🛠️ JEE➗ Maths

Let \({P}_{n}\) denote the total number of triangles formed by joining the vertices of an \(n -\)side regular polygon. I…

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Let \({P}_{n}\) denote the total number of triangles formed by joining the vertices of an \(n -\)side regular polygon. If \({P}_{n+1}-{P}_{n}=66\), then the sum of all distinct prime divisors of \(n\) is:

[JEE Main 2026, 2 Apr (Shift 2)]

a

\(7\)

b

\(8\)

c

\(5\)

d

\(6\)

✓ Correct answer: c)

\(5\)

Explanation

\( P_n=\) number of triangles formed by joining the vertices of an \(n -\)side regular polygon.

\( P_n={ }^n C_3 \)
\( { }^{n+1} C_3-{ }^n C_3=66\)
\( \Rightarrow \frac{(\mathrm{n}+1)(\mathrm{n})(\mathrm{n}-1)}{6}-\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)}{6}=66\)
\( \Rightarrow \mathrm{n}(\mathrm{n}-1)[(\mathrm{n}+1)-(\mathrm{n}-2)]=396 \)
\( \Rightarrow 3 \mathrm{n}(\mathrm{n}-1)=396\)
\( \mathrm{n}^2-\mathrm{n}-132=0 \)
\( \Rightarrow(\mathrm{n}-12)(\mathrm{n}+11)=0\)
\( \Rightarrow \mathrm{n}=12,-11 \)
\( 12=2^2 3^1\)
divisor are \(=1,2,3,4,6,12\)
prime divisor are \(=3,2 \)
sum \(=5\)

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