Question

Interior angle of a polygon are in AP. If the smallest angle is 120° and common difference is 5°, find the number of sides of polygon.

Answer 1

hello,
the AP can be written as 
120,125,130..........n
sum of all interior angles of a polygon is (n-2)180
therefore sum of n terms of the AP will be (n-2)180
S=n/2[2a+(n-1)d]
(n-2)180=n/2[2·120+(n-1)5]
180n-360=n/2[240+5n-5]
180n-360=n/2[235+5n]
360n-720=235n+5n²
5n²-125n+720=0
n²-25n+144=0
n²-16n-9n+144=0
n(n-16)-9(n-16)=0
(n-9)(n-16)=0
n=9 or n=16
this means that sum of terms from 9 to 16 is zero
therefore 16th term will be greater than 180
i.e. if n=16  16th angle will 2.120-(16-1)5=195° which is greater than 180°
therefore n=16 is not possible
hence n=9 i.e no. of side is 9

Answer 2

Answer:

$n\;=\;9$

Step-by-step explanation:

We have,

$a\;=\;120$

$d\;=\;5$

Sum of interior angles of a polygon of n sides is (n - 2)180°.

So,

$\frac{n}{2}[2(120)+(n-1)5]\;=\;(n-2)180$

$n[5n+235]\;=\;(n-2)360$

$n^2+47n\;=\;(n-2)72$

$n^2-25n+144\;=\;0$

$(n-9)(n-16)\;=\;0$

∴ $n\;=\;9\;or\;16$

Neglect 16 since the largest angle is $[120+(15)5]\;=\;195$ which is not possible no longer angle of a polygon is more than 180.

∴ $n\;=\;9$