If in an A.P, $a_{9} = 0, a_{19} = k$ , then $a_{29}$ is

k^{2}

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10 k

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9 k

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2 k

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Solution

The correct option is D $2 k$
Let $a$ and $d$ be the first term and common difference of given AP.
Since, $a_{9} = 0 \Rightarrow a + 8 d = 0$ ....(1) [using $a_{n} = a + (n - 1) d$ ]
Similarly,
$a_{19} = k \Rightarrow a + 18 d = k$ ....(2)
On subtracting (1) from (2). we get
$a + 18 d - a - 8 d = k$
$\Rightarrow 10 d = k$
$\Rightarrow d = \frac{k}{10}$
By substituting $d = \frac{k}{10}$ in (1), we get
$a + \frac{8 k}{10} = 0$
$\Rightarrow a = - \frac{8 k}{10}$
Now, $a_{29} = a + 28 d = - \frac{8 k}{10} + \frac{28 k}{10} = 2 k$
Hence, option D is correct.

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