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 $16$
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
The Fibonacci sequence is defined by
$1 = {a_1} = {a_2}{\rm{ }}$ and ${a_n} = {a_{n - 1}} + {a_{n - 2}},n\, > \,2$
Find $\frac{a_{n+1}}{a_{n}},$ for $n=1,2,3,4,5$
If $\frac{a^{n}+b^{n}}{a^{n-1}+b^{n-1}}$ is the $A.M.$ between $a$ and $b,$ then find the value of $n$.
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
Let ${a_1},{a_2},\;.\;.\;.\;.,{a_{49}}$ be in $A.P.$ such that $\mathop \sum \limits_{k = 0}^{12} {a_{4k + 1}} = 416$ and ${a_9} + {a_{43}} = 66$. If $a_1^2 + a_2^2 + \ldots + a_{17}^2 = 140m,$ then $m = \;\;..\;.\;.\;.\;$