Question

the interior angle of a polygon are in A.P the smallest angle is 120 and the common difference is 5 find the no. of side of polygon

Answer 1

we know that exterior angles of any polygon sum up to 360°.
exterior angles will be
180-(120)=60
180-(120+5)=55
180-(120+2×5)=50
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180-[120+(n-1)5]=65-5n

see that the exterior angles are in decreasing A.P.
we know the sum of an A.P.
$sn = \frac{n}{2} (2a + (n - 1)d)$
sn is the sum of A.P.
a is the 1st term of A.P.
n is the number of terms in A.P.
d is the common difference of A.P.

putting values in the formula
$360 = \frac{n}{2}(2 \times 60 + (n - 1)( - 5)) \\ 720 = 120n - 5 {n}^{2} + 5n \\ 144 = 24n - {n}^{2} + n \\ {n}^{2} - 25n + 144 = 0 \\ {n}^{2} - 16n - 9n + 144 = 0 \\ n(n - 16) - 9(n - 16) = 0 \\ (n - 9)(n - 16) = 0 \\ n = 9 \: \: or \: \: 16$
● Let us check the last exterior angle
=65-5n
put n=9
= 65 - 5×9 = 20°
put n=16
= 65 - 5×16 = -15°

You know that any exterior angle cannot be negetive so n=16 is rejected answer .

Hence the number of sides of the required polygon is '9'.