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

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(C) Since $2a+b, a-b, a+3b$ are in $G.P.$,we have $(a-b)^2 = (2a+b)(a+3b)$.Expanding both sides: $a^2 - 2ab + b^2 = 2a^2 + 6ab + ab + 3b^2$.Rearrangin

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(C) Since $2a+b, a-b, a+3b$ are in $G.P.$,we have $(a-b)^2 = (2a+b)(a+3b)$.Expanding both sides: $a^2 - 2ab + b^2 = 2a^2 + 6ab + ab + 3b^2$.Rearranging the terms: $a^2 + 9ab + 2b^2 = 0$.Dividing by $a^2$ (since $a 
eq 0$): $2(\frac{b}{a})^2 + 9(\frac{b}{a}) + 1 = 0$.Let $x = \frac{b}{a}$. Then $2x^2 + 9x + 1 = 0$. The roots are $x_1 = \frac{b_1}{a_1}$ and $x_2 = \frac{b_2}{a_2}$.From the quadratic equation,$x_1 + x_2 = -\frac{9}{2}$ and $x_1x_2 = \frac{1}{2}$.We need to find $2(a_1b_2 + a_2b_1) + 9a_1a_2$.Divide the expression by $a_1a_2$: $\frac{2(a_1b_2 + a_2b_1) + 9a_1a_2}{a_1a_2} = 2(\frac{b_2}{a_2} + \frac{b_1}{a_1}) + 9 = 2(x_1 + x_2) + 9$.Substituting the sum of roots: $2(-\frac{9}{2}) + 9 = -9 + 9 = 0$.

There are two such pairs of non-zero real values of $a$ 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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