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

(i) $\dfrac{a^3 + b^3 + c^3}{b^3 + c^3 + d^3} = \dfrac{a}{d}$

(ii) (a² - b²)(c² - d²) = (b² - c²)²

(iii) (a + d)(b + c) - (a + c)(b + d) = (b - c)²

(iv) a : d = triplicate ratio of (a - b) : (b - c)

(v) $\big(\dfrac{a - b}{c} + \dfrac{a - c}{b}\big)^2 - \big(\dfrac{d - b}{c} + \dfrac{d - c}{b}\big)^2 = (a - d)^2\big(\dfrac{1}{c^2} - \dfrac{1}{b^2}\big).$

(i) Since, a, b, c, d are in continued proportion

$\therefore \dfrac{a}{b} = \dfrac{b}{c} = \dfrac{c}{d} = k \\[1em] \therefore c = dk, b = ck = (dk)k = dk^2 \text{ and } a = bk = (ck)k = (dk)k^2 = dk^3. \\[1em] \text{L.H.S.} = \dfrac{a^3 + b^3 + c^3}{b^3 + c^3 + d^3} \\[1em] = \dfrac{(dk^3)^3 + (dk^2)^3 + (dk)^3}{(dk^2)^3 + (dk)^3 + d^3} \\[1em] = \dfrac{d^3k^9 + d^3k^6 + d^3k^3}{d^3k^6 + d^3k^3 + d^3} \\[1em] = \dfrac{d^3k^3(k^6 + k^3 + 1)}{d^3(k^6 + k^3 + 1)} \\[1em] = k^3. \\[1em] \text{R.H.S.} = \dfrac{a}{d} \\[1em] = \dfrac{dk^3}{d} \\[1em] = k^3. \\[1em]$

Since, L.H.S. = R.H.S. hence, proved that,

$\dfrac{a^3 + b^3 + c^3}{b^3 + c^3 + d^3} = \dfrac{a}{d}$.

(ii) Since, a, b, c, d are in continued proportion

$\therefore \dfrac{a}{b} = \dfrac{b}{c} = \dfrac{c}{d} = k \\[1em] \therefore c = dk, b = ck = (dk)k = dk^2 \text{ and } a = bk = (ck)k = (dk)k^2 = dk^3. \\[1em] \text{L.H.S.} = (a^2 - b^2)(c^2 - d^2) \\[1em] = [(dk^3)^2 - (dk^2)^2][(dk)^2 - (d^2)] \\[1em] = (d^2k^6 - d^2k^4)(d^2k^2 - d^2) \\[1em] = d^2k^4(k^2 - 1)d^2(k^2 - 1) \\[1em] = d^4k^4(k^2 - 1)^2. \\[1em] \text{R.H.S.} = (b^2 - c^2)^2 \\[1em] = ((dk^2)^2 - (dk)^2)^2 \\[1em] = (d^2k^4 - d^2k^2)^2 \\[1em] = (d^2k^4- d^2k^2)(d^2k^4 - d^2k^2) \\[1em] = d^2k^2(k^2 - 1)d^2k^2(k^2 - 1) \\[1em] = d^4k^4(k^2 - 1)^2. \\[1em]$

Since, L.H.S. = R.H.S. hence, proved that,

(a² - b²)(c² - d²) = (b² - c²)².

(iii) Since, a, b, c, d are in continued proportion

$\therefore \dfrac{a}{b} = \dfrac{b}{c} = \dfrac{c}{d} = k \\[1em] \therefore c = dk, b = ck = (dk)k = dk^2 \text{ and } a = bk = (ck)k = (dk)k^2 = dk^3. \\[1em] \text{L.H.S.} = (a + d)(b + c) - (a + c)(b + d) \\[1em] = (dk^3 + d)(dk^2 + dk) - (dk^3 + dk)(dk^2 + d) \\[1em] = d(k^3 + 1)dk(k + 1) - dk(k^2 + 1)d(k^2 + 1) \\[1em] = d^2k[(k^3 + 1)(k + 1) - (k^2 + 1)^2] \\[1em] = d^2k[(k^4 + k^3 + k + 1) - (k^4 + 1 + 2k^2)] \\[1em] = d^2k(k^4 - k^4 + k^3 - 2k^2 + k + 1 - 1) \\[1em] = d^2k(k^3 - 2k^2 + k) \\[1em] = d^2k(k(k^2 - 2k + 1)) \\[1em] = d^2k^2(k^2 - 2k + 1) \\[1em] = d^2k^2(k - 1)^2. \\[1em] \text{R.H.S.} = (b - c)^2 \\[1em] = (dk^2 - dk)^2 \\[1em] = [(dk)^2(k - 1)^2] \\[1em] = d^2k^2(k - 1)^2.$

Since, L.H.S = R.H.S, hence proved that,

(a + d)(b + c) - (a + c)(b + d) = (b - c)².

