 1+ 2a.

Rewriting this...
x^2 = -1 - 2a
x = 1 + a.

Now: x^2 + 2x - 1 = -1 - 2a + 2(1 + a) - 1 = -1 - 2a + 2 + 2a - 1 = 0... (thus is it proved...)

Yeah...and from the given equation x^2 + 2x - 1 = 0, we have
x = -1 + root(8)/2 = -1 + root(2) since x is positive.

From y = x^2 we now have
y= (-1 + root(2))^2 = 1 + 2 - 2root(2) = 3 - 2root(2) = 3 - root(8) however you want it...

Hope this helps...

Answer 3

Geometric sequence: x^0=1, x^1=1, x^2=y

Arithmetic sequence: each successive difference is equal:
x-1 = -y-x

x-1=-y-x
2x-1=-y
y= -2x+1

x^2 = -2x+1
x^2+2x-1=0

x = +.414 and x = -2.414

So for the positive x value, y = -2x+1 = -2(.414)+1 =
-.828+1 = .172

Answer 4

By definiton: 1/x=x/y due to geo. sequence,
and 1-x=x-(-y) due to arith. sequence.
That gives y=x^2 and 2x=1-y. Substitute y=x^2.
2x=1-x^2 or x^2+2x-1=0. (1st result)
Solving this quadratic, x= {-2+/- sq.root (4+4)}/2,
2x=-2+/-sq.root 8, 2x= -2+/-2*sq. root 2, x=-1+/-sq. root 2.
As x is positive, the answer would be x= (sq.root 2)-1.
Hence y=x^2={(sq. root 2)-1}^2=2+1-2*sq. root 2=3-2*sq. root 2.
(2nd result)

Answer 5

you're on your own buddy! lol