Solution :
We know that the (r + 1)th term or general term in the expansion of \((x + a)^n\) is given by
\(T_{r + 1}\) = \(^{n}C_r x^{n – r} a^r\)
In the expansion of \(({x\over a} – {3a\over x^2})^{12}\), we have
\(T_{9}\) = \(T_{8 + 1}\) = \(^{12}C_8 ({x\over a})^{12 – 8} ({-3a\over x^2})^8\)
\(\implies\) \(T_{9}\) = \(^{12}C_8 ({x\over a})^4 ({-3a\over x^2})^8\)
\(\implies\) \(T_{9}\) = \(^{12}C_4\) \(3^8 a^4\over x^{12}\) = \((^{12}C_4 x^{-12} a^4)3^8\)
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