Question

The interior angle of a polygon are in a.p the smaller angle is 120 and the common difference is 5 then find the no.of the sides of polygon?

Answer 1

Let the number of sides of polygon be 'n'.

As we know,

$\sf Sum \ of \ interior \ angles \ of \ a \ polygon \having\ n \ sides = (n - 2) \times 180 \degree \ \ \ \ \ \ ......(1)$

Consider

a : First term of A.P.

d : Common difference of A.P.

Its given that smallest angle of polygon is 120°

Therefore,

a = 120° and d = 5°

Sum of interior angles of polygon = 120° + 125° + 130° + ..... + n terms

So,

$\implies (n-2) \times 180^{\circ}=\frac{n}{2} [2 \times 120^{\circ} + (n-1)\times 5^{\circ}] \\ \\ \implies 2 \times 180(n-2)= n[2 \times 120 + (n-1)\times 5] \\ \\ \implies360(n-2)=n[240 +5n -5] \\ \\ \implies 360(n-2)=240n +5n^{2} -5n \\ \\ \implies 360n - 720 = 240n + 5n^{2} -5n \\ \\ \implies 240n + 5n^{2} -5n=360n - 720 \\ \\ \implies 5n^{2} -5n+240n-360n +720=0 \\ \\ \implies 5n^{2} -125n+720=0 \\ \\ \implies (5n^{2} -125n+720=0) \times \frac{1}{5} \\ \\ \implies n^{2} -25n+144=0$

$\implies n^{2}-16n-9n+144=0 \\ \\ \implies n(n-16)-9(n-16)=0 \\ \\ \implies(n-9)(n-16)=0\\ \\ \implies n-9=0 \ \ \ \ \ or \ \ \ \ \ n-16=0 \\ \\ \implies n=9 \ \ \ \ \ or \ \ \ \ \ n=16$

When n = 16,

Last angle of the polygon;

$a_{n} =120 \times(160-1)5=120 \times 75 = 195$

Which is not possible.

Therefore,

n = 9

Answer 2

Answer:

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