Question
If $a, b, c, d$ are in continued proportion, prove that:
$
\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a}{d}
$

Answer

$a, b, c, d$ are in continued proportion
$
\begin{aligned}
& \therefore \frac{a}{b}=\frac{b}{c}=\frac{c}{d}= k \text { (say) } \\
& \therefore c = dk _{ t } b = ck = dk . k = dk ^2, \\
& a = bk = dk ^2 \cdot k = dk ^3 \\
& \text { L.H.S. }=\frac{a^3+b^3+c^3}{b^3+c^3+d^3} \\
& =\frac{(d k)^3+\left(d k^2\right)^3+(d k)^3}{\left(d k^2\right)^3+(d k)^3+d^3} \\
& =\frac{d^3 k^9+d^3 k^6+d^3 k^3}{d^3 k^6+d^3 k^3+d^3} \\
& =\frac{d^3 k^3\left(k^6+k^3+1\right)}{d^3\left(k^6+k^3+1\right)} \\
& = k ^3 \\
& \text { R.H.S. }=\frac{a}{d} \\
& =\frac{d k^3}{d} \\
& = k ^3 \\
& \therefore \text { L.H.S. }=\text { R.H.S. }
\end{aligned}
$

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