Find a, b and n in the expansion of

Question:

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.

Solution:

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, $\frac{n b}{a}=10$

$\Rightarrow b=5$

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