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Composite Function

If A1A2A3.....

Question

If $A_{1} A_{2} A_{3} . . . A_{n}$ be a regular polygon of $n$ sides and
$\frac{1}{A_{1} A_{2}} = \frac{1}{A_{1} A_{3}} + \frac{1}{A_{1} A_{4}},$ then

A

n = 5

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B

n = 6

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C

n = 7

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D

none of these.

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Solution

The correct option is C $n = 7$
If radius of circle is $r$ then
$A_{1} A_{2} = 2 r sin (\frac{π}{n})$
$A_{1} A_{3} = 2 r sin (\frac{2 π}{n})$
$A_{1} A_{4} = 2 r sin (\frac{3 π}{n})$
$∵ \frac{1}{A_{1} A_{2}} = \frac{1}{A_{1} A_{3}} + \frac{1}{A_{1} A_{4}}$
$\Rightarrow \frac{1}{2 r sin (\frac{π}{n})} = \frac{1}{2 r sin (\frac{2 π}{n})} + \frac{1}{2 r sin (\frac{3 π}{n})}$
$\Rightarrow sin (\frac{2 π}{n}) sin (\frac{3 π}{n}) = sin (\frac{3 π}{n}) sin (\frac{π}{n}) + sin (\frac{2 π}{n}) sin (\frac{π}{n})$
$\Rightarrow sin (\frac{2 π}{n}) [sin (\frac{3 π}{n}) - sin (\frac{π}{n})] = sin (\frac{3 π}{n}) sin (\frac{π}{n})$
Using transformation angle formula, we get
$\Rightarrow sin (\frac{2 π}{n}) .2 cos (\frac{2 π}{n}) sin (\frac{π}{n}) = sin (\frac{3 π}{n}) sin (\frac{π}{n})$
$\Rightarrow 2 sin (\frac{2 π}{n}) cos (\frac{2 π}{n}) = sin (\frac{3 π}{n})$
Using multiple angle formula, $2 sin A cos A = sin 2 A$ we get
$sin (\frac{4 π}{n}) = sin (\frac{3 π}{n})$
$∴ \frac{4 π}{n} = r + {(- 1)}^{r} \frac{3}{n}$ for $r = 1, n = 7$

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a_{r} > 0; \forall r, n \in N

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are in A.P, then

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\frac{1}{a_{1} a_{2}} + \frac{1}{a_{2} a_{3}} + \frac{1}{a_{3} a_{4}} + . . . \frac{1}{a_{n - 1} a_{n}} = \frac{n - 1}{a_{1} a_{n}}

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If $a_{1}, a_{2}, a_{3} . . . . . a_{n}$ are in A.P. Where $a_{i} > 0$ for all i, then the value of
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