Question

If $a,b,c$ are in continued proportion then the value of $\frac{a^2+ab+b^2}{b^2+bc+c^2}$ is equal to __.

• A. $0$
• B. $1$
• C. $\frac{b}{c}$
• D. $\frac{a}{c}$

Answer 1

The correct option is D $\frac{a}{c}$

Given $a,b,c$ are in continued proportion which means $\frac{a}{b}=\frac{b}{c}$.

Let $\frac{a}{b}=\frac{b}{c}$ = k.

$\Rightarrow a=bk,b=ck,a=c{k}^2$ ...(i)

Now $\frac{a^2+ab+b^2}{b^2+bc+c^2}$ $=$

$\frac{(c{k}^2{)}^2+k^{2}c\times ck+(ck{)}^2}{(ck{)}^2+kc\times c+(c{)}^2}$

$\Rightarrow \frac{c^2{k}^4+c^2{k}^3+c^2{k}^2}{c^2{k}^2+c^{2}k+c^2}=\frac{c^2{k}^2(k^2+k+1)}{c^2(k^2+k+1)}=k^2$

From the equation (i) $\frac{a}{c}=k^2$

Hence, $\frac{a^2+ab+b^2}{b^2+bc+c^2}=\frac{a}{c}$