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Mathematics

If a, b, c, d are in continued proportion, prove that :

(a−bc+a−cc)2−(d−bc+d−cb)2=(a−d)2(1c2+1b2)\Big(\dfrac{a - b}{c} + \dfrac{a - c}{c}\Big)^2 - \Big(\dfrac{d - b}{c} + \dfrac{d - c}{b}\Big)^2 = (a - d)^2 \Big(\dfrac{1}{c^2} + \dfrac{1}{b^2}\Big)

Ratio Proportion

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Answer

Given a, b, c, d are in continued proportion.

∴ a : b = b : c = c : d

ab=bc=cd\dfrac{a}{b} = \dfrac{b}{c} = \dfrac{c}{d} = k (let)

⇒ c = dk, b = ck = (dk)k = dk2, a = bk = (dk2)k = dk3.

Substituting values of a, b and c in L.H.S. of (a−bc+a−cc)2−(d−bc+d−cb)2=(a−d)2(1c2+1b2)\Big(\dfrac{a - b}{c} + \dfrac{a - c}{c}\Big)^2 - \Big(\dfrac{d - b}{c} + \dfrac{d - c}{b}\Big)^2 = (a - d)^2 \Big(\dfrac{1}{c^2} + \dfrac{1}{b^2}\Big) , we get :

⇒(dk3−dk2dk+dk3−dk2dk2)2−(d−dk2dk+d−dkdk2)2⇒(k(dk3−dk2)+dk3−dkdk2)2−(k(d−dk2)+d−dkdk2)2⇒(dk4−dk3+dk3−dkdk2)2−(kd−dk3+d−dkdk2)2⇒(dk4−dkdk2)2−(d−dk3dk2)2⇒(dk(k3−1)dk2)2−(d(1−k3)dk2)2⇒(d2k2(k3−1)2d2k4)−(d2(1−k3)2d2k4)⇒((k3−1)2k2)−((1−k3)2k4)⇒(k6+1−2k3k2)−(1+k6−2k3k4)⇒(k2(k6+1−2k3)−(1+k6−2k3)k4)⇒(k8+k2−2k5−1−k6+2k3k4).\Rightarrow \Big(\dfrac{dk^3 - dk^2}{dk} + \dfrac{dk^3 - dk^2}{dk^2}\Big)^2 - \Big(\dfrac{d - dk^2}{dk} + \dfrac{d - dk}{dk^2}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{k(dk^3 - dk^2) + dk^3 - dk}{dk^2}\Big)^2 - \Big(\dfrac{k(d - dk^2) + d - dk}{dk^2}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{dk^4 - dk^3 + dk^3 - dk}{dk^2}\Big)^2 - \Big(\dfrac{kd - dk^3 + d - dk}{dk^2}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{dk^4 - dk}{dk^2}\Big)^2 - \Big(\dfrac{d - dk^3}{dk^2}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{dk(k^3 - 1)}{dk^2}\Big)^2 - \Big(\dfrac{d(1 - k^3)}{dk^2}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{d^2k^2(k^3 - 1)^2}{d^2k^4}\Big) - \Big(\dfrac{d^2(1 - k^3)^2}{d^2k^4}\Big) \\[1em] \Rightarrow \Big(\dfrac{(k^3 - 1)^2}{k^2}\Big) - \Big(\dfrac{(1 - k^3)^2}{k^4}\Big) \\[1em] \Rightarrow \Big(\dfrac{k^6 + 1 - 2k^3}{k^2}\Big) - \Big(\dfrac{1 + k^6 - 2k^3}{k^4}\Big) \\[1em] \Rightarrow \Big(\dfrac{k^2(k^6 + 1 - 2k^3) - (1 + k^6 - 2k^3)}{k^4}\Big) \\[1em] \Rightarrow \Big(\dfrac{k^8 + k^2 - 2k^5 - 1 - k^6 + 2k^3}{k^4}\Big).

Substituting values of a, b and c in R.H.S. of (a−bc+a−cc)2−(d−bc+d−cb)2=(a−d)2(1c2+1b2)\Big(\dfrac{a - b}{c} + \dfrac{a - c}{c}\Big)^2 - \Big(\dfrac{d - b}{c} + \dfrac{d - c}{b}\Big)^2 = (a - d)^2 \Big(\dfrac{1}{c^2} + \dfrac{1}{b^2}\Big) , we get :

=(a−d)2(1c2+1b2)=(dk3−d)2(1(dk)2+1(dk2)2)=(dk3−d)2(1d2k2+1d2k4)=d2d2k2(k3−1)2(1−1k2)=(k3−1)2(k2−1)k4=(k6+1−2k3)(k2−1)k4=k8−k6+k2−1+2k3−2k5k4.= (a - d)^2 \Big(\dfrac{1}{c^2} + \dfrac{1}{b^2}\Big) \\[1em] = (dk^3 - d)^2 \Big(\dfrac{1}{(dk)^2} + \dfrac{1}{(dk^2)^2}\Big) \\[1em] = (dk^3 - d)^2 \Big(\dfrac{1}{d^2k^2} + \dfrac{1}{d^2k^4}\Big) \\[1em] = \dfrac{d^2}{d^2k^2}(k^3 - 1)^2 \Big(1 - \dfrac{1}{k^2} \Big) \\[1em] = \dfrac{(k^3 - 1)^2(k^2 - 1)}{k^4} \\[1em] = \dfrac{(k^6 + 1 - 2k^3)(k^2 - 1)}{k^4} \\[1em] = \dfrac{k^8 - k^6 + k^2 - 1 + 2k^3 - 2k^5}{k^4}.

Since, L.H.S. = R.H.S.

Hence, proved that (a−bc+a−cc)2−(d−bc+d−cb)2=(a−d)2(1c2+1b2)\Big(\dfrac{a - b}{c} + \dfrac{a - c}{c}\Big)^2 - \Big(\dfrac{d - b}{c} + \dfrac{d - c}{b}\Big)^2 = (a - d)^2 \Big(\dfrac{1}{c^2} + \dfrac{1}{b^2}\Big) .

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