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8. Sequences and Series
hard
The interior angles of a polygon are in $A.P.$ If the smallest angle be ${120^o}$ and the common difference be $5^o$, then the number of sides is
A
$8$
B
$10$
C
$9$
D
$6$
(IIT-1980)
Solution
(c) Let the number of sides of the polygon be $n$.
Then the sum of interior angles of the polygon
$ = (2n – 4)\frac{\pi }{2} = (n – 2)\pi $
Since the angles are in $A.P. $ and $a = {120^o},\;d = 5$,
therefore $\frac{n}{2}[2 \times 120 + (n – 1)5] = (n – 2)180$
==> ${n^2} – 25n + 144 = 0$
==> $(n – 9)(n – 16) = 0$
==> $n = 9,\;16$
But $n = 16$ gives ${T_{16}} = a + 15d = {120^o} + {15.5^o} = {195^o}$,
which is impossible as interior angle cannot be greater than ${180^o}$.
Hence $n = 9$.
Standard 11
Mathematics
