If $a, b, c, d$ are in continued proportion, prove that
$2 a + 3 d : 3 a - 4 d = 2 a^{3} + 3 b^{3} : 3 a^{3} - 4 b^{3}$ .

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Solution

We know $\frac{a}{b} = \frac{b}{c} = \frac{c}{d} = k$ , then
$a = b k = c k^{2} = d k^{3} and b = c k = d k^{2} and c = d k$
L.H.S. $= \frac{2 a + 3 d}{3 a - 4 d} = \frac{2 d k^{3} + 3 d}{3 d k^{3} - 4 d} = \frac{2 k^{3} + 3}{3 k^{3} - 4}$
R.H.S. $= \frac{2 a^{3} + 3 b^{3}}{3 a^{3} - 4 b^{3}} = \frac{2 d^{3} k^{9} + 3 d^{3} k^{6}}{3 d^{3} k^{9} - 4 d^{3} k^{6}} = \frac{d^{3} k^{6} (2 k^{3} + 3)}{d^{3} k^{6} (3 k^{3} - 4)} = \frac{2 k^{3} + 3}{3 k^{3} - 4}$
$∴$ L.H.S. $=$ R.H.S.

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