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Find the 9th term in the expansion of \(\Big(\frac{a}{b}-\frac{b}{2a^2}\Big)^{12}\)

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To find: 9th term in the expansion of \(\Big(\frac{a}{b}-\frac{b}{2a^2}\Big)^{12}\) 

Formula used: (i)

 nCr\(\frac{n!}{(n-r)!(r)!}\)

For 9th term, r+1=9 

⇒ r = 8

In, \(\Big(\frac{a}{b}-\frac{b}{2a^2}\Big)^{12}\) 

9th term = T8+1

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