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The correct Answer is:
`n = 7`

Let a be the side of n sided regular polgon `A_(1) A_(2) A_(3) A_(4)..A_(n)`

`:.` Angle subtended by each side at centre `= (2pi)/(n)`
Also `OA_(1) = OA_(2) = OA_(3) = .... = OA_(n) = r` (say)
In `DeltaA_(1) OA_(2)`, using cosine rule
`A_(1) A_(2)^(2) = r^(2) + r^(2) -2r^(2) cos.(2pi)/(n)`
`=2r^(2) (1- cos.(2pi)/(n))`
`=4r^(2) sin.(pi)/(n)`
`:. A_(1) A_(2) = 2r sin.(pi)/(n)`
Similarly in `DeltaA_(1) OA_(3), A_(1) A_(3) = 2r sin.(2pi)/(n)`
and in `DeltaA_(1) OA_(4), A_(1) A_(4) = 2r sin.(3pi)/(n)`
But given that `(1)/(A_(1)A_(2)) = (1)/(A_(1)A_(3)) + (1)/(A_(1)A_(4))`
`rArr (1)/(2sin.(pi)/(n)) = (1)/(2sin.(2pi)/(n)) + (1)/(2sin.(3pi)/(n))`
`rArr 2sin.(2pi)/(n) sin.(3pi)/(n) = 2 sin.(pi)/(n) sin.(3pi)/(n) + 2 sin.(pi)/(n) sin.(pi)/(n)`
`rArr cos.(pi)/(n) - cos.(5pi)/(n) = cos.(2pi)/(n) + cos.(4pi)/(n) - cos.(pi)/(n) - cos.(3pi)/(n)`
`rArr cos.(5pi)/(n) + cos.(2pi)/(n) = cos.(4pi)/(n) + cos.(3pi)/(n)`
`rArr 2 cos.(7pi)/(n) cos.(3pi)/(2n) = 2 cos.(7pi)/(2n) cos.(pi)/(2n)`
`rArr cos.(7pi)/(2n) (cos.(3pi)/(2n) - cop.(pi)/(2n)) = 0`
`rArr 2 cos.(7pi)/(2n) sin.(pi)/(n) sin.(pi)/(2n) = 0`
`rArr (7pi)/(2n) = (2k + 1) (pi)/(2) " or " (pi)/(2) " or " (pi)/(n) = kpi " or " (pi)/(2n) = k pi, k in Z`
`rArr n = (7)/(2k + 1) " or " n = (1)/(k)` (not possible)
or `n = (1)/(2k)` (not possible)
`:. n = 7 " for " k = 0`
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