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 $16$
Find the sum of all numbers between $200$ and $400$ which are divisible by $7.$
If $a_1, a_2, a_3 …………$ an are in $A.P$ and $a_1 + a_4 + a_7 + …………… + a_{16} = 114$, then $a_1 + a_6 + a_{11} + a_{16}$ is equal to
If $a$ and $b$ are the roots of $x^{2}-3 x+p=0$ and $c, d$ are roots of $x^{2}-12 x+q=0$ where $a, b, c, d$ form a $G.P.$ Prove that $(q+p):(q-p)=17: 15$
If the sum and product of the first three term in an $A.P$. are $33$ and $1155$, respectively, then a value of its $11^{th}$ tern is
Write the first five terms of the sequences whose $n^{t h}$ term is $a_{n}=(-1)^{n-1} 5^{n+1}$