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(B) The general term in the expansion of $\left(a x^2+\frac{b}{x^3}\right)^{15}$ is given by $T_{r+1} = {}^{15}C_r (a x^2)^{15-r} (b x^{-3})^r = {}^{1

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$(9: 4)$

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(B) The general term in the expansion of $\left(a x^2+\frac{b}{x^3}\right)^{15}$ is given by $T_{r+1} = {}^{15}C_r (a x^2)^{15-r} (b x^{-3})^r = {}^{15}C_r a^{15-r} b^r x^{30-5r}$.For $x^{10}$,$30-5r = 10$ $\Rightarrow 5r = 20$ $\Rightarrow r = 4$. Thus,$l = {}^{15}C_4 a^{11} b^4$.For the constant term,$30-5r = 0 \Rightarrow r = 6$. Thus,$m = {}^{15}C_6 a^9 b^6$.For $x^{-10}$,$30-5r = -10$ $\Rightarrow 5r = 40$ $\Rightarrow r = 8$. Thus,$n = {}^{15}C_8 a^7 b^8$.Given $\frac{l}{m} + \frac{m}{n} = \frac{26}{11}$.$\frac{l}{m} = \frac{{}^{15}C_4 a^{11} b^4}{{}^{15}C_6 a^9 b^6} = \frac{1365}{5005} \cdot \frac{a^2}{b^2} = \frac{3}{11} \cdot \frac{a^2}{b^2}$.$\frac{m}{n} = \frac{{}^{15}C_6 a^9 b^6}{{}^{15}C_8 a^7 b^8} = \frac{5005}{6435} \cdot \frac{a^2}{b^2} = \frac{7}{9} \cdot \frac{a^2}{b^2}$.Substituting these into the equation: $\frac{a^2}{b^2} \left( \frac{3}{11} + \frac{7}{9} \right) = \frac{26}{11}$.$\frac{a^2}{b^2} \left( \frac{27 + 77}{99} \right) = \frac{26}{11}$ $\Rightarrow \frac{a^2}{b^2} \left( \frac{104}{99} \right) = \frac{26}{11}$.$\frac{a^2}{b^2} = \frac{26}{11} \cdot \frac{99}{104} = \frac{9}{4}$.Therefore,$a^2 : b^2 = 9 : 4$.

Suppose $l, m, n$ respectively represent the coefficient of $x^{10}$,the constant term,and the coefficient of $x^{-10}$ in the expansion of $\left(a x^2+\frac{b}{x^3}\right)^{15}$. If $\frac{l}{m}+\frac{m}{n}=\frac{26}{11}$,then $a^2: b^2=$

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