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If $T_n$ denotes no. of triangles formed by the vertices of a regular $n$ sided polygon and if $T_{n+1}-T_n=21$, then $n=?$

$\begin{array}{1 1} 4 \\ 5 \\ 6\\ 7 \end{array}$

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Since 3 points are reguired to form a triangle,
no. of $\Delta^s$ using $n$ points (vertices of the polygon) = $T_n=^nC_3$
$\therefore\:T_{n+1}=^{n+1}C_3$
$Given:\:T_{n+1}-T_n=^{n+1}C_3-^nC_3=21$
$\Rightarrow\:\large\frac{(n+1)n(n-1)}{3!}$$-\large\frac{n(n-1)(n-2)}{3!}$$=21$
$\Rightarrow\:n^2-n=42$
$\Rightarrow\:(n+6)(n-7)=0$
$\Rightarrow\:n=7\:or\:-6$ (which is not possible)
$\therefore\:n=7$
answered Aug 13, 2013 by rvidyagovindarajan_1
 

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