Q57 / 160 · TS EAMCET 2020 (11 Sep Shift 1) · Mathematics

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 $a^2: b^2=$

  • A
    $16: 9$
  • Correct answer
    B
    $9: 4$
  • C
    $4: 1$
  • D
    $1: 25$

Solution

Given binomial $\left(a x^2+\frac{b}{x^3}\right)^{15}$, so the general term, $T_{r+1}={ }^{15} C_r a^{15-r} b^r x^{30-5 r}$ $\therefore$ The coefficient of $x^{10}$ is (at $r=4$ ) $={ }^{15} C_4 a^{11} b^4=l$ (given) Similarly, $m =\text { the constant term (at } r=6) \\ ={ }^{15} C_6 a^9 b^6$ and $n=$ coefficient of $x^{-10}$ is (at $\left.r=8\right)={ }^{15} C_8 a^7 b^8$ $\because \quad \frac{l}{m}+\frac{m}{n}=\frac{26}{11}$ $\Rightarrow \quad \frac{{ }^{15} C_4 a^{11} b^4}{{ }^{15} C_6 a^9 b^6}+\frac{{ }^{15} C_6 a^9 b^6}{{ }^{15} C_8 a^7 b^8}=\frac{26}{11}$ $\Rightarrow \quad \frac{6 ! 9 ! a^2}{4 ! 11 ! b^2}+\frac{8 ! 7 ! a^2}{6 ! 9 ! b^2}=\frac{26}{11}$ $\Rightarrow \frac{6 \times 5}{11 \times 10}\left(\frac{a^2}{b^2}\right)+\frac{7}{9}\left(\frac{a^2}{b^2}\right)=\frac{26}{11} \Rightarrow \frac{a^2}{b^2}\left(\frac{3}{11}+\frac{7}{9}\right)=\frac{26}{11}$ $\Rightarrow \quad \frac{a^2}{b^2}\left(\frac{27+77}{99}\right)=\frac{26}{11} \Rightarrow \frac{a^2}{b^2}=\frac{9 \times 26}{104}=\frac{9}{4}$

Hence, the correct answer is option: 'B'.

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