Question

if abcd are in continued proportion than show
√a5+b2c2+a3c2/b4c+d4+b2cd2=a/d​

Answer 1

Given:

$a$, $b$, $c$ and $d$ are in continued proportion

To prove:

$\sqrt{\dfrac{a^{5}+b^{2}c^{2}+a^{3}c^{2}}{b^{4}c+d^{4}+b^{2}cd^{2}}}=\dfrac{a}{d}$

Step-by-step explanation:

Since $a$, $b$, $c$ and $d$ are in continued proportion,

$\quad\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=k$ (say), $k\neq 0$

Then, we have

$\quad a=bk,b=ck,c=dk$

$\Rightarrow a=dk^{3},b=dk^{2},c=dk$

Now, Left Hand Side

$=\sqrt{\dfrac{a^{5}+b^{2}c^{2}+a^{3}c^{2}}{b^{4}c+d^{4}+b^{2}cd^{2}}}$

$=\sqrt{\dfrac{(dk^{3})^{5}+(dk^{2})^{2}(dk)^{2}+(dk^{3})^{3}(dk)^{2}}{(dk^{2})^{4}(dk)+d^{4}+(dk^{2})^{2}(dk)d^{2}}}$

$=\sqrt{\dfrac{d^{5}k^{15}+d^{2}k^{4}.d^{2}k^{2}+d^{3}k^{9}.d^{2}k^{2}}{d^{2}k^{8}.dk+d^{4}+d^{2}k^{4}.dk.d^{2}}}$

$=\sqrt{\dfrac{d^{5}k^{15}+d^{4}k^{6}+d^{5}k^{11}}{d^{5}k^{9}+d^{4}+d^{5}k^{5}}}$

$=\sqrt{\dfrac{d^{4}k^{6}(dk^{9}+1+dk^{5})}{d^{4}(dk^{9}+1+dk^{5})}}$

$=\sqrt{k^{6}}$

$=k^{3}$

and Right Hand Side

$=\dfrac{a}{d}$

$=\dfrac{dk^{3}}{d}$

$=k^{3}$

So, Right Hand Side = Left Hand Side

Conclusion:

$\sqrt{\dfrac{a^{5}+b^{2}c^{2}+a^{3}c^{2}}{b^{4}c+d^{4}+b^{2}cd^{2}}}=\dfrac{a}{d}$

Proved.

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