Question

please tell the answer of 25 (c).
It will be very helpful to me ​

Answer 1

✿$\underline{\textbf{\textsf{ \purple{Solution}:- }}}$

$\sf{\\}$

As $\sf{a,\:b,\:c,\:d}$ are in continued proportion,

$\sf{ \dfrac{a}{b} = \dfrac{b}{c} = \dfrac{c}{d} }$

Let it be equal to $\sf k$

$\longrightarrow \sf \dfrac{a}{b} = \dfrac{b}{c} = \dfrac{c}{d} = k\quad\dots(1)$

$\sf{\\}$

Triplicate ratio: The triplicate ratio of $\sf a : b$ is $\sf a^3 : b^3$

$\sf{\\}$

$\underline{\underline{\tt{ \red{To\:prove}:- }}}$

$\sf \: \dfrac{ {(a - b)}^{3} }{ {(b - c)}^{3} } = \dfrac{a}{d}$

$\underline{\underline{\tt{ \red{Proof}:- }}}$

From (1),

$\sf a = bk$

$\sf{b = ck}$

$\sf{c = dk}$

On substituting the value of c in b and then the value of b in a, we get,

$\quad\quad\bullet \sf a = dk^3$

$\quad\quad\bullet \sf b = dk^2$

$\quad\quad\bullet \sf c = dk$

$\sf{\\}$

$\underline{\underline{\bf{ \pink{L.H.S}:- }}}$

$\sf \: \dfrac{ {(a - b)}^{3} }{(b - c) ^{3} }$

$\longrightarrow \sf{\Bigg[ \dfrac{(dk^3 - dk^2)}{(dk^2 - dk)} \Bigg]^3}$

$\longrightarrow \sf{\Bigg[ \dfrac{(dk^2)(k - 1)}{(dk)(k - 1)} \Bigg]^3}$

$\longrightarrow \sf{\Bigg[ \dfrac{( \cancel{d}{k}^{\cancel{2}})\cancel{(k - 1)}}{\cancel{(dk)}\cancel{(k - 1)}} \Bigg]^3}$

$\longrightarrow \sf k^3$

$\underline{\underline{\bf{ \pink{R.H.S}:- }}}$

$\sf \dfrac{a}{d}$

$\longrightarrow \sf \dfrac{dk^3}{d}$

$\longrightarrow \sf \dfrac{\cancel{d} k^3 }{ \cancel{d}}$

$\longrightarrow \sf k^3$

$\sf{\\}$

$\leadsto \underline{\boxed{\mathsf{ \pink{L.H.S = R.H.S} }}}$

Hence proved.