If the $9$ th term of an A.P. is zero, then prove that $29$ th term is double of $19$ th term.

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Solution

Ninth term $=$ $a + (n - 1) d$ $= 0$
$a + 8 d = 0$
Twenty ninth term $=$ $a + 28 d$ $=$ $- 8 d + 28 d$ $= 20 d$
$2 \times$ ninetinth term $=$ $2 a + 36 d$ $= - 16 d + 36 d$ $= 20 d$
Hence, Twenty ninth term $=$ $2 \times$ ninetinth term

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