# Geometric Progression Problem

1. Jul 14, 2008

### crays

Hi guys,
In a geometric progression, the second term exceeds the first term by 20 and the fourth term exceeds the second term by 15. Find the possible values of the first term.

ar - a = 20
ar3 - ar = 15

ar = 20 + a

(20 + a)3 - (20 + a) = 15

But this method seem to be wrong. Any help?

2. Jul 14, 2008

### HallsofIvy

Staff Emeritus
Looks about as good as you could expect. it eleminates r. If you let x= 20+ a, then you have x3- x= 15. By the "rational root theorem", the only possible rational roots are integer divisors of 15: 1, 3, 5, and 15. It is easy to see that none of those work (13- 1= 0, 33- 3= 24, 53-5= 120, and 153- 15= 3360) and so x, and therefore a must be an irrational number. Probably a numerical solution is the best you could do.

3. Jul 14, 2008

### crays

but the answer says -40 and -8. I don't get it @_@.

4. Jul 14, 2008

### HallsofIvy

Staff Emeritus
Did you check that solution? if a= -40 and r= -8, then ar= 320 which is NOT 20 more than -40.

Or if you meant a= -8 and r= -40, then ar= 320 again and the second terms is still NOT 20 more than the first term.

Either the answer given is wrong or you are mis-reading the problem.

5. Jul 14, 2008

### crays

it says, find the possible values of the first term. Values, so i assume, there is 2 possibility?

6. Jul 14, 2008

### rock.freak667

$$a(r-1)=20$$

$$ar(r^2-1)=15$$

try dividing them

7. Jul 14, 2008

### HallsofIvy

Staff Emeritus
Ouch! I think I will go back and delete my post so I can pretend I never wrote it!
That's very nice, rockfreak667.

crays, your original mistake (and mine, since I didn't notice it) was thinking that since ar= 20+ a, ar3- ar= 15 would be (20+r)3- (20+r)= 15. That wrong because it is only r that is cubed in ar3. It is NOT
(ar)3.

Last edited: Jul 15, 2008