can someone who is good at math help me with this problem that stumped me?

Question

w, x, y, and z are consecutive terms of an arithmetic sequence with common difference 1998. find z^2-w^2 / y^2-x^2.

* z^2 means z-squared, w^2 is w-squared, and so on.

I will give 10 points to the person who has the best explanation and correct answer.

Answer 1

Let w= the first term of the arithmetic sequence
a=the common difference =1998

Then x= w+a and
y= x+a

Substituting the value of x, we get

y= w+a+a
y= w+2a

z=y+a

Subsituting the value of y, we get

z=w+2a+a
z=w+3a

Now w,x,y,and z are all expressed in terms of w. Note that "a" is not a variable but is given as equals 1998. We use "a" just to simplify the calculations.

To find the value of the expression (z^2-w^2)/(y^2-x^2), we substitute the value of w,x,y,and z in the given expression as follows:

[(w+3a)^2-w^2]/
[w+2a]^2-(w+a)^2}

=(w^2+6aw+3^2a^2-w^2)/
[(w^2+4aw+2^2a^2-(w^2+2aw+a^2)]

= (6aw+9a^2)/
(w^2+4aw+4a^2-w^2-2aw-a^2)

=(6aw+9a^2)/(2aw+3a^2)

Factor out 3a from the numerator and a from the denominator:

3a(2w+3a)/a(2w+3a)

The a(2w+3a) will cancel out from the numerator and denominator leaving 3.

So the answer is 3.

What about 1998? Never mind. It only means that whatever is the amount of the common difference, the expression given will always yield the number 3. If you try to work the expressions using 1998, you'll still get 3. Of course. But you will likely be groggy after doing this exercise.That's the reason why we substituted a=1998. To simplify our calculations.

Answer 2

I think by consecutive terms of an arithmetic sequence with common difference 1998 you mean
x = w + 1998
y = x + 1998
z = y + 1998
and so forth. Yes?

If so,
y = x + 1998 = w + 2*1998
z = y + 1998 = w + 3*1998

Note that if you meant to evaluate (z^2-w^2)/ (y^2-x^2) you would get a different answer from what I am now going to tell you. By
z^2-w^2/ y^2-x^2 I am assuming you mean z^2-(w^2/ y^2)-x^2, which is the standard way to evaluate the expression you wrote though I am afraid you might have meant the other.

z^2-(w^2/ y^2)-x^2 = (w+3*1998)^2 - [w^2/(w + 2*1998)] - (w+1998)^2
= (w^2 + 2*3*1998*w + 3*3*1998^2) - [w^2/(w + 2*1998)] - (w^2 + 2*1998*w + 1998^2)
= [1/(w + 2*1998)] * [(w + 2*1998)*(w^2 + 2*3*1998*w + 3*3*1998^2) - w^2 - (w + 2*1998)(w^2 + 2*1998*w + 1998^2)]

Now it is necessary to multiply all this out and present it as powers of w.

I believe the other expression you might have meant would have come out simpler.

Answer 3

Since they are in arithmetic progression,
x = w + 1998
y = w + 1998 +1998 = w + 2*1998
z = w + 1998 +1998 +1998 = w + 3*1998
for our convenience we could write 1998 = a
then ,
x = w +a, y = w +2a, z= w +3a
now consider the given expression
z^2 - w^2 / y^2 - x^2
using the identity (a^2 - b^2 ) = (a-b) (a+b) on both numerator and denominator
(z -w )(z+w) / (y-x)(y +x)
z - w = 3a, z + w = w+3a +w = 2w +3a
y - x = a , y + x = w+a +w+2a = 2w +3a

substituting these values in the above equation
(3a) (2w+3a) / a(2w+3a)
all others will cancel out leaving just 3 which is the required answer.

Answer 4

Just to make sure you notation is correct because if this is interpreted wrong then this will be all wrong. Since you're not using parenthesis to group I am assuming this.
z^2-(w^2/y^2) - x^2 since that would be order of operations. We know the comman diff is 1998 and since an arith. seq we know the common diff is obtained by x-w = 1998, y-x=1998 and z-y=1998 so solving for each variable Z=1998+y, x=-1998+y, w=-1998+x
filling out the equation:
(y+1998)^2 - [(-1998+x)/(-1998+z)]^2 -(-1998+y)^2
so we can relate x and z in terms of y
(y+1998)^2 - [(-1998+(-1998+y))/(-1998+(1998+y)]^2 -(-1998+y)^2
And now everything is in like terms and reduce. Hopefully that clarifies.
Once you get Y plug back in Z=1998+Y to get Z and so fourth to get all other variables

Answer 5

If the common difference is 1998, then you can rewrite each variable in terms of w:
x = w + 1998
y = w + 3996
z = w + 5994

(z^2-w^2) / (y^2-x^2) can now be rewritten as:
((w + 5994)^2 - w^2) / ((w + 3996)^2 - (w + 1998)^2)
(w^2 + 11998w + 5994 - w^2) / (w^2 + 7992w + 3996 - w^2 - 3996w - 1998)
(11998w + 5994) / (3996w + 1998)
5994(2w + 1) / 1998 (2w + 1)
5994 / 1998 = 3 (solution!)

Answer 6

if i understood the question correctly that would be for example w=1, x=1999 (the two numbers having a difference of 1998),y=3997, and z=5995

this would then be (5995^2 - 1^2)/(3997^2-1999^2) = 3

Answer 7

I'm sorry I'm not god in math, but I hope this site helps you......

http://www.mathforum.org/dr.math/