If four quantities are in continued proportion,show that difference between the first and last is at least 3 times the difference between the other 2.

Question

I found this question in Higher Algebra by Hall and Knight. I am not able to attack the question in any way. Any help would be appreciated. Thank you.

Answer 1

I think in modern language we would speak of four terms of a geometric progression, so $a, ar, ar^2, ar^3$. Then the question is to prove that $|a-ar^3|\geq 3|ar-ar^2|$.

It's probably easiest to split into cases depending on the values of $a$ and $r$, in order to avoid the modulus signs. For example, if $a\geq 0$ and $r\geq 1$ then the sequence is increasing, so in this case it is sufficient to prove $a(r^3-1)>3a(r^2-r)$, i.e. $a(r^3-3r^2+3r-1)\geq 0$ - can you see why this is true?

[edit: it seems the question must expect you to assume $r\geq 0$ since otherwise the result is not true, e.g. for $1,-2,4,-8$.]

Answer 2

Without loss of generality, assume $a/b = b/c = c/d = k $, k is positive. Then I can write $(a, b, c, d)$ as $ (k^3 d, k^2d, kd, d) $. Then $a-d = d(k^3 - 1)$ and $b-c = (k^2 - k)d. $ Since $k^3-1 = (k-1)(k^2 + k +1) , (a-d)/(b-c) = (k^2 +k+1)/k >= 3$ as $k^2 - 2k +1 = (k-1)^2>= 0.$ Equality exists when $k=1$, i.e. when all are equal.