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If in the expansion of $(x+a)^{15},$ the $11^{th}$ term is $G.M.$ of $8^{th} $ & $12^{th}$ terms, then which term is the greatest term?

$\begin{array}{1 1} T_6 \\ T_7 \\ T_8 \\ T_9 \end{array} $

1 Answer

Given: $(T_{11})^2=T_8.T_{12}$ in the expansion of $(x+a)^{15}$
$\Rightarrow\:\large\frac{a}{x}=\sqrt {\frac{75}{77}}$
Let the $(r+1)^{th}$ term be the highest term.
Then $ \large\frac{T_{r+1}}{T_r}\geq 1$
$\Rightarrow\:\large\frac{^{15}C_r.x^{15-r}.a^r}{^{15}C_{r-1}.x^{16-r}.a^{r-1}}$$\geq 1$
$\Rightarrow\:\large\frac{16-r}{r}.\frac{a}{x}\geq 1$
$i.e.,$ $8^{th}$ term is biggest term.
answered Nov 24, 2013 by rvidyagovindarajan_1

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