MAH-CET 2025 — Mathematics PYQ

Solving the Equation:

1. $n$ se divide karne par:

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

2. $6$ se multiply karne par:

   $$6(n-1)=6+(n-1)(n-2)$$
   $$6n-6=6+(n^2-3n+2)$$
   $$6n-6=n^2-3n+8$$

3. Standard Quadratic form:

   $$n^2-9n+14=0$$

4. Factorization:

   $$n^2-7n-2n+14=0$$
   $$(n-7)(n-2)=0$$

Final Values:

Kyunki 4th term ke liye $n\ge 3$ hona chahiye, isliye: