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${(a - b)^n},\,n \ge 5,$ ના દ્રીપદી વિતરણમાં $5^{th}$ અને $6^{th}$ પદનો સરવાળો શૂન્ય હોય તો $\frac{a}{b}$ મેળવો.
$\frac{1}{6}(n - 5)$
$\frac{1}{5}(n - 4)$
$\frac{5}{{(n - 4)}}$
$\frac{6}{{(n - 5)}}$
Solution
(b) ${T_{r + 1}}\,\, = \,{}^n\,{C_r}{(a)^{n – r}}{( – \,b)^r}.$
${T_5} = {T_{4 + 1}} = {}^n{C_4}{a^{n – 4}}{( – \,b)^4}$ $ = {\,^{\,n}}{C_4}{a^{n – 4}}{b^4}$
and $6^{th}$ term
= $({T_6}) = {T_{5 + 1}} = {}^n{C_5}\,{a^{n – 5}}{( – \,b)^5}$ $ = – {\,^n}{C_5}\,{a^{n – 5}}\,{b^5}$
Since ${T_5} + {T_6} = 0$, therefore
${}^n{C_4}\,{a^{n – 4}}\,{b^4} – {}^n{C_5}\,{a^{n – 5}}\,{b^5} = 0$ ==> $\frac{{{a^{n – 4}}\,{b^4}}}{{{a^{n – 5}}\,{b^5}}} = \frac{{{}^n{C_5}}}{{{}^n{C_4}}}$
==> $\frac{a}{b} = \frac{{n!}}{{(n – 5)!\,\,5!}}\,.\,\frac{{4!\,\,(n – 4)!}}{{n!}}$ ==> $\frac{a}{b} = \frac{{n – 4}}{5}$.
