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7.Binomial Theorem
medium
જો ${(3 + ax)^9}$ ના વિસ્તરણમાં ${x^2}$ અને ${x^3}$ ના સહગુણક સમાન હોય તો $a$ ની કિમંત મેળવો.
A
$ - \frac{7}{9}$
B
$ - \frac{9}{7}$
C
$\frac{7}{9}$
D
$\frac{9}{7}$
Solution
(d) ${T_{r + 1}} = {}^{\rm{9}}{C_r}{(3)^{9 – r}}{(ax)^r} = {}^{\rm{9}}{C_r}{(3)^{9 – r}}{a^r}{x^r}$
$\therefore $ Coefficient of ${x^r}$= ${}^9{C_r}{3^{9 – r}}{a^r}$
Hence, coefficient of ${x^2} = {}^9{C_2}{3^{9 – 2}}{a^2}$ and coefficient of ${x^{\rm{3}}}$
= ${}^9{C_3}{3^{9 – 3}}{a^3}$
So, we must have ${}^9{C_2}{3^7}{a^2} = {}^9{C_3}{3^6}{a^3}$
==> $\frac{{9.8}}{{1.2}}.3 = \frac{{9.8.7}}{{1.2.3}}.a\,\,\, $
$\Rightarrow \,\,a = \frac{9}{7}.$
Standard 11
Mathematics
