i am getting 16 as the answer, what am i doing wrong

Let $n$ = number of sides. The angles in AP are: $120, 123, 126, \ldots$ with last term $= 120 + (n-1) \cdot 3$.

Sum of interior angles of a polygon $= (n-2) \cdot 180$

Sum of the AP $= \dfrac{n}{2} \cdot (\text{first} + \text{last}) = \dfrac{n}{2} \cdot (120 + 120 + 3(n-1)) = \dfrac{n}{2} \cdot (237 + 3n)$

Setting them equal:

$\frac{n}{2}(237 + 3n) = 180(n-2)$

$n(237 + 3n) = 360(n-2)$

$237n + 3n^2 = 360n - 720$

$3n^2 - 123n + 720 = 0$

$n^2 - 41n + 240 = 0$

Using the quadratic formula:

$n = \frac{41 \pm \sqrt{1681 - 960}}{2} = \frac{41 \pm \sqrt{721}}{2} = \frac{41 \pm 26.85}{2}$

This gives $n \approx 33.9$ or $n \approx 7.07$.

Now check the convexity constraint (all angles $< 180°$):

• For $n = 7$: largest angle $= 120 + 6 \times 3 = 138°$. Valid.
• For $n = 34$: largest angle $= 120 + 33 \times 3 = 219°$. Exceeds $180°$, so NOT a convex polygon.

Answer: $n = 7$

You likely got 16 by either making a sign error in the quadratic or forgetting to check the convexity condition and picking the larger root. The key step people miss is rejecting $n = 34$ because it violates the convex polygon requirement.