A sequence $a_{1}, a_{2}, a_{3} . . . . . . . a_{n}$ is an A.P. if and only if for any three consecutive terms $a_{k - 1}, a_{k}, a_{k + 1}$ the middle term is equal to the half-sum of its neighbors.
$a_{k} = . . . . . . . . . . . . . . . . .$

\frac{a_{k - 1} + a_{k + 1}}{2}

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\frac{a_{k + 1} - a_{k - 1}}{2}

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\frac{a_{k + 2} + a_{k + 1}}{2}

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a_{k} + a_{k - 1}

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Solution

The correct option is A $\frac{a_{k - 1} + a_{k + 1}}{2}$
For any three consecutive terms:
$a_{k - 1}, a_{k}, a_{k + 1}$ to be in A.P., common difference should be same.
$∴ a_{k} - a_{k - 1} = a_{k + 1} - a_{k}$
$\Rightarrow 2 a_{k} = a_{k - 1} + a_{k + 1}$
$∴ a_{k} = \frac{a_{k - 1} + a_{k + 1}}{2}$
$(A n s \to A)$

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a_{k} = cos (\frac{k π}{7}) + i sin (\frac{k π}{7})

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a_{k + 1} = a_{k} + 3

a_{1} = 7

Let $a_{0} = \frac{5}{2}$ and $a_{k} = a_{k - 1}^{2} - 2 f o r k \geq 1,$ then the value of $\infty Π k = 0 (1 - \frac{1}{a_{k}})$ is

a_{k + 1} = a_{k} + 3

a_{1} = 7

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