JEE MAIN - Mathematics (2013 (Offline) - No. 14)
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 :
7
5
10
8
Explanation
Number of possible triangle using n vertices = nC3
$$ \therefore $$ Tn = nC3
then Tn + 1 = n + 1C3
Given, $${T_{n + 1}} - {T_n}$$ = 10
$$ \Rightarrow $$ n + 1C3 - nC3 = 10
$$ \Rightarrow $$ $${{\left( {n + 1} \right)n\left( {n - 1} \right)} \over 6} - {{n\left( {n - 1} \right)\left( {n - 2} \right)} \over 6}$$ = 10
$$ \Rightarrow $$ 3n(n - 1) = 60
$$ \Rightarrow $$ n(n - 1) = 20
$$ \Rightarrow $$ n2 - n - 20 = 0
$$ \Rightarrow $$ (n - 5)(n + 4) = 0
$$ \therefore $$ n = 5
$$ \therefore $$ Tn = nC3
then Tn + 1 = n + 1C3
Given, $${T_{n + 1}} - {T_n}$$ = 10
$$ \Rightarrow $$ n + 1C3 - nC3 = 10
$$ \Rightarrow $$ $${{\left( {n + 1} \right)n\left( {n - 1} \right)} \over 6} - {{n\left( {n - 1} \right)\left( {n - 2} \right)} \over 6}$$ = 10
$$ \Rightarrow $$ 3n(n - 1) = 60
$$ \Rightarrow $$ n(n - 1) = 20
$$ \Rightarrow $$ n2 - n - 20 = 0
$$ \Rightarrow $$ (n - 5)(n + 4) = 0
$$ \therefore $$ n = 5
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