Find a, b and n in the expansion of (a + b)n, if the first three terms in the expansion are 729, 7290 and 30375 respectively.
We have :
$T_{1}=729, T_{2}=7290$ and $T_{3}=30375$
Now,
${ }^{n} C_{0} a^{n} b^{0}=729$
$\Rightarrow a^{n}=729$
$\Rightarrow a^{n}=3^{6}$
${ }^{n} C_{1} a^{n-1} b^{1}=7290$
${ }^{n} C_{2} a^{n-2} b^{2}=30375$
Also,
$\frac{{ }^{n} C_{2} a^{n-2} b^{2}}{{ }^{n} C_{1} a^{n-1} b^{1}}=\frac{30375}{7290}$
$\Rightarrow \frac{n-1}{2} \times \frac{b}{a}=\frac{25}{6} \quad \ldots(\mathrm{i})$
$\Rightarrow \frac{(n-1) b}{a}=\frac{25}{3}$
And,
$\frac{{ }^{n} C_{1} a^{n-1} b^{1}}{{ }^{n} C_{0} a^{n} b^{0}}=\frac{7290}{729}$
$\Rightarrow \frac{n b}{a}=\frac{10}{1}$ ...(ii)
On dividing $($ ii $)$ by $(i)$, we get
$\frac{\frac{n b}{a}}{\frac{(n-1) b}{a}}=\frac{10 \times 3}{25}$
$\Rightarrow \frac{n}{n-1}=\frac{6}{5}$
$\Rightarrow n=6$
Since, $a^{6}=3^{6}$
Hence, $a=3$
Now, $\frac{n b}{a}=10$
$\Rightarrow b=5$