If four positive quantities are in continued proportion, show that the difference between the first and last is at least three times as great as the difference between the other two.

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Solution

Let the four quantities are $a, b, c, d$

$\Rightarrow \frac{a}{b} = \frac{b}{c} = \frac{c}{d} = \frac{1}{r}$ (say)

$\Rightarrow b = a r, c = a r^{2}, d = a r^{3}$

$\frac{d - a}{c - b} = \frac{a r^{3} - a}{a r^{2} - a r} \frac{d - a}{c - b} = \frac{r^{3} - 1}{r (r - 1)} \frac{d - a}{c - b} = \frac{(r - 1) (r^{2} + r + 1)}{r (r - 1)} \frac{d - a}{c - b} = \frac{r^{2} + r + 1}{r} = \frac{r^{2} - 2 r + 1 + 3 r}{r} \frac{d - a}{c - b} = \frac{3 r + {(r - 1)}^{2}}{r} \frac{d - a}{c - b} = 3 + \frac{{(r - 1)}^{2}}{r} \Rightarrow \frac{d - a}{c - b} \geq$

Hence proved.

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If four positive quantities are in continued proportion, show that the difference between the first and last is at least three times as great as the difference between the other two.

Let the four quantities are a,b,c,d ⇒ab=bc=cd=1r (say) ⇒b=ar,c=ar2,d=ar3 d−ac−b=ar3−aar2−ard−ac−b=r3−1r(r−1)d−ac−b=(r−1)(r2+r+1)r(r−1)d−ac−b=r2+r+1r=r2−2r+1+3rrd−ac−b=3r+(r−1)2rd−ac−b=3+(r−1)2r⇒d−ac−b≥ Hence proved.