Question

If a, b, c, d are in proportion then prove that $\frac{{11a}^2+9ac}{{11b}^2+9bd}=\frac{a^2+3ac}{b^2+3bd}$

Answer 1

Given a,b,c and d are in proportion

$\frac{a}{b}=\frac{c}{d}--------\left(1\right)$

ad=bc

$a=\frac{bc}{d}$

$\frac{11a^2+9ac}{11b^2+9bd}=\frac{a^2+3ac}{b^2+3bd}$

$\Rightarrow \frac{a(11a+9c)}{b(11b+9d)}=\frac{a(a+3c)}{b(b+3d)}$

$\Rightarrow \frac{a\left(11\right(\frac{bc}{d})+9c)}{b(11b+9d)}=\frac{a(\frac{bc}{d}+3c)}{b(b+3d)}$

$\Rightarrow \frac{ac(11b+9d)}{bd(11b+9d)}=\frac{ac(b+3d)}{bd(b+3d)}$

$\Rightarrow \frac{ac}{bd}=\frac{ac}{bd}$