Let T_n denote the number of triangles which can be formed usingn the vertices of a regular polygon of n sides. Ifn T_n + 1 - T_n = 21, then n equals

n n (b) : T_n= ^n C_3 Therefore T_n+1-T_n=21 n ^n+1 C_3- ^n C_3=21 n ^n C_2+ ^n C_3- ^n C_3=21 => ^n C_2=21 n (1 / 2) n(n-1)=21 => n^2-n-42=0 n (n-7)(n+6)=0n As n >= 1 , we get n=7n\

Let T_{n} denote the number of triangles which can be f... | Sequence & Series (Mathematics)

\n\begin{aligned}\n& \text { (b) }: T_n={ }^n C_3 \therefore T_{n+1}-T_n=21 \\\n& { }^{n+1} C_3-{ }^n C_3=21 \\\n& { }^n C_2+{ }^n C_3-{ }^n C_3=21 \Rightarrow{ }^n C_2=21 \\\n& \frac{1}{2} n(n-1)=21 \Rightarrow n^2-n-42=0 \\\n& (n-7)(n+6)=0\n&\text { As } n \geq 1 \text {, we get } n=7\n\end{aligned}

Explanation

Let a_{1},a_{2},,a_{n} be a given A.P. whose common difference is an integer and S_{n} = a_{1} + a_{2} + + a_{n}. If a_{1} = 1,a_{n} = 300 and 15 \leq n \leq 50, then the ordered pair ( S_{n - 4},a_{n - 4} ) is equal to

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Number of 4 -digit numbers that are less than or equal to 2800 and either divisible by 3 or by 11 , is equal to

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