There are two such pairs of non-zero real val | vedclass

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Question

$(a-b)^{2}=(2 a+b)(a+3 b)$

$a^{2}+2 b^{2}+9 a b=0$

$2.\left(\frac{b}{a}\right)^{2}+9 \cdot\left(\frac{b}{a}\right)+1=0$

$\left(\frac{b_{1}}{a

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$(a-b)^{2}=(2 a+b)(a+3 b)$

$a^{2}+2 b^{2}+9 a b=0$

$2.\left(\frac{b}{a}ight)^{2}+9 \cdot\left(\frac{b}{a}ight)+1=0$

$\left(\frac{b_{1}}{a_{1}}+\frac{b_{2}}{a_{2}}ight)=-\frac{9}{2}$

$	herefore 9 a_{1} a_{2}+2\left(a_{1} b_{2}+a_{2} b_{1}ight)=9 a_{1} a_{2}+2 a_{1} a_{2}$

$\left(\frac{b_{1}}{a_{1}}+\frac{b_{2}}{a_{2}}ight)=9 a_{1} a_{2}-9 a_{1} a_{2}=0$

There are two such pairs of non-zero real valuesof $a$ and $b$ i.e. $(a_1,b_1)$ and $(a_2,b_2)$ for which $2a+b,a-b,a+3b$ are three consecutive terms of a $G.P.$, then the value of $2(a_1b_2 + a_2b_1) + 9a_1a_2$ is-

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