Question

If b is the mean proportional of a and c,then prove that


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Answer 1

Given :-

• b is the mean proportional between a and c.

To prove :-

$\rm :\longmapsto\:{\bigg(\dfrac{ab + bc + ca}{a + b + c}\bigg) }^{3} = abc$

Solution :-

Since,

b is the mean proportion between a and c

$\bf\implies \:a,b,b,c \: are \: in \: proportion$

$\rm :\implies\:a : b :: b : c$

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

$\bf\implies \: {b}^{2} = ac$

Now,

Consider,

$\rm :\longmapsto\:{\bigg(\dfrac{ab + bc + ca}{a + b + c}\bigg) }^{3}$

$\rm : = \: \:{\bigg(\dfrac{ab + bc + {b}^{2} }{a + b + c}\bigg) }^{3} \: \: \: \: \{ \because \: {b}^{2} = ac \}$

$\rm : = \: \:{\bigg(\dfrac{b \: \: \cancel{(a + c + b)}}{ \cancel{a + b + c}}\bigg) }^{3} \:$

$\rm \: = \: \: {b}^{3}$

$\rm \: = \: \: b \times {b}^{2}$

$\rm \: = \: \: b \times ac\: \: \: \: \: \: \{ \because \: {b}^{2} = ac \}$

$\rm \: = \: \: abc$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

To find third proportional to a and b,

Third proportional (x) to a and b is

$\bf\implies \:a,b,b,x \: are \: in \: proportion$

$\rm :\implies\:a : b :: b : x$

$\rm :\implies\:\dfrac{a}{b} = \dfrac{b}{x}$

$\bf\implies \:x = \dfrac{ {b}^{2} }{a}$