If $a + b + c$ = 15 and $a^{2} + b^{2} + c^{2}$ = 77,
then ( $a b + b c + c a)$ is :

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144

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148

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Solution

The correct option is A 74
Given, $a + b + c = 15 a n d a^{2} + b^{2} + c^{2} = 77$ .
Using the identity $(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a$ ,
$\Rightarrow 15^{2} = 77 + 2 \times (a b + b c + c a)$
$\Rightarrow 2 \times (a b + b c + c a) = 225 - 77 = 148$
$\Rightarrow (a b + b c + c a) = 74$ .

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