Question
If a, b, c, d are in continued proportion, prove that $(b-c)^2+(c-a)^2+(d-b)^2=(d-a)^2$
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Answer
Since $a, b, c, d$ are in continued proportion, we have
$\frac{a}{b}=\frac{b}{c}=\frac{c}{d}= K \text { (say) } $
$\therefore c = dK , b = cK = dK ^2 \text { and } a = bK = dK ^3 . $
$\text { L.H.S. }$
$=(b-c)^2+(c-a)^2+(d-b)^2 $
$=\left(d K^2-d K\right)^2+\left(d K-d K^3\right)^2+\left(d-d K^2\right)^2 $
$=d^2 K^2(K-1)^2+d^2 K^2\left(1-K^2\right)^2+d^2\left(1-K^2\right)^2 $
$=d^2\left[K^2(K-1)^2+K^2\left(K^2-1\right)^2+d^2\left(k^2-1\right)^2\right] $
$=d^2\left[K^2(K-1)^2+K^2(K-1)^2(K+1)^2+(K-1)^2(K+1)^2\right] $
$=d^2(K-1)^2\left[K^2+K^2(K+1)^2+(K+1)^2\right] $
$=d^2(K-1)^2\left[K^2+K^2\left(K^2+2 K+1\right)+K^2+2 K+1\right] $
$=d^2(k-1)^2\left[K^4+2 K^3+3 K^2+2 K+1\right] $
$=d^2(K-1)^2\left(K^2+K+1\right)^2 $
$=d^2\left[(K-1)\left(K^2+K+1\right)^2\right] $
$=d^2\left(K^3-1\right)^2=\left(d K^3-d\right)^2=(a-d)^2=(d-a)^2=\text { R.H.S. }$
Hence, $(b-c)^2+(c-a)^2+(d-b)^2=(d-a)^2$.
$\frac{a}{b}=\frac{b}{c}=\frac{c}{d}= K \text { (say) } $
$\therefore c = dK , b = cK = dK ^2 \text { and } a = bK = dK ^3 . $
$\text { L.H.S. }$
$=(b-c)^2+(c-a)^2+(d-b)^2 $
$=\left(d K^2-d K\right)^2+\left(d K-d K^3\right)^2+\left(d-d K^2\right)^2 $
$=d^2 K^2(K-1)^2+d^2 K^2\left(1-K^2\right)^2+d^2\left(1-K^2\right)^2 $
$=d^2\left[K^2(K-1)^2+K^2\left(K^2-1\right)^2+d^2\left(k^2-1\right)^2\right] $
$=d^2\left[K^2(K-1)^2+K^2(K-1)^2(K+1)^2+(K-1)^2(K+1)^2\right] $
$=d^2(K-1)^2\left[K^2+K^2(K+1)^2+(K+1)^2\right] $
$=d^2(K-1)^2\left[K^2+K^2\left(K^2+2 K+1\right)+K^2+2 K+1\right] $
$=d^2(k-1)^2\left[K^4+2 K^3+3 K^2+2 K+1\right] $
$=d^2(K-1)^2\left(K^2+K+1\right)^2 $
$=d^2\left[(K-1)\left(K^2+K+1\right)^2\right] $
$=d^2\left(K^3-1\right)^2=\left(d K^3-d\right)^2=(a-d)^2=(d-a)^2=\text { R.H.S. }$
Hence, $(b-c)^2+(c-a)^2+(d-b)^2=(d-a)^2$.
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