If second, third and fourth term in expansion of {(x + a)^n} are 240, 720 and 1080 respectively, the

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7.Binomial Theorem

hard

If the second, third and fourth term in the expansion of ${(x + a)^n}$ are $240, 720$ and $1080$ respectively, then the value of $n$ is

A

$15$

B

$20$

C

$10$

D

$5$

Solution

(d) ${T_2} = n\,{(x)^{n – 1}}{(a)^1} = 240$ …..$(i)$

${T_3} = \frac{{n\,(n – 1)}}{{1.2}}{x^{n – 2}}{a^2} = 720$…..$(ii)$

${T_4} = \frac{{n\,(n – 1)(n – 2)}}{{1.2.3}}{x^{n – 3}}{a^3} = 1080$ …..$(iii)$

To eliminate $x$,$\frac{{{T_2}\,.\,{T_4}}}{{{T_3}^2}} = \frac{{240\,.\,1080}}{{720\,.\,720}} = \frac{1}{2}$

==> $\frac{{{T_2}}}{{{T_3}}}\,\,.\,\,\frac{{{T_4}}}{{{T_3}}} = \frac{1}{2}$

Now, $\frac{{{T_{r + 1}}}}{{{T_r}}} = \frac{{{}^n{C_r}}}{{{}^n{C_{r – 1}}}} = \frac{{n – r + 1}}{r}$

Putting $r = 3$ and $2$ in above expression,we get

$ \Rightarrow \frac{{n – 2}}{3}\,.\,\frac{2}{{n – 1}} = \frac{1}{2}$

$\Rightarrow n = 5$.

Standard 11

Mathematics

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