Question
If $a, b, c, d, e$ are in continued proportion, prove that: $a : e = a^4 : b^4.$
✓
Answer
$a, b, c, d, e$ are in continued proportion
$\Rightarrow \frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{c}= k \text { (say) }$
$ d = ek _{ c } \ c = ek ^2, b = ek ^3 \text { and } a = ek ^4$
Now $ \text{L.H.S.} =\frac{a}{e}$
$=\frac{e k^4}{e}$
$ =k^4$
$ \text{R.H.S.}$ $\frac{a^4}{b^4}$
$=\frac{\left(e k^4\right)^4}{\left(e k^3\right)^4}$
$ =\frac{e^4 k^{16}}{e^4 k^{12}}$
$= k ^{16-12}$
$ = k ^4$
$ \therefore \text { L.H.S. = R.H.S. }$
$\Rightarrow \frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{c}= k \text { (say) }$
$ d = ek _{ c } \ c = ek ^2, b = ek ^3 \text { and } a = ek ^4$
Now $ \text{L.H.S.} =\frac{a}{e}$
$=\frac{e k^4}{e}$
$ =k^4$
$ \text{R.H.S.}$ $\frac{a^4}{b^4}$
$=\frac{\left(e k^4\right)^4}{\left(e k^3\right)^4}$
$ =\frac{e^4 k^{16}}{e^4 k^{12}}$
$= k ^{16-12}$
$ = k ^4$
$ \therefore \text { L.H.S. = R.H.S. }$
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