I didn't come up with this problem, but I still think it's a good one.
There are 4 positive integers in order from least to greatest, such that the first three make an aritmetic sequence, and the last three make a geometric sequence. If the difference between the largest and smallest term is 30, what are the terms?
(In reply to
solution by Charlie)
Good job. I'm getting tired of finding complete solutions already posted, and felt I ought to contribute something once in a while, so here it goes.
I don't dispute the correctness of your solution, but it isn't obvious that your final equation
w = (4a^2 - 30a)/(30-3a)
has only 9 as a possible value for a. I did my algebra a bit differently.
Using your notation and starting from here:
(w+2a)^2 - w^2 - wa = 30 (w+a)
I got:
w^2+4wa+4a^2-w^2-wa =30 (w+a)
a(3w+4a)/(w+a) = 30
Now, since w and a are both positive integers,
3(w+a) < 3w+4a < 4(w+a)
so 3a < 30 < 4a
The only integer possibilities are then 8 and 9. 8 gives a non-integral value
for w, as you know. The end result is of course the same, but your final leap was not obvious to me at all.
re: solution