(iv) Since, a, b, c, d are in continued proportion

$\therefore \dfrac{a}{b} = \dfrac{b}{c} = \dfrac{c}{d} = k \\[1em] \therefore c = dk, b = ck = (dk)k = dk^2 \text{ and } a = bk = (ck)k = (dk)k^2 = dk^3. \\[1em] \text{L.H.S.} = a : d \\[1em] = \dfrac{a}{d} \\[1em] = \dfrac{dk^3}{d} \\[1em] = k^3. \\[1em]$

R.H.S. = triplicate ratio of (a - b) : (b - c)

$= (a - b)^3 : (b - c)^3 \\[1em] = \dfrac{(a - b)^3}{(b - c)^3} \\[1em] = \dfrac{(dk^3 - dk^2)^3}{(dk^2 - dk)^3} \\[1em] = \dfrac{[k(dk^2 - dk)]^3}{(dk^2 - dk)^3} \\[1em] = \dfrac{k^3(dk^2 - dk)^3}{(dk^2 - dk)^3} \\[1em] = k^3.$

Since, L.H.S. = R.H.S. hence proved that,

a : d = triplicate ratio of (a - b) : (b - c).

(v) Since, a, b, c, d are in continued proportion

$\therefore \dfrac{a}{b} = \dfrac{b}{c} = \dfrac{c}{d} = k \\[1em] \therefore c = dk, b = ck = (dk)k = dk^2 \text{ and } a = bk = (ck)k = (dk)k^2 = dk^3. \\[1em] \text{L.H.S.} = \big(\dfrac{a - b}{c} + \dfrac{a - c}{b}\big)^2 - \big(\dfrac{d - b}{c} + \dfrac{d - c}{b}\big)^2. \\[1em] = \big(\dfrac{dk^3 - dk^2}{dk} + \dfrac{dk^3 - dk}{dk^2}\big)^2 - \big(\dfrac{d - dk^2}{dk} + \dfrac{d - dk}{dk^2}\big)^2. \\[1em] = \big(\dfrac{k(dk^3 - dk^2) + dk^3 - dk}{dk^2} \big)^2 - \big(\dfrac{k(d - dk^2) + d - dk}{dk^2}\big)^2. \\[1em] = \big(\dfrac{dk^4 - dk^3 + dk^3 - dk}{dk^2} \big)^2 - \big(\dfrac{kd - dk^3 + d - dk}{dk^2}\big)^2. \\[1em] = \big(\dfrac{dk^4 - dk}{dk^2} \big)^2 - \big(\dfrac{d - dk^3}{dk^2}\big)^2. \\[1em] = \big(\dfrac{dk(k^3 - 1)}{dk^2}\big)^2 - \big(\dfrac{d(1 - k^3)}{dk^2}\big)^2 \\[1em] = \dfrac{d^2k^2(k^3 - 1)^2}{d^2k^4} - \dfrac{d^2(1 - k^3)^2}{d^2k^4} \\[1em] = \dfrac{(k^3 - 1)^2}{k^2} - \dfrac{(1 - k^3)^2}{k^4} \\[1em] = \dfrac{k^6 + 1 -2k^3}{k^2} - \dfrac{1 + k^6 - 2k^3}{k^4} \\[1em] = \dfrac{k^2(k^6 + 1 - 2k^3) - (1 + k^6 - 2k^3)}{k^4} \\[1em] = \dfrac{k^8 + k^2 - 2k^5 - 1 - k^6 + 2k^3}{k^4} \\[1em] \text{R.H.S.} = (a - d)^2\big(\dfrac{1}{c^2} - \dfrac{1}{b^2}\big) \\[1em] = (dk^3 - d)^2\big(\dfrac{1}{(dk)^2} - \dfrac{1}{(dk^2)^2}\big) \\[1em] = d^2(k^3 - 1)^2 \big(\dfrac{1}{d^2k^2} - \dfrac{1}{d^2k^4}\big) \\[1em] = \dfrac{d^2}{d^2k^2}(k^3 - 1)^2\big(1 - \dfrac{1}{k^2}\big) \\[1em] = \dfrac{(k^3 - 1)^2(k^2 - 1)}{k^4} \\[1em] = \dfrac{(k^6 + 1 - 2k^3)(k^2 - 1)}{k^4} \\[1em] = \dfrac{k^8 - k^6 + k^2 - 1 + 2k^3 - 2k^5}{k^4}$

Since, L.H.S = R.H.S hence proved that,

$\big(\dfrac{a - b}{c} + \dfrac{a - c}{b}\big)^2 - \big(\dfrac{d - b}{c} + \dfrac{d - c}{b}\big)^2 = (a - d)^2\big(\dfrac{1}{c^2} - \dfrac{1}{b^2}\big).$