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

(i) $\dfrac{a + b}{b + c} = \dfrac{a^{2}(b - c)}{b^{2}(a - b)}$

(ii) $\dfrac{a + b + c}{a - b + c} = \dfrac{(a + b + c)^{2}}{(a^{2} + b^{2} + c^{2})}$

(iii) $\dfrac{a^{2} + ab + b^{2}}{b^{2} + bc + c^{2}} = \dfrac{a}{c}$

(iv) (a + b + c)(a − b + c) = (a² + b² + c²)

(v) a²b²c²(a⁻³ + b⁻³ + c⁻³) = (a³ + b³ + c³)

(vi) ad(c² + d²) = c³(b + d)

(i) Given,

⇒ a, b, c are in continued proportion

∴ a : b = b : c

$\Rightarrow \dfrac{a}{b} = \dfrac{b}{c}$ = k (let)

⇒ b = ck, a = bk = (ck)k = ck².

Substituting values of a and b in L.H.S. of equation $\dfrac{a + b}{b + c} = \dfrac{a^{2}(b - c)}{b^{2}(a - b)}$, we get :

$\Rightarrow \dfrac{a + b}{b + c} \\[1em] \Rightarrow \dfrac{ck^2 + ck}{ck + c} \\[1em] \Rightarrow \dfrac{c k(k + 1)}{c(k + 1)} \\[1em] \Rightarrow k.$

Substituting values of a and b in R.H.S. of equation $\dfrac{a + b}{b + c} = \dfrac{a^{2}(b - c)}{b^{2}(a - b)}$, we get :

$\Rightarrow \dfrac{a^2(b - c)}{b^2(a - b)} \\[1em] \Rightarrow \dfrac{(ck^2)^2\big(ck - c\big)}{(ck)^2\big(ck^2 - ck\big)} \\[1em] \Rightarrow \dfrac{c^2k^4(ck - c)}{c^2k^2(ck^2 - ck)} \\[1em] \Rightarrow \dfrac{c^3 k^4 (k - 1)}{c^3k^3(k - 1)} \\[1em] \Rightarrow k.$

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

Hence, proved that $\dfrac{a + b}{b + c} = \dfrac{a^{2}(b - c)}{b^{2}(a - b)}$.

(ii) Given,

⇒ a, b, c are in continued proportion

∴ a : b = b : c

$\Rightarrow \dfrac{a}{b} = \dfrac{b}{c}$ = k (let)

⇒ b = ck, a = bk = (ck)k = ck².

Substituting values of a and b in L.H.S. of equation $\dfrac{a + b + c}{a - b + c} = \dfrac{(a + b + c)^{2}}{(a^{2} + b^{2} + c^{2})}$, we get :

$\Rightarrow \dfrac{a + b + c}{a - b + c} \\[1em] \Rightarrow \dfrac{ck^2 + ck + c}{ck^2 - ck + c} \\[1em] \Rightarrow \dfrac{c(k^2 + k + 1)}{c(k^2 - k + 1)} \\[1em] \Rightarrow \dfrac{k^2 + k + 1}{k^2 - k + 1}.$

Substituting values of a and b in R.H.S. of equation $\dfrac{a + b + c}{a - b + c} = \dfrac{(a + b + c)^{2}}{(a^{2} + b^{2} + c^{2})}$, we get :

$\Rightarrow \dfrac{(a + b + c)^2}{a^2 + b^2 + c^2} \\[1em] \Rightarrow \dfrac{(ck^2 + ck + c)^2}{(ck^2)^2 + (ck)^2 + c^2} \\[1em] \Rightarrow \dfrac{c^2(k^2 + k + 1)^2}{c^2(k^4 + k^2 + 1)} \\[1em] \Rightarrow \dfrac{(k^2 + k + 1)^2}{(k^4 + k^2 + 1)} \\[1em] \Rightarrow \dfrac{(k^2 + k + 1)^2}{(k^2 + k + 1)(k^2 - k + 1)} \\[1em] \Rightarrow \dfrac{k^2 + k + 1}{k^2 - k + 1}.$

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

Hence, proved that $\dfrac{a + b + c}{a - b + c} = \dfrac{(a + b + c)^{2}}{(a^{2} + b^{2} + c^{2})}$.

(iii) Given,

⇒ a, b, c are in continued proportion

∴ a : b = b : c

$\Rightarrow \dfrac{a}{b} = \dfrac{b}{c}$ = k (let)

⇒ b = ck, a = bk = (ck)k = ck².

