JEE Main 2025 (Online) 28th January Evening ShiftQuestion 1
The interior angles of a polygon with $n$ sides, are in an A.P. with common difference $6^{\circ}$. If the largest interior angle of the polygon is $219^{\circ}$, then $n$ is equal to $\_\_\_\_\_\_\_\_$ .
Sequence and Series · Arithmetic Progression
Answer$20$
Solution
Formulas used. (1) Sum of interior angles of an $n$-gon $=(n-2)\,180^\circ$(2) Sum of an A.P.: $S_n=\frac n2\left[2a+(n-1)d\right]$Step 1 - Set up the A.P. Start from the largest angle and go down by $6^\circ$: $219,\ 213,\ 207,\dots$ $$\text{Sum}=\frac n2\left[2(219)-6(n-1)\right]=219n-3n(n-1).$$ Step 2 - Equate. $$219n-3n^2+3n=180n-360\ \Longrightarrow\ 3n^2-42n-360=0\ \Longrightarrow\ n^2-14n-120=0 .$$ Step 3 - Solve. $$(n-20)(n+6)=0\ \Longrightarrow\ n=20 .$$ Check. The smallest angle is $219-19\times6=105^\circ$; no angle equals $180^\circ$. Answer: 20