Question

Solve it guys please ​

Answer 1

Hope it help

Mark me brainliest :-)

Answer 2

$\large\underline{\sf{Given- }}$

$\rm :\longmapsto\:\dfrac{a}{b} = \dfrac{b}{c}$

$\large\underline{\sf{To\:prove - }}$

$\rm :\longmapsto\:\dfrac{ {(a + b)}^{2} }{ {b}^{2} } = \dfrac{ {(b + c)}^{2} }{ {c}^{2} }$

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:\dfrac{a}{b} = \dfrac{b}{c}$

Let assume that

$\rm :\longmapsto\:\dfrac{a}{b} = \dfrac{b}{c} = k$

So,

$\rm :\longmapsto\:b = ck$

and

$\rm :\longmapsto\:a = bk = ck \times k = {ck}^{2}$

Now,

Consider LHS

$\rm :\longmapsto\:\dfrac{ {(a + b)}^{2} }{ {b}^{2} }$

$\rm \: = \: \: \dfrac{ {( {ck}^{2} + ck)}^{2} }{ {(ck)}^{2} }$

$\rm \: = \: \: \dfrac{ {( {ck}(k + 1))}^{2} }{ {c}^{2} {k}^{2} }$

$\rm \: = \: \: \dfrac{ {c}^{2} {k}^{2} {(k + 1)}^{2} }{ {c}^{2} {k}^{2} }$

$\rm \: = \: \: {(k + 1)}^{2}$

Now, Consider RHS

$\rm :\longmapsto\:\dfrac{ {(b + c)}^{2} }{ {c}^{2} }$

$\rm \: = \: \: \:\dfrac{ {(ck + c)}^{2} }{ {c}^{2} }$

$\rm \: = \: \: \:\dfrac{ {(c(k + 1))}^{2} }{ {c}^{2} }$

$\rm \: = \: \: \:\dfrac{ { {c}^{2} (k + 1)}^{2} }{ {c}^{2} }$

$\rm \: = \: \: {(k + 1)}^{2}$

Thus,

$\rm :\longmapsto\:\dfrac{ {(a + b)}^{2} }{ {b}^{2} } = \dfrac{ {(b + c)}^{2} }{ {c}^{2} }$

Additional Information :-

If a : b = c : d, then

$\rm :\longmapsto\:\dfrac{a}{c} = \dfrac{b}{d} \: is \: called \: alternendo.$

$\rm :\longmapsto\:\dfrac{b}{a} = \dfrac{d}{c} \: is \: called \: invertendo.$

$\rm :\longmapsto\:\dfrac{a + b}{b} = \dfrac{c + d}{d} \: is \: called \: componendo.$

$\rm :\longmapsto\:\dfrac{a - b}{b} = \dfrac{c - d}{d} \: is \: called \: dividendo.$