If the second, third and fourth term in the e | vedclass

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(d) ${T_2} = n\,{(x)^{n - 1}}{(a)^1} = 240$ .....$(i)$

${T_3} = \frac{{n\,(n - 1)}}{{1.2}}{x^{n - 2}}{a^2} = 720$&hellip;..$(ii)$

${T_4} = \frac

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(d) ${T_2} = n\,{(x)^{n - 1}}{(a)^1} = 240$ .....$(i)$

${T_3} = \frac{{n\,(n - 1)}}{{1.2}}{x^{n - 2}}{a^2} = 720$&hellip;..$(ii)$

${T_4} = \frac{{n\,(n - 1)(n - 2)}}{{1.2.3}}{x^{n - 3}}{a^3} = 1080$ &hellip;..$(iii)$

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}$

$\Rightarrow n = 5$.

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

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