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

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Similar Questions

The number of integral terms in the expansion of $(7^{1/3} + 11^{1/9})^{6561}$ is :-

Suppose $2-p, p, 2-\alpha, \alpha$ are the coefficient of four consecutive terms in the expansion of $(1+x)^n$. Then the value of $p^2-\alpha^2+6 \alpha+2 p$ equals

If the second term of the expansion ${\left[ {{a^{\frac{1}{{13}}}}\,\, + \,\,\frac{a}{{\sqrt {{a^{ - 1}}} }}} \right]^n}$ is $14a^{5/2}$ then the value of $\frac{{^n{C_3}}}{{^n{C_2}}}$ is :

The positive value of $\lambda $ for which the co-efficient of $x^2$ in the expression ${x^2}{\left( {\sqrt x + \frac{\lambda }{{{x^2}}}} \right)^{10}}$ is $720$ is

In the expansion of ${\left( {\frac{{x\,\, + \,\,1}}{{{x^{\frac{2}{3}}}\,\, - \,\,{x^{\frac{1}{3}}}\,\, + \,\,1}}\,\, - \,\,\frac{{x\,\, - \,\,1}}{{x\,\, - \,\,{x^{\frac{1}{2}}}}}} \right)^{10}}$, the term which does not contain $x$ is :