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Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem
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Find $a,b$ and $n$ in the expansion of $(a+b)^n$ if the first three terms of the expansion are $729,7290$ and $30375$ respectively.

$\begin{array}{1 1}(A)\;a=2,b=4,n=7\\(B)\;a=3,b=5,n=6\\(C)\;a=4,b=5,n=8\\(D)\;a=4,b=6,n=10\end{array} $

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1 Answer

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  • $C(n,r)=\large\frac{n!}{r!(n-r)!}$
$T_1$ of $(a+b)^n=a^n=729$------(1)
$T_2$ of $(a+b)^n=nC_1a^{n-1}b=7290$------(2)
$T_3$ of $(a+b)^n=nC_2a^{n-2}b^2=30375$------(3)
Dividing (1) by (2)
$\large\frac{a^n}{nC_1a^{n-1}b}=\frac{729}{7290}=\frac{1}{10}$
$\large\frac{a}{nb}=\frac{1}{10}$-----(4)
Dividing (2) by (3)
$\large\frac{nC_1a^{n-1}b}{nC_2a^{n-2}b^2}=\frac{7290}{30375}$
Or $\large\frac{na^{n-1}b}{\Large\frac{n(n-1)}{2}a^{n-2}b^2}=\frac{7290}{30375}$
$\Rightarrow \large\frac{6}{25}$
Or $\large\frac{n}{n-1}\times \frac{a}{b}=\frac{6}{25}$-----(5)
Dividing (4) by (5)
$\large\frac{a}{nb}\times \frac{(n-1)b}{2a}=\frac{1}{10}\times \frac{25}{6}$
$\Rightarrow \large\frac{8}{12}$
Or $\large\frac{n-1}{2n}=\frac{5}{12}$
$12(n-1)=5(2n)$
$12n-12=10n$
$2n=12$
$n=6$
Also putting $n=6,a=3$ in (4)
$\large\frac{3}{6b}=\frac{1}{10}$
$\therefore b=\large\frac{3\times 10}{6}$$=5$
Thus $a=3,b=5,n=6$
Hence (B) is the correct answer.
answered Jun 23, 2014 by sreemathi.v
 

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