Question:
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
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
Show Hint
The number of triangles formed from an \(n\)-sided polygon is
\[
{}^nC_3
\]
because any \(3\) vertices determine one triangle.
Updated On: Jun 26, 2026
- \(5\)
- \(3\)
- \(7\)
- \(4\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Understand the meaning of \(T_n\).
A triangle is formed by choosing any \(3\) vertices from an \(n\)-sided polygon.
Therefore, \[ T_n={}^{n}C_3 \]
Step 2: Write \(T_{n+1}\).
For an \((n+1)\)-sided polygon, \[ T_{n+1}={}^{n+1}C_3 \]
Step 3: Use the given condition.
Given, \[ T_{n+1}-T_n=10 \] So, \[ {}^{n+1}C_3-{}^nC_3=10 \]
Step 4: Apply the combination identity.
Using the identity, \[ {}^{n+1}C_3-{}^nC_3={}^nC_2 \] Therefore, \[ {}^nC_2=10 \]
Step 5: Simplify the combination.
Now, \[ {}^nC_2=\frac{n(n-1)}{2} \] Hence, \[ \frac{n(n-1)}{2}=10 \]
Step 6: Solve for \(n\).
Multiplying both sides by \(2\), \[ n(n-1)=20 \] \[ n^2-n-20=0 \] Factorizing, \[ (n-5)(n+4)=0 \] So, \[ n=5 \quad \text{or} \quad n=-4 \] Since \(n\) represents the number of sides of a polygon, \[ n=5 \]
Step 7: Final conclusion.
Therefore, \[ \boxed{5} \]
A triangle is formed by choosing any \(3\) vertices from an \(n\)-sided polygon.
Therefore, \[ T_n={}^{n}C_3 \]
Step 2: Write \(T_{n+1}\).
For an \((n+1)\)-sided polygon, \[ T_{n+1}={}^{n+1}C_3 \]
Step 3: Use the given condition.
Given, \[ T_{n+1}-T_n=10 \] So, \[ {}^{n+1}C_3-{}^nC_3=10 \]
Step 4: Apply the combination identity.
Using the identity, \[ {}^{n+1}C_3-{}^nC_3={}^nC_2 \] Therefore, \[ {}^nC_2=10 \]
Step 5: Simplify the combination.
Now, \[ {}^nC_2=\frac{n(n-1)}{2} \] Hence, \[ \frac{n(n-1)}{2}=10 \]
Step 6: Solve for \(n\).
Multiplying both sides by \(2\), \[ n(n-1)=20 \] \[ n^2-n-20=0 \] Factorizing, \[ (n-5)(n+4)=0 \] So, \[ n=5 \quad \text{or} \quad n=-4 \] Since \(n\) represents the number of sides of a polygon, \[ n=5 \]
Step 7: Final conclusion.
Therefore, \[ \boxed{5} \]
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