MAH-CET 2026 — Mathematics PYQ

Step 1: Formula for the number of triangles

A regular polygon with $n$ sides has $n$ vertices. To form a single triangle, we need to choose any 3 distinct vertices from these $n$ vertices. No 3 vertices of a regular polygon are collinear.

Therefore, the number of triangles $T_n$ that can be formed is given by the combination formula:

$$T_n=\left(\genfrac{}{}{0ex}{}{n}{3}\right)=\frac{n(n-1)(n-2)}{3\cdot 2\cdot 1}$$

Similarly, for a polygon with $n+1$ sides, the number of triangles $T_{n+1}$ is:

$$T_{n+1}=\left(\genfrac{}{}{0ex}{}{n+1}{3}\right)=\frac{(n+1)(n)(n-1)}{3\cdot 2\cdot 1}$$

Step 2: Use the given algebraic condition

We are given the relation:

$$T_{n+1}-T_n=21$$

Substitute the combination expressions into the equation:

$$\left(\genfrac{}{}{0ex}{}{n+1}{3}\right)-\left(\genfrac{}{}{0ex}{}{n}{3}\right)=21$$

Using Pascal's Identity rule ($\left(\genfrac{}{}{0ex}{}{n}{r}\right)+\left(\genfrac{}{}{0ex}{}{n}{r-1}\right)=\left(\genfrac{}{}{0ex}{}{n+1}{r}\right)$), we can rewrite $\left(\genfrac{}{}{0ex}{}{n+1}{3}\right)-\left(\genfrac{}{}{0ex}{}{n}{3}\right)$ directly as $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$.

Alternatively, expanding it out:

$$\frac{(n+1)n(n-1)}{6}-\frac{n(n-1)(n-2)}{6}=21$$

Factor out the common term $\frac{n(n-1)}{6}$:

$$\frac{n(n-1)}{6}\left[(n+1)-(n-2)\right]=21$$

$$\frac{n(n-1)}{6}[n+1-n+2]=21$$

$$\frac{n(n-1)}{6}[3]=21$$

Simplify the fraction:

$$\frac{n(n-1)}{2}=21$$

Step 3: Solve for $n$

Multiply both sides by 2:

$$n(n-1)=42$$

Expand and form a quadratic equation:

$$n^2-n-42=0$$

Factorize the quadratic equation:

$$n^2-7n+6n-42=0$$

$$n(n-7)+6(n-7)=0$$

$$(n-7)(n+6)=0$$

This gives two possible values for $n$:

$$n=7\text{or}n=-6$$

Since the number of sides of a polygon must be a positive integer ($n\ge 3$), we discard $n=-6$.

Therefore, $n=7$.

Correct Answer:

(d) 7