# Arithmetic Sequence

1. Jan 9, 2004

### Derivative86

The first four terms in an arithmetic sequence are x+y, x-y, xy, x/y, in that order. What is the fifth term?
Can anyone show me how to do it?
O btw, the fifth term is a number, is not a variable
Thx

Last edited: Jan 9, 2004
2. Jan 9, 2004

### Hurkyl

Staff Emeritus
Well, you know the first term: x + y.

You know the second term: x - y.

So, you can find the difference between successive terms.

Thus, you can write down the third and fourth terms in a different way, and you'll have two equations in two unknowns.

3. Jan 9, 2004

### Derivative86

No, it would be 3 variable, becuz of the common difference between them, and the system is very complicated, cuz is in terms of x and y

4. Jan 9, 2004

### Hurkyl

Staff Emeritus
Yes, but you can solve for the difference in terms of x and y, so there are only two unknowns!

And it won't be too complicated; have a little faith and try it. If you get stuck, post how far you got and I can nudge you the rest of the way.

5. Mar 12, 2009

### K Sengupta

Let us substitute x = py. Then,((p+1)y, (p-1)y, p*y^2 and p (in this order) are first four terms in an arithmetic sequence with a common difference (c.d) of -2y.

Thus, (p-3)y = p*y^2 …(i), and:
(p-5)y = p …(ii)

From (i), we note that y must be nonzero, otherwise x/y becomes indeterminate. Therefore, (i) forces y = (p-3)/p, while from (ii), we obtain: y = p/(p-5)

Hence, (p-3)/p = p/(p-5), whereupon in terms of cross multiplication, we obtain:

p^2 = p^2 – 8p+ 15, so that: p = 15/8, giving: y = -15/25 = -3/5, and:
x = (15/8)*(-3/5) = -9/8

So, the c.d = -2y = 6/5, and accordingly:

The required fifth term = p – 2y = 15/8 + 6/5 = 123/40

Last edited: Mar 12, 2009