Substituting values of a and b in L.H.S. of equation $\dfrac{a^{2} + ab + b^{2}}{b^{2} + bc + c^{2}} = \dfrac{a}{c}$, we get :

$\Rightarrow \dfrac{a^2 + ab + b^2}{b^2 + bc + c^2} \\[1em] \Rightarrow \dfrac{(ck^2)^2 + (ck^2)(ck) + (ck)^2}{(ck)^2 + (ck)c + c^2} \\[1em] \Rightarrow \dfrac{c^2k^4 + c^2k^3 + c^2k^2}{c^2k^2 + c^2k + c^2} \\[1em] \Rightarrow \dfrac{c^2(k^4 + k^3 + k^2)}{c^2(k^2 + k + 1)} \\[1em] \Rightarrow \dfrac{k^2(k^2 + k + 1)}{(k^2 + k + 1)} \\[1em] \Rightarrow k^2.$

Substituting values of a and b in R.H.S. of equation $\dfrac{a^{2} + ab + b^{2}}{b^{2} + bc + c^{2}} = \dfrac{a}{c}$, we get :

$\Rightarrow \dfrac{a}{c} \\[1em] \Rightarrow \dfrac{ck^2}{c} \\[1em] \Rightarrow k^2.$

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

Hence, proved that $\dfrac{a^{2} + ab + b^{2}}{b^{2} + bc + c^{2}} = \dfrac{a}{c}$.

(iv) Given,

⇒ a, b, c are in continued proportion

∴ a : b = b : c

$\Rightarrow \dfrac{a}{b} = \dfrac{b}{c}$ = k (let)

⇒ b = ck, a = bk = (ck)k = ck².

Substituting values of a and b in L.H.S. of (a + b + c)(a − b + c) = (a² + b² + c²), we get:

⇒ (a + b + c)(a − b + c)

⇒ ((ck²) + (ck) + c)((ck²) − (ck) + c)

⇒ c(k² + k + 1). c(k² - k + 1)

⇒ c²(k⁴ + k² + 1)

Substituting values of a and b in R.H.S. of (a + b + c)(a − b + c) = (a² + b² + c²), we get:

⇒ (a² + b² + c²)

⇒ (ck²)² + (ck)² + c²

⇒ (c²k⁴ + c²k² + c²)

⇒ c²(k⁴ + k² + 1)

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

Hence, proved that (a + b + c)(a − b + c) = (a² + b² + c²).

(v) Given,

⇒ a, b, c are in continued proportion

∴ a : b = b : c

$\Rightarrow \dfrac{a}{b} = \dfrac{b}{c}$ = k (let)

⇒ b = ck, a = bk = (ck)k = ck².

Substituting values of a and b in L.H.S. of a²b²c²(a⁻³ + b⁻³ + c⁻³) = (a³ + b³ + c³), we get:

$\Rightarrow a^2 b^2 c^2\big(a^{ - 3} + b^{ - 3} + c^{ - 3}\big) \\[1em] \Rightarrow c^6 k^6\Big(\dfrac{1}{c^3 k^6} + \dfrac{1}{c^3 k^3} + \dfrac{1}{c^3}\Big) \\[1em] \Rightarrow c^3\big(1 + k^3 + k^6\big).$

Substituting values of a and b in R.H.S. of a²b²c²(a⁻³ + b⁻³ + c⁻³) = (a³ + b³ + c³), we get:

$\Rightarrow a^3 + b^3 + c^3 \\[1em] \Rightarrow (ck^2)^3 + (ck)^3 + c^3 \\[1em] \Rightarrow c^3k^6 + c^3k^3 + c^3 \\[1em] \Rightarrow c^3(k^6 + k^3 + 1).$

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

Hence, proved that a²b²c²(a⁻³ + b⁻³ + c⁻³) = (a³ + b³ + c³).

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

$\therefore \dfrac{a}{b} = \dfrac{b}{c} = \dfrac{c}{d} = k$ (let).

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

Substituting values in L.H.S. of the equation ad(c² + d²) = c³(b + d), we get :

L.H.S = ad(c² + d²)

= dk³.(d).[(dk)² + d²]

= d²k³.[d²(k² + 1)]

= d⁴k³(k² + 1).

Substituting values in R.H.S. of the equation ad(c² + d²) = c³(b + d), we get :

R.H.S = c³(b + d)

= (dk)³.(dk² + d)

= d³k³[d(k² + 1)]

= d⁴k³(k² + 1).

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

Hence, proved that ad(c² + d²) = c³(b + d).