# Find the middle terms in the expansion of    Q8.    $\left(\frac{x}{3} + 9y \right )^{10}$

P Pankaj Sanodiya

As we know that the middle term in the expansion of  $(a+b)^n$ when n is even is,

$\left ( \frac{n}{2}+1 \right )^{th}\:term\:$,

Hence the middle term of the expansion     $\left(\frac{x}{3} + 9y \right )^{10}$   is,

$\left ( \frac{10}{2}+1 \right )^{th}\:term\:$

Which is $6^{th}\:term$

Now,

As we know that the general  $(r+1)^{th}$ term  $T_{r+1}$ in the binomial expansion of  $(a+b)^n$  is given by

$T_{r+1}=^nC_ra^{n-r}b^r$

So the $6^{th}$ term of the expansion of   $\left(\frac{x}{3} + 9y \right )^{10}$ is

$\\\Rightarrow T_6= T_{5+1}\\=^{10}C_5\left ( \frac{x}{3} \right )^{10-5}\left ( 9y \right )^5\\$

$=\left ( \frac{1}{3} \right )^5\times9^5\times^{10}C_5\times x^5y^5$

$=\left ( \frac{1}{3} \right )^5\times9^5\times\left ( \frac{10!}{5!5!} \right )\times x^5y^5$

$=\left ( \frac{1}{3^5} \right )\times9^5\times\left ( \frac{10\times9\times8\times7\times6}{5\times4\times3\times2} \right )\times x^5y^5$

$=61236x^5y^5$

Hence the middle term of the expansion of  $\left(\frac{x}{3} + 9y \right )^{10}$ is  $61236x^5y^5$.

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