Question
In the binomial expansion of $${\left( {a - b} \right)^n},n \geqslant 5,$$ the sum of $${5^{th}}$$ and $${6^{th}}$$ terms is zero, then $$\frac{a}{b}$$ equals
A.
$$\frac{{n - 5}}{6}$$
B.
$$\frac{{n - 4}}{5}$$
C.
$$\frac{5}{{n - 4}}$$
D.
$$\frac{6}{{n - 5}}$$
Answer :
$$\frac{{n - 4}}{5}$$
Solution :
$${T_{r + 1}} = {\left( { - 1} \right)^r}.{\,^n}{C_r}{\left( a \right)^{n - r}}.{\left( b \right)^r}$$ is an expansion of $${\left( {a - b} \right)^n}$$
$$\eqalign{
& \therefore \,\,{5^{th}}\,{\text{term}}\,{\text{ = }}\,{t_5} = {t_{4 + 1}} \cr
& = \,{\left( { - 1} \right)^4}{.^n}{C_4}{\left( a \right)^{n - 4}}.{\left( b \right)^4} = {\,^n}{C_4}.{a^{n - 4}}\,.\,{b^4} \cr
& {6^{th}}{\text{ term }} = {t_6} = {t_{5 + 1}} \cr
& = {\left( { - 1} \right)^5}.{\,^n}{C_5}{\left( a \right)^{n - 5}}{\left( b \right)^5} \cr
& {\text{Given }}\,{t_5} + {t_6} = 0 \cr
& \therefore \,{\,^n}{C_4}.{a^{n - 4}}.{b^4} + \left( { - {\,^n}{C_5}.{a^{n - 5}}.{b^5}} \right) = 0 \cr
& \Rightarrow \,\,\frac{{n!}}{{4!\left( {n - 4} \right)!}}.\frac{{{a^n}}}{{{a^4}}}.{b^4} - \frac{{n!}}{{5!\left( {n - 5} \right)!}}.\frac{{{a^n}{b^5}}}{{{a^5}}} = 0 \cr
& \Rightarrow \,\,\frac{{n!.{a^n}{b^4}}}{{4!\left( {n - 5} \right)!.{a^4}}}\left[ {\frac{1}{{\left( {n - 4} \right)}} - \frac{b}{{5.a}}} \right] = 0 \cr
& {\text{or, }}\frac{1}{{n - 4}} - \frac{b}{{5a}} = 0 \cr
& \Rightarrow \,\,\frac{a}{b} = \frac{{n - 4}}{5} \cr} $$