The difference between any two consecutive interior angles of a polygon is $5^{\circ}$ If the smallest angle is $120^{\circ},$ find the number of the sides of the polygon.
The angles of the polygon will form an $A.P.$ with common difference $d$ as $5^{\circ}$ and first term $a$ as $120^{\circ}$
It is known that the sum of all angles of a polygon with $n$ sides is $180(n-2)$
$\therefore S_{n}=180^{\circ}(n-2)$
$\Rightarrow \frac{n}{2}[2 a+(n-1) d]=180^{\circ}(n-2)$
$\Rightarrow \frac{n}{2}\left[240^{\circ}+(n-1) 5^{\circ}\right]=180^{\circ}(n-2)$
$\Rightarrow n[240+(n-1) 5]=360(n-2)$
$\Rightarrow 240 n+5 n^{2}-5 n=360 n-720$
$\Rightarrow 5 n^{2}-125 n+720=0$
$\Rightarrow n^{2}-25 n+144=0$
$\Rightarrow n^{2}-16 n-9 n+144=0$
$\Rightarrow n(n-16)-9(n-16)=0$
$\Rightarrow(n-9)(n-16)=0$
$\Rightarrow n=9$ or $16$
In an $A.P.,$ the first term is $2$ and the sum of the first five terms is one-fourth of the next five terms. Show that $20^{th}$ term is $-112$
If the angles of a quadrilateral are in $A.P.$ whose common difference is ${10^o}$, then the angles of the quadrilateral are
If $x_1 , x_2 , ..... , x_n$ and $\frac{1}{{{h_1}}},\frac{1}{{{h^2}}},......\frac{1}{{{h_n}}}$ are two $A.P' s$ such that $x_3 = h_2 = 8$ and $x_8 = h_7 = 20$, then $x_5. h_{10}$ equals
Let $3,6,9,12, \ldots$ upto $78$ terms and $5,9,13,17, \ldots$ upto $59$ terms be two series. Then, the sum of the terms common to both the series is equal to
If the sum of first $11$ terms of an $A.P.$, $a_{1} a_{2}, a_{3}, \ldots$is $0\left(\mathrm{a}_{1} \neq 0\right),$ then the sum of the $A.P.$, $a_{1}, a_{3}, a_{5}, \ldots, a_{23}$ is $k a_{1},$ where $k$ is equal to