Question

$Ifa+b+c=m\text{and}{a}^2+b^2+c^2=n,\text{then}(ab+bc+ca)is$

• A. $\frac{m^2-n}{2}$
• B. $\frac{n^2-m}{2}$
• C. $\frac{m^2-n}{4}$
• D. $\frac{n^2-m}{4}$

Answer 1

The correct option is A $\frac{m^2-n}{2}$

Given, $a+b+c=m\text{and}{a}^2+b^2+c^2=n$.
Using the identity $(a+b+c{)}^2=a^2+b^2+c^2+2ab+2bc+2ca$
$\Rightarrow m^2=n+2(ab+bc+ca)$
$\Rightarrow 2(ab+bc+ca)=m^2-n$
$\Rightarrow (ab+bc+ca)=\frac{m^2-n}{2}$