Solution :
The $$7^{th}$$ term from the end $$= 5^{th}$$ term from beginning
$${T_5} = {\,^{10}}{C_4}{x^6}{\left( { - \frac{2}{{{x^2}}}} \right)^4} = {\,^{10}}{C_4} \cdot {2^4}\left( {\frac{1}{{{x^2}}}} \right)$$
Releted MCQ Question on Algebra >> Binomial Theorem
Releted Question 1
Given positive integers $$r > 1, n > 2$$ and that the co - efficient of $${\left( {3r} \right)^{th}}\,{\text{and }}{\left( {r + 2} \right)^{th}}$$ terms in the binomial expansion of $${\left( {1 + x} \right)^{2n}}$$ are equal. Then
If in the expansion of $${\left( {1 + x} \right)^m}{\left( {1 - x} \right)^n},$$ the co-efficients of $$x$$ and $${x^2}$$ are $$3$$ and $$- 6\,$$ respectively, then $$m$$ is