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Let $T_n$ be the number of all possible triangles formed by joining vertices of an $n$-sided regular polygon. If $T_{n+1}-T_n=10$, then the value of $n$ is
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The correct answer is:
5
A triangle can be formed from 3 vertices, so select 3 vertices out of $x$ vertices.
The number of ways of selecting 3 vertices out of $n$ is ${ }^n C_3$.
$$
\begin{array}{ll}
\therefore & T_{n+1}-T_n=10 \\
\Rightarrow & { }^{n+1} C_3-{ }^n C_3=10 \\
\Rightarrow & { }^n C_2=10 \left[\because{ }^n C_r+{ }^n C_{r+1}={ }^{n+1} C_{r+1}\right]
\\
\Rightarrow & \frac{n(n-1)}{2}=10 \Rightarrow n^2-n-20=0 \\
\Rightarrow & n^2-5 n+4 n-20=0 \\
\Rightarrow & n(n-5)+4(n-5)=0 \\
\Rightarrow & (n-5)(n+4)=0 \Rightarrow n=5, n \neq-4
\end{array}
$$
The number of ways of selecting 3 vertices out of $n$ is ${ }^n C_3$.
$$
\begin{array}{ll}
\therefore & T_{n+1}-T_n=10 \\
\Rightarrow & { }^{n+1} C_3-{ }^n C_3=10 \\
\Rightarrow & { }^n C_2=10 \left[\because{ }^n C_r+{ }^n C_{r+1}={ }^{n+1} C_{r+1}\right]
\\
\Rightarrow & \frac{n(n-1)}{2}=10 \Rightarrow n^2-n-20=0 \\
\Rightarrow & n^2-5 n+4 n-20=0 \\
\Rightarrow & n(n-5)+4(n-5)=0 \\
\Rightarrow & (n-5)(n+4)=0 \Rightarrow n=5, n \neq-4
\end{array}
$$
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