Q46 / 160 · TS EAMCET 2013 · Mathematics
If $t_n$ denotes the number of triangles formed with $n$ points in a plane, no three of which are collinear and if $t_{n+1}-t_n=36$, then $n$ is equal to
- A$7$
- B$8$
- Correct answerC$9$
- D$10$
Solution
$t_n=$ The number of triangles formed with $n$ points in a plane, no three of which are collinear.
i.e.,
$$
\begin{aligned}
& t_n={ }^n C_3 \\
& t_{n+1}={ }^{n+1} C_3
\end{aligned}
$$
$$
\Rightarrow \quad t_{n+1}={ }^{n+1} C_3
$$
Now, $t_{n+1}-t_n=36$
$$
\begin{array}{crrl}
\Rightarrow & { }^{n+1} C_3-{ }^n C_3=36 \\
\Rightarrow & \frac{(n+1) !}{(n-2) ! 3 !}-\frac{n !}{(n-3) ! 3 !}=36 \\
\Rightarrow & \frac{(n+1) n(n-1)}{6}-\frac{n(n-1)(n-2)}{6}=36 \\
\Rightarrow & & n(n-1)(n+1-n+2)=36 \times 6 \\
\Rightarrow & & 3 n(n-1)=36 \times 6 \\
\Rightarrow & & n^2-n-72=0 \\
\Rightarrow & & n^2-9 n+8 n-72=0 \\
\Rightarrow & & n(n-9)+8(n-9)=0 \\
\Rightarrow & & (n-9)(n+8)=0 \\
\therefore & & n=9 \quad(\because n \neq-3)
\end{array}
$$
Hence, the correct answer is option: 'C'.