Question

Find a, b and n in the expansion of $(a+b{)}^n$ if the first three terms of the expansion are $729,7290and30375,$ respectively. Insert a number $k$ so that $a,k,,b,n$ forms an AP.

Answer 1

Solution - we know that -

$(a+b{)}^n{=}^n{C}_o{a}^n{b}^o{+}^n{C}_1{a}^{n-1}b{+}^n{C}_2{a}^{n-2}{b}^2+---{+}^n{C}_{n-1}a{b}^{n-1}{+}^n{C}_n{a}^o{n}^n$

$=a^n{+}^n{C}_1{a}^{n-1}{b}^1{+}^n{C}_2{a}^{n-2}{b}^2+...{+}^n{C}_n+b^{n-1}+b^n$

So, first 3 terms are $a^n{,}^n{C}_1{a}^{n-1}b$ and ${}^n{C}_2{a}^{n-2}{b}^2$

and given, $a^n=\mathrm{729....}\left(1\right)$

${}^n{C}_{1}b=\mathrm{7290.....}\left(2\right)$

${}^n{C}_2{a}^{n-2}{b}^2=\mathrm{30375....}\left(3\right)$

diving (1) and (2)

$\frac{a^n}{{}^n{C}_1{a}^{n-1}b}=\frac{729}{7290}\Rightarrow \frac{a}{nb}=\frac{1}{10}...\left(4\right)$

$[{\because }^n{C}_r=\frac{\angle n}{\angle n-r\angle r}]$

dividing (2) and (3)

$\frac{{}^n{C}_1{a}^{n-1}b}{{}^n{C}_1{a}^{n-2}{b}^2}=\frac{7290}{30375}$

$\frac{29}{(n-1)b}=\frac{6}{25}....\left(5\right)$

dividing (4) and (5)

$\frac{\frac{a}{nb}}{\frac{2a}{(n-1)b}}=\frac{\frac{1}{10}}{\frac{6}{25}}\Rightarrow \frac{n-1}{2n}=\frac{5}{12}\Rightarrow 12n-12=10n$

$\Rightarrow 2n=12$

$\Rightarrow n=6$ putting $n=6$ in eq${}^n$ (1)

and $a^n=729\Rightarrow a^6=729=a^6=(3{)}^6$

$\Rightarrow a=3$

putting $a=3$ and $n=6$ in eq${}^n$ (5)

$\frac{2a}{(n-1)b}=\frac{6}{25}$

$\Rightarrow \frac{6}{5b}=\frac{6}{25}\Rightarrow b=5$

Here $a=3,b=5,n=6$

$a,k,b,n$ forms a A.P.

So, $k=4$