Question

If $\frac{a}{b}=\frac{c}{d}=\frac{e}{f}$ and $\frac{2a^4{b}^2+3a^2{c}^2-5e^{4}f}{2b^6+3b^2{d}^2-5f^5}={\left(\frac{a}{b}\right)}^n$ then the value if $n$ is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is D 4

Let $\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=r$

$\Rightarrow a=br;c=dr;e=fr$

Given that $\frac{2a^4{b}^2+3a^2{c}^2-5e^{4}f}{2b^6+3b^2{d}^2-5f^5}=(\frac{a}{b}{)}^n$

Substitute the above obtained values in this expression, we get

Given that $\frac{2(br{)}^4{b}^2+3(br{)}^2(dr{)}^2-5(fr{)}^{4}f}{2b^6+3b^2{d}^2-5f^5}=(\frac{a}{b}{)}^n$

$\Rightarrow \frac{2r^4{b}^6+3b^2{d}^2{r}^4-5f^5{r}^4}{2b^6+3b^2{d}^2-5f^5}=(\frac{a}{b}{)}^n$

$\Rightarrow \frac{r^4(2b^6+3b^2{d}^2-5f^5)}{2b^6+3b^2{d}^2-5f^5}=(\frac{a}{b}{)}^n$

$\Rightarrow r^4=(\frac{a}{b}{)}^n$

But $r=\frac{a}{b}$

$\Rightarrow (\frac{a}{b}{)}^4=(\frac{a}{b}{)}^n$

Therefore, $r=4